Lecture 0 - alove7 2 1. A Harnack inequality implies Cα regularity for 0 < α < 1. 2. A positive...

Lecture 0

Course overview

In this course, we will mainly be concerned with the following problems: 1) Harmonic functions Δu = 0, i.e. i uii = 0.

Dirichlet problem: (Ω ⊂ Rn)

Δu = 0 , x ∈ Ω, u = ϕ , x ∈ ∂Ω.

2) Heat equation: ut = Δu, u : Rn+1 R.→Boundary value problem: cylinder domain Ω × [0, T ),Ω ⊂ Rn .

ut = Δu , (x, t) ∈ Ω × [0, T ), u = ϕ , (x, t) ∈ Ω × 0 ∪ ∂Ω × [0, T ).

3)Poisson Equation: (Ω ⊂ Rn)

Δu = f , x ∈ Ω, u = ϕ , x ∈ ∂Ω.

For which f, ϕ, Ω can we solve? Parabolic:

ut −Δu = f(x, t) , (x, t) ∈ Ω × [0, T ), u = ϕ(x, t) , (x, t) ∈ Ω × 0 ∪ ∂Ω × [0, T ).

We will prove existence theorems by method of priori estimates.For Δu = f , when does certain regularity of f imply regularity of u?• If f continuous, is u ∈ C2? NO!

We will always consider in the Holder spaces Cα(Ω), C0,α(Ω). The norm is

f(x) − f(y)= sup

|.fCα (Ω)

x,y∈Ω,x=y

||x− y|α

Thus f ∈ Cα(Ω) = f(x) − f(y) Cα(Ω) x− y .⇒ | | ≤ f | |α

When α = 1, f is just Lipstitz continuous functions.

For Δu = f in Ω, we will get Interior Estimates

uC2,α(Ω) ≤ C(fCα(Ω) + uC0 (Ω)),

1

where Ω ⊂⊂ Ω, C = C(Ω, Ω).

Notion of weak solution: Δu = f weakly on Ω if Ω uΔϕ = Ω ϕf, ∀ϕ ∈ Cc

2(Ω), here u ∈ L1 loc(Ω).

Regularity theorem: If u is a weak solution, then u should has as much regularity as the priori estimates.

In practical problems, it’s usually easy to prove existence of weak solutions. The harder problem: prove weak solution is regular, and therefore solves the original

equation strongly.

In general, the global estimates should depend on ∂Ω and ϕ:

Δu = f , x ∈ Ω, u = ϕ , x ∈ ∂Ω.

f ∈ Cα(Ω). Assume ϕ is the restriction of a C2,α function on Rn to ∂Ω, i.e. ϕ has a C2,α extension, and ∂Ω is C2,α smooth. Then u ∈ C2,α(Ω) and

uC2,α(Ω) ≤ C(fC2,α(Ω) + uC0 (Ω) + ϕC2,α (∂Ω)).

1

Lp theory: Δu = f, f ∈ Lp(Ω).(( Ω |fp) p < ∞)|

If u is a weak solution. Does 2nd order derivation of u belong to Lp, i.e.

1

( |D2 u|p) p < ∞ ? 1 < p < ∞Ω

We can get uW 2,p(Ω) ≤ C(fLp + uLp ).

We just look at Δ. The next is more general elliptic operators:

Lu = aij (x)Dij u + bi(x)Diu + c(x)u = f.i,j i

We also consider the following problems:

Lu = f , x ∈ Ω, u = ϕ , x ∈ ∂Ω.

ut − Lu = f(x, t) , (x, t) ∈ Ω × [0, T ),u = ϕ(x, t) , (x, t) ∈ Ω × 0 ∪ ∂Ω × [0, T ).

2

We call L is uniformly elliptic if ΛI ≥ (aij ) ≥ λI, λ > 0. Schauder Theory: L is uniformly elliptic, aij , bi, c ∈ Cα(Ω), then

uC2,α(Ω) ≤ C(fCα(Ω) + uC0 (Ω)).

Idea: Assuming coefficients are all Cα, locally L is close to a constant coefficients operator.

Maximum principle: Bound C0 norm of solution in terms of boundary data of f .

uC0(Ω) ≤ C(fC0(Ω) + sup ϕ ).| |

This is an A Priori estimate:I) Assume solution exists;II) Prove solutions satisfies a priori bounds;III) Therefore the solution exists.

Motivation: If you want to completely understand Perelman’s proof of Poincare conjecture, you have to know this stuff.

∂ g = −2Ric

∂t∂

gij ∼ Δg gij + lower terms. ∂t

Fundamental Result: (M3, g) compact 3manifold, then ∃ε > 0 s.t. Ricci flow system has a smooth solution on M × [0, ε). (This is called short time existence theorem.)

Examples of harmonic functions in Rn

a) Constant.b) linear functions.c) Homogeneous harmonic polynomials: Hk (Rn).

dimHk (Rn) = (2k + n − 2) (k+n−3)! .k!(n−2)!

d) n = 2, the real or image part of holomorphic functions is harmonic. They are C∞. Even more, they are Cω .

e) Fundamental solution

C , n > 2,rn−2 u(x) = C ln r , n = 2.

3

is harmonic on Rn − 0.

Fundamental solutions for Laplacian and heat operator

Definition 1 , n > 2,

Γ(x, y) =

n(2−

1 n)ωn

|x − y|2−n

1 2π log x − y , n = 2.| |

Γ is harmonic: ∂Γ(x, y)

= 1

(x − y i) x −

1 y n

i

∂xi nωn | |i∂2Γ(x, y)

=1 1

δij − n(x − yi)(xj − yj )

= ⇒ ∂xi∂xj nωn

x − y|n |x − y n+2 | |

= ΔxΓ(x, y) = 0 ⇒

= Δy Γ(x, y) = 0.⇒

Definition 2 21 |x−y|

e 4(t0−t) .Λ(x, y, t, t0) = (4π t − t0 )n/2| |

We have Λt = ΔΛ:

2

Λt = n 1 x − y|

Λ− 2 (t − t0)

Λ + |

4(t − t0)2

i i

Λ i = x − y

Λx 2(t0 − t)

= Λ i i =(xi − yi)2 1

Λ + Λx x⇒ 4(t0 − t) 2(t0 − t)

2x − yn 1 Λ +

|Λ.= ⇒ ΔΛ = −

2 (t − t0) 4(t − t0

|)2

Heat Kernel: 2

4tK(x, y, t) = (4πt

1)n/2

e−|x−y|

.

It’s easy to check Kt = ΔK. Suppose u : Rn → R is bounded and C0, then

u(x, t) = K(x, y, t)u(y)dy Rn

is C∞ on Rn × (0, ∞) and ut = Δu.

4

Lecture 1

Mean Value Theorem

Theorem 1 Suppose Ω ⊂ Rn , u ∈ C2(Ω), Δu = 0 in Ω, and B = B(y, R) ⊂⊂ Ω, then

1 1 u(y) = uds = udx

RnnωnRn−1 ∂B ωn B

∂u Proof:By Green’s formula, for r ∈ (0, R), ds = Δudx = 0. Thus ∂Br ∂ν Br

∂u ∂u 0 = ds = (y + rω)ds

∂Br ∂ν ∂Br

∂r

= r n−1 ∂u (y + rω)dω

Sn−1 ∂r

n−1 ∂ = r u(y + rω)dω

∂r Sn−1

= r n−1 ∂ (r 1−n uds)

∂r ∂Br

= 1 uds = const for any r. rn−1 ∂Br

⇒

But we also have

1nωnumin(Br) ≤

rn−1 uds ≤ nωnumax(Br), ∂Br

taking limit as r → ∞, we get for any r

1 u(y) = uds.

nωnrn−1 ∂Br

Integral it, we get the solid mean value thm.

Remark 1 We have u ≥ 0 = 1 uds, and we call such u subnωnRn−1 ∂B ⇒ u(y) ≤

harmonic, i.e. u lies below hamonic function sharing the same boundary values. Also we have u ≤ 0 = 1 uds and we call u superharmonic.

nωnRn−1 ∂B ⇒ u(y) ≥

Application: Maximum principle and uniqueness.

1

Theorem 2 Ω ⊂ Rn, u ∈ C2(Ω), Δu ≥ 0, If ∃p ∈ Ω s.t.

u(p) = max u, Ω

then u is constant.

Proof: Let M = sup u, ΩM = x ∈ Ω|u(x) = M .

Ω

ΩM is not empty because p ∈ M , ΩM is closed by continuity, ΩM is open by mean value inequality. Thus ΩM = M , i.e. u is constant function.

Corollary 1 u ∈ C2(Ω) C0(Ω), Δu = 0, then if Ω bounded, we have

inf u ≤ sup, x ∈ Ω. ∂Ω ∂Ω

Corollary 2 u, v ∈ C2(Ω) C0(Ω), Δu = Δv in Ω, u = v on ∂Ω = ⇒ u ≡ v on ∂Ω.

Corollary 3 Δu ≥ 0, Δv = 0, u ≡ v on ∂Ω = ⇒ u ≤ v in Ω. (Hence ”subharmonic” )

Application: Harnack Inequality.

Theorem 3 Suppose Ω domain, u ∈ C2(Ω), Δu = 0, Ω ⊂⊂ Ω, u ≥ 0 in Ω, then ∃constant C = C(n, Ω, Ω) s.t.

sup u ≤ C inf u. Ω Ω

Proof: Let y ∈ Ω, B(y, 4R) ⊂ Ω. Take x1, x2 ∈ B(y, R), we have

1 1 u(x1) = udx ≤ udx,

ωnRnB(x1,R) ωnRn

B(y,2R)

1 1 udx, u(x2) =

ωn(3R)nB(x2,3R)

udx ≥ ωn(3R)n

B(y,2R)

= ⇒ u(x1) ≤ 3n u(x2),

= ⇒ sup ≤ 3n inf . B(y,R) B(y,R)

Choose R little enough s.t. B(y, 4R) ⊂ Ω for ∀y ∈ Ω. Let x1, x2 ∈ Ω s.t. to be maximal and minimal point of u in Ω respectively. We can cover Ω by N balls of radius R since Ω is compact, so we have

sup u ≤ u(x1) ≤ 3n u(x u.1) ≤ · · · ≤ 3nN inf Ω Ω

This completes our proof.

2

Remark 2 1. A Harnack inequality implies Cα regularity for 0 < α < 1. 2. A positive (or more generally bounded above or below) harmonic function on Rn is constant.

A Priori Estimate for harmonic function.

Theorem 4 u ∈ C∞, Δu = 0, Ω ⊂ Ω. Then for multiindex α, there exists constant C = C(n, α, Ω, Ω) s.t.

sup Dα u ≤ C sup u .| | Ω | |

Ω

Proof: Since ∂ Δ = Δ ∂ , Du is also harmonic. So by mean value theorem and ∂xi ∂xi

divergence theorems, we have for B(y, R) ⊂ Ω,

1 1 Du(y) = Dudx = u−ν ds →

RnωnRnB(y,R) ωn ∂B

n = ⇒ |Du(y) sup |u|| ≤

R ∂B

n = ⇒ |Du(y) u .| ≤

d(y, ∂Ω) sup | |Ω

By induction, we get the stated estimate for higher order derivatives.

Remark 3 We can weaken the assumptions to u ∈ C2(Ω): u ∈ C2(Ω) and Δu = 0 = ⇒ u analytic. We will do this next time.

Green’s Representation Formula. Suppose Ω is C1 domain, u, v ∈ C2(Ω).

Green’s 1st identity:

∂u vΔudx + Du · Dvdx = v ds.

∂ν Ω Ω ∂Ω

Green’s 2nd identity:

∂u ∂v (vΔu − uΔv)dx = )ds. (v

∂ν − u

∂ν Ω ∂Ω

Find solution for Laplacian:

, n > 2,Γ(x) =

n(2−

1 n)ωn

|x|2−n

1 2π log |x| , n = 2.

Note that away from origin, ΔΓ(x) = 0.

3

Theorem 5 Suppose u ∈ C2(Ω), then for y ∈ Ω, we have

∂Γ ∂u u(y) = (u (x − y) − Γ(x − y) )dσ + Γ(x − y)Δudx.

∂Ω ∂ν ∂ν Ω

Proof: Take ρ small enough s.t. Bρ = Bρ(y) ⊂ Ω. Apply Green’s 2nd formula to u and v(x) = Γ(x − y), which is harmonic in Ω \ y, on the domain Ω \Bρ, we get

∂u ∂Γ ∂u ∂ΓΓ(x−y)Δudx = (Γ(x−y) u (x−y))dσ+ (Γ(x−y u (x−y)))dσ.

∂ν −

∂ν ∂Br ho ∂ν −

∂ν ∂ΩΩ\Bρ

Let ρ → 0, notice that as ρ 0→

∂u | Γ(x − y) dσ ≤ Γ(ρ) sup Du nωnρn−1 0,∂ν

| | | →∂Bρ Bρ

∂Γ

∂Br ho ∂ν ∂Bρ nωnρn−1

∂Bρ

udσ → −u(y),u (x − y)dσ = −Γ(ρ) udσ = −1

thus we get the Green’s Representation Formula.

Application of Green’s Formula:

Theorem 6 Let B = BR(0) and ϕ is continuous function on ∂B. Then R2 2 ϕ(y)−|x |

∂B x−y n ds , x ∈ B, u(x) = nωnR

ϕ(x) | |

, x ∈ ∂B.

belongs to C2(B) ∩ C0(B) and satisfies Δu = 0 in B.

4

Lecture 10

Interior C2,α estimate for Newtonian potential (continued)

Proposition 1 Consider B1 = BR(x0) ⊂ B2R(x0) = B2, f ∈ Cα(B2R), where 0 < α < 1. Let ω(x) = B2

Γ(x − y)f (y)dy, the Newtonian Potential of f in B2. Then ω ∈ C2,α(BR(x0)) and we have estimate

D2ω0;BR + Rα D2ω α;BR ≤ C(f 0;B2R + Rα f α;B2R ),| | | |

where C = C(n, α) is constant.

Proof:(continued)

(V I) = (f (x) − f (x)) Dij Γ(x − y)dy B2\Bδ (ξ)

≤ |f |Cα(x) x − x|α| ∂(B2\Bδ (ξ))

DiΓ(x − y)νj dsy| |

≤ |f |Cα(x)δα( DiΓ(x − y)νj dsy + DiΓ(x − y)νj dsy )| | | |∂B2 ∂Bδ (ξ)

1 1 δ 1 ≤ c|f |Cα(x)δα(

∂B2 x − y n−1 dsy + y − ξ y − x ) | | ∂Bδ (ξ) |x − y|n−1 dsy ) (

2 ≤

2 | | ≤ | |

1 1 ≤ c|f |Cα(x)(δα nωn(2R)n−1 + nωn(δ)n−1)Rn−1 (δ/2)n−1

≤ c|f |Cα(x)δα .

(V ) = (Dij Γ(x − y) − Dij Γ(x − y))(f (y) − f (x))dy B2\Bδ (ξ)

α

B2\Bδ (ξ) |DDij Γ(ˆ Cα(x) x − y dy≤ x − y)||x − x||f | | |

1 ≤ cδ f Cα(x) x − y n+1 |x − y|αdy | ||y−ξ|≥δ |ˆ |

Since δ 3

x − y x − ξ + ξ − y + ξ − y ξ − y| | ≤ | | | | ≤ 2

| | ≤ 2 | |

and δ 1

y − ξ y − ˆ x − ξ y − ˆ x| + y − ξx| + ˆ x| + y − ˆ| | ≤ | | | ≤ |2 ≤ |

2 | |

1= y − ξ y − ˆx|,⇒

2 | | ≤ |

1

We thus get 3R 1(V ) ≤ cδ f Cα(x)

dy n+1−α ≤ cδ|f |Cα(x) rn+1−α r n−1dr

δ| |

|y−ξ|≥δ |y − ξ| 3R 1≤ cδ f Cα(x) r α−2dr ≤ cδ f Cα(x) α − 1((3R)α−1 − δα−1)| | | |

δ c ≤

1 − α |f |Cα (x)δ

α .

Combine all the results, we have shown

α|Dij ω(x) − Dij ω(x) ≤ C( |f (x)|

+ f Cα(x) + f Cα(x)) x − x ,| Rα | | | | | |

thus Rα |Dij ω(x) − Dij ω(x)| ≤ C( f (x) + Rα f Cα(x) + Rα f Cα(x)), |x − x|α | | | | | |

i.e. Rα D2ω α;BR ≤ C(f C0;B2R

+ Rα f| | | |Cα(B2R )).

Since we have already known that

+ Rα|f |α),|D2ω|0 ≤ c(f C0

we finally get

D2ω0;BR + Rα D2ω α;BR ≤ C(f 0;B2R + Rα f α;B2R ). | | | |

Exercise: 1) Find a continuous function f s.t. Δu = f does not have a C2 solution. 2) Find g ∈ C1 and Δu = g but u is not C2,1 .

Interior C2,α estimates for Poisson’s equation.

Application: u ∈ C2(B2R(x0)), Δu = f, f ∈ Cα(B2R(x0)).

Theorem 1 C uC2,α (BR) ≤ Rα (uC0 (B2R) + f Cα(B2R)).

Proof: Since Δ(u − N f ) = 0, where N f is the Newtonian potential of f , thus from the C2,α estimate of N f we can get the C2,α of u.

2

Theorem 2 Let u ∈ C2(Ω), Δu = f, f ∈ Cα(Ω), then for any Ω ⊂⊂ Ω, we have

uC2,α(Ω) ≤ C(uC0(Ω) + f Cα (Ω)).

Proof: Take x ∈ Ω, choose R s.t.B2R ⊂ Ω, then

C u C2,α(x) ≤ u C2,α(BR(x)) ≤

Rα (uC0;B2R + f| | | | | |Cα (B2R))

C Rα (uC0;Ω + f |Cα(Ω)).≤ |

Taking superior over all x and using previous estimate, we get

uC2,α(Ω) ≤ C(uC0(Ω) + f Cα(Ω)).

Boundary estimate on Newtonian potential: C2,α estimate up to the boundary for domain with flat boundary portion.

Suppose BR ⊂ B2R, with flat boundary portion B+ ⊂ B2+ .1

Lemma 1 Let f ∈ Cα(B2+), ω(x) = Γ(x − y)f (y)dy be the Newtonian potential of B+

2

f in B2+ . Then ω ∈ C2,alpha(B1

+) and

|D2ω 0;B1 + Rα D2ω α;B+ ≤ C( f 0;B2 + Rα f α;B+ ).| | |1

| | | |2

Proof: Examine the proof form last time. From C2 estimate have

Dij ω(x) = Dij Γ(x − y)(f (y) − f (x))dy − f (x) DiΓ(x − y)νj (y)dsy . B+ ∂B+

2 2

(Note DiΓ(x − y)νj (y)dsy = Dj Γ(x − y)νi(y)dsy .)∂B+ ∂B+ 2 2

So if either i = n or j = n, the integral on the lower boundary portion of B+ vanishes. 2 (Since ν = (0, 0, · · · , −1).)

If i = j = n, then

Dnnω = f (x) (DnΓ(x − y) −DnΓ(x − y))(−1)dσ∂B+ 2

≤ |f (x) |DDnΓ(ˆ dσ | x − y)||x − x|∂B+ 2

1 ≤ |f (x) δ dσ|∂B+ x − y n

2 |ˆ |

1 ≤ |f (x) δ nωRn−1|Rn

≤ c|f (x) δα .|

Since we know Δω = f , thus ωnn = f − ω11 − ω22 − · · · , so we see that ωnn ∈ Cα , and we can get estimate for ωnn from estimates for ωii, i < n.

3

Lecture 11

Review of Green’s functions. G : Ω × Ω −→ R.Given x ∈ Ω, let hx(y) : Ω −→ R be s.t. Δy hx(y) = 0 and hx(y) = −Γ( x − y ) for| |

y ∈ ∂Ω. By definition, G(x, y) = Γ( x − y ) + hx(y).| |If Green’s function exists, then for u ∈ C1(Ω) ∩ C2(Ω), y ∈ Ω, we have

∂G(x, y) u(y) = u(x) dσ + G(x, y)Δu(x)dx.

∂Ω ∂ν Ω

Thus we can see: If u = 0 on ∂Ω, then u(y) = Ω G(x, y)Δu(x)dx = G ∗Δu.

(Compare) By Green’s formula, we have If u ∈ Cc

2(Rn), then u(y) = Γ ∗Δu.

Proposition 1 a) G(x, y) = G(y, x); b) G(x, y) < 0, for x, y ∈ Ω, x = y. c) Ω G(x, y)f (y)dy → 0 as x ∂Ω, where f is bounded and integrable. →

Proof of c): From definition, G(x, y) = 0 if x ∈ Ω, y ∈ ∂Ω.

By a), G(x, y) = 0 for y ∈ Ω, x ∈ ∂Ω. Thus G : Ω × Ω − diag −→ R.

|G(x, y)f (y) |G(x, y) dy| |dy ≤ f L∞ Ω |

Ω

C | |n−2 dy≤ f L∞

Ω x − y

≤ Cf L∞ .

By dominate convergence, we can change limit and integral.

Example. Green’s function for Rn +

Given y = (y1 , · · · , yn), let y∗ = (y1 , · · · , yn−1 , −yn). It is easy to check that G(x) = Γ(x −y) −Γ(x −y∗) = Γ(x −y) −Γ(x∗ −y) is Green’s

function for Rn :+

•hx(y) = G(x, y) − Γ(x − y) is harmonic in Ω; •G(x, y) = 0 on ∂Ω.

Review of Schwartz reflection. First we go back to harmonic functions.

1

Theorem 1 A C0(Ω) function u is harmonic if and only if for every ball BR(y) ⊂⊂ Ω, we have

1 u(y) = uds.

nωnRn−1 ∂B

Proof: = is just mean value theorem. ⇒

⇐=: Use the Poisson kernel: Given any Ball BR(y) ⊂ Ω, Define R2−|x2| u(y)

n ds , x ∈ BR,nωnR ∂B x−yh(x) = u(x)

| |, x ∈ ∂B.

Then h ∈ C2(BR) ∩ C0(BR) and satisfies Δu = 0. So h satisfies the mean value property. Therefore u − h satisfies the mean value property and u = h on ∂BR.

But recall the uniqueness theorem for solutions of Poisson’s equation – we only used the mean value property. Therefore u = h, so u is harmonic.

+, T = Ω+ ∩ ∂Rn is a domain in ∂RnNow suppose Ω+ ⊂ Rn +. Let Ω− = (Ω+)∗, i.e. +

Ω− = .(x1, · · · , xn) ∈ Rn|(x1, · · · , −xn) ∈ Ω+

Suppose we have u harmonic in Ω+ , u ∈ C0(Ω+ ∪ T ), and u = 0 on T . Define

, xΩ+ ∪ T, u(x1, · · · , xn) =

u(x1, · · · , xn) , x ∈ Ω−.u(x1, · · · , −xn)

Theorem 2 The function u defined above is harmonic in Ω+ ∪ T ∪ Ω−.

Proof: Obviously u is in C0Ω+ ∪ T ∪ Ω−. If one examines the above proof, one only requires that for each point y ∈ Ω, ∃R > 0

so that mean value property holds in Br (y), r < R. Also remember in the proof of maximum principle, we assumed that the function has a interior max, then use mean value theorem in small ball around this point.

Certainly here we have this property in Ω+ ∪ Ω−, and on T if follows from the definition of u, ∂BR(x∈T ) u = 0.

C2,α boundary estimate for Poisson’s equation with flat boundary portion.

Theorem 3 Let u ∈ C2(B2+) ∩ C0(B2

+), f ∈ Cα(B2+), and Δu = f in B2

+ , u = 0 on T . Then u ∈ C2,α(B1

+) and

uC2,α(B+2 ) + fCα (B+

1 ) ≤ C(uC0(B+2 )

).

2

Proof: Reflect f with respect to T , i.e.

, xn ≥ 0,f ∗(x) = f ∗(x1, · · · , xn) =

f (x1, · · · , xn) , xn ≤ 0.f (x1, · · · , −xn)

Let D = B2+ ∪ B2

− ∪ (B2 ∩ T ), then f ∗ ∈ Cα(D) and f Cα(D) ≤ 2f Cα(B+ . Let 2 )

G(x, y) be the Green’s function of upper half space. Define

ω(x) = G(x, y)f (y)dy B+ 2

= (Γ(x − y) − Γ(x − y∗))f (y)dy B+ 2

= (Γ(x − y) − Γ(x∗ − y))f (y)dy B+ 2

Γ(x − y)f (y)dy − Γ(x − y)f ∗(y)dy. = B+

2 B−2

Then Δω = f . It’s easy to check that ω(x) = 0 on T . Thus

−2

Γ(x − y)f ∗(y)dy = Γ(x − y)f ∗(y)dy − Γ(x − y)f (y)dy, D B+

2B

so ω(x) = 2 Γ(x − y)f (y)dy − Γ(x − y)f ∗(y)dy.

B+ D2

We did estimates for the first term earlier. For the second term, think of B1+ ⊂ D

and just use interior estimates from last week. We thus get

+2 ).ω

C2,α(B1 ) ≤ Cf C0,α(B+

Let v = u − ω in B2+, then on B+ we have Δv = Δu −Δω = f − f = 0 and v = 0 2

on T . We may reflect v, then by Schwartz reflection we know that v∗ is harmonic in D.

Now use the interior estimates for harmonic functions, we get

1 ) ≤ Cv∗C0(D) ≤ 2vC0(D).vC2,α(B+

So

uC2,α(B+ ,α(B+ + ωC2,α(B+2 )).

1 ) ≤ vC21 ) 1 ) ≤ C(uC0(B+ + f Cα(B+

2 )

Application: Global C2,α Regularity Theorem for Dirichlet problem in a ball with zero boundary data.

3

Theorem 4 Suppose B is a ball in Rn , u ∈ C2(B) ∩ C0(B), f ∈ Cα(B), Δu = f in B and u = 0 on ∂B. Then u ∈ C2,α(B).

1Proof: By dilation and translation, we can assume B = B1/2(0, · · · , 0, 2 ). Look at the inversion x Ix = x

2 , then the ball B is mapped to a half space x

→ | |B∗ = xn ≥ 1 while ∂B is mapped onto ∂B∗ = xn = 1.x|

The Kelvin Transform of u is v(x) = |x|2−nu( x 2 ) ∈ C2(B∗) ∩ C0(B∗) and we have

x| |

Δy v(y) = |y|−n−2Δxu(x) = |y|−n−2f( |yy |2 ) ∈ Cα(B∗).

By the previous theorem, u ∈ C2,α up to the boundary. By rotation, we could do this for any boundary point, so u ∈ C2,α.

Corollary 1 Suppose ϕ ∈ C2,α(B), f ∈ Cα(B). Then the Dirichlet problem

Δu = f , x ∈ B, u = ϕ , x ∈ ∂B.

is uniquely solvable for u ∈ C2,α(B).

Proof:The existence of u comes from Perron’s method. Since Δϕ ∈ Cα(B), so let v be the unique solution of Δv = f −Δϕ in B with v = 0

on ∂B. Then v ∈ C(B) ∩ C0(∂B). By above result, v ∈ C2,α(B). But u − ϕ solves the problem also: Δ(u − ϕ) = Δu −Δϕ = f −Δϕ in B; u − ϕ = 0

on ∂B. By uniqueness, v = u − ϕ. So u ∈ C2,α(B).

4

Lecture 2

Definition of Green’s function for general domains. Suppose u ∈ C2(Ω) ∩ C1(Ω), then for y ∈ Ω, the Green Representation formula tells

us ∂Γ ∂u

u(y) = (u (x − y) − Γ(x − y) )dσ + Γ(x − y)Δudx. ∂Ω ∂ν ∂ν Ω

Definition 1 For integrable f , Ω Γ(x − y)f(x)dx is called Newtonian Potential with density f .

Remark 1 If u ∈ C02(Rn), i.e. compact supported, then have

u(y) = Γ(x − y)Δudx. Ω

If u is harmonic, then we have

∂Γ ∂u u(y) = (u (x − y) − Γ(x − y) )dσ.

∂ν ∂ν ∂Ω

Thus harmonic functions are analytic.

Now let h be harmonic, by Green’s 2nd identity, we get

∂u ∂h hΔu = (h

∂ν − u

∂ν )ds

Ω ∂Ω

i.e. ∂h ∂u

0 = (u∂ν

− h∂ν

)ds + hΔu ∂Ω Ω

Adding Green’s representation formula, we get

∂ ∂h ∂u u(y) =

∂Ω(u(x)( Γ(x−y)+ )−(Γ(x−y)+h(x)) )ds+ (Γ(x−y)+h(x))Δudx.

∂νx ∂νx ∂νx Ω

Now fix x, we choose hy (x) s.t. Δhy (x) = 0 in Ω and hy (x) = −Γ(x − y) on ∂Ω. Let G(x, y) = Γ(x − y) + hy (x), then we have

∂ u(y) = u(x) G(x, y)ds + G(x, y)Δudx.

∂Ω ∂νx Ω

Definition 2 Such a function G(x,y), defined for x ∈ Ω, y ∈ Ω, x = y which satisfies G(x, y) = 0 for x ∈ ∂Ω and h(x, y) = G(x, y) − Γ(x − y) is harmonic in x ∈ Ω, is called a Green function for domain Ω.

1

Remark 2 1. By Maximum Principle, G is unique if exists. 2. If G exists for a domain Ω and u is harmonic in Ω, then we can get an explicit formula for u in terms of boundary values:

∂G u(y) = u ds.

∂ν ∂Ω

Green’s function for ball B(0, R)

Proposition 1 The Green’s function for the ball B(0, R) is

1 R x|R|y|2−n) , n ≥ 3,

G(x, y) = n1(2−n)ωn

(|x − y|2−n − | |x| x − R x

2π (log x − y − log |R|y|) , n = 2.| | | |x| x −

xRemark 3 G(x, y) = Γ(x − y) − Γ( R |x| x − |R

|y), thus ΔyG(x, y) = 0 and G(x, y) = Γ(x − y)+ a harmonic function on boundary.

Claim 1 G(x, y) = G(y, x), G(x, y) ≤ 0.

R y R xProof: By squaring, we can get | |y| y − |R|x| = |

R|y|, thus G(x, y) = G(y, x).| |x| x −

This implies ΔxG(x, y) = 0 by previous remark. R xFor x, y ∈ B(0, R), we have x − y| ≤ | |x| x −

|R|y , thus G(x, y) ≤ 0 since the function | |

Γ is decreasing as a real function.

2∂G 1Proposition 2 ∂νx = R2−|y|

|n , x ∈ ∂B(0, R)nωnR x−y|

1 RProof:By symmetry, G(x, y) = n(2−n)ωn ( x − y 2−n |

R|x|2−n). Thus | | − | |y| y − y

( xi − yi

n ) − | | − | |xi)(

−|y| )∂G 1 ( Ryi yy R R = .

∂xi nωn x − y x − y n| | | |

So 2∂G ∂G xi 1 1 1 2 y

∂νx =<

∂xi , |x|

> = nωn x − y|n ( )(|x| − < x, y > + < x, y > −|

R

|2 |x|

2) x| ||

1 =

nωnR x − y n (R2 − |y|2)

| |

This completes the proof.

Corollary 1 If u ∈ C2(BR) ∩ C0(BR) and Δu = 0, then

2 u(x) u(y) =

R2 − |y|dσx.

nωnR ∂BR x − y n| |

2

|

Remark 4 Previously, we regarded u ∈ C2(Ω) ∩ C1(Ω). Under the assumption of this corollary, we can get formula holds for r < R. Since u ∈ C0(Ω), just take limit as r R.→Again, we see that harmonic functions are analytic.

Poisson Integral Formula

Theorem 1 Let ϕ : ∂B(0, R) → R be continuous, then R2−|x|2 ϕ(y)

u(x) = nωnR ∂B(0,R) |x−y|n dσy , x ∈ B(0, R), ϕ(x) , x ∈ ∂B(0, R).

satisfies Δu = 0 in B(0, R) and u ∈ C2(B) ∩ C0((B))

ϕ(y) ∂G Proof: For x ∈ B(0, R), the definition of u gives u(x) = (x, y)dσy,∂B(0,R) ∂νy

thus ∂G

Δxu(x) = ϕ(y)Δx (x, y)dσy ∂B(0,R) ∂νy

∂ = ϕ(y) ΔxG(x, y)dσy = 0.

∂B(0,R) ∂νy

so Δu(x) = 0 in B and u ∈ C2(B)We have known that for harmonic function ω ∈ C2(B) ∩ C1(BR),

2 ω(x)ω(y) =

R2 − |y|dσx.

nωnR ∂B(0,R) x − y n| |

Take ω ≡ 1, we get 1 = R2−|y|2 1

n dσx, i.e. nωnR ∂B(0,R) x−y| |

2 11 =

R2 − |y|x − y n dσx = K(x, y)dσy.

∂B(0,R) nωnR | ∂B |1Here K(x, y) = fracR2 − |y|2nωnR x−y n is called Poisson Kernel. | |

Now consider x0 ∈ ∂B. For any > 0, there ∃δ > 0 s.t. ϕ(x) − ϕ(x0) < for any | | δ x − x0 < δ. Choose M large enough such that ϕ(x) < M∀x ∈ ∂B. For x − x0 < 2 ,| | | |

we have

u(x) − u(x0) = K(x, y)(ϕ(y) − ϕ(x0))dσy| | | ∂B

K(x, y) (ϕ(y) − ϕ(x0))|dσy + y−x0 >δ

K(x, y) (ϕ(y) − ϕ(x0)) dσy≤ || |

| ||y−x0|≤δ

R2 − |x|2 1 nωnRn−1≤ + 2M

nωnR (δ/2)n

≤ + 2C(R2 − |x|2).

3

Thus for x close to ∂B, u(x) − u(x0) ≤ 2, i.e. x ∈ C0(B) | |

Mean Value Property (MVP)

Theorem 2 If a C0(Ω) function u satisfies

1 u(y) = udσ

nωnRn−1 ∂B

for every ball B = B(y, R) ⊂⊂ Ω (MVP), then u is harmonic. In particular, u is analytic.

Proof: Take any B(y, R) ⊂⊂ Ω, u ∈ C0(∂B(y, R)). Thus by Poisson integral formula, there is harmonic function h on B(y, R) s.t. h = u on ∂B(y, R).

Consider ω = h − u. Obviously ω satisfies MVP on any ball ⊂ B(y, R). Recall that our maximum principle and uniqueness proof only need MVP, so ω has zero boundary value implies ω = 0 in B(y, R). So u = h in B, i.e. u is harmonic.

Remark 5 The proof just need ”for each x ∈ Ω, ∃B(x, R) ⊂ Ω s.t. MVP is satisfied on all balls in B(x, R)”.

counterexample (NOT C0):Take u on plane, u(x, y) = 1 for y > 0, u(x, y) = −1 for y < 0, u(x, y) = 0 for y = 0. Obviously u is not harmonic.

Corollary 2 The limit of a uniformly convergent sequence of harmonic functions is harmonic.

Proof: The limit is continuous and still satisfies MVP.

4

Lecture 3

MVP + integrable ⇔ harmonic

Theorem 1 Suppose u ∈ L1 loc, then u is harmonic u satisfies MVP on Ω.⇔

Proof: Take C∞ function ρ on Rn with properties: (a) Supp(ρ) ⊂ B(0, 1); (b) ρ ≥ 0; (c) ρ is radical, i.e. ρ(x) = ρ(|x|); and (d) ρ(x)dx = 1. B(0,1)

By these properties, we have 1 1

1 = (0, 1)ρ(x)dx = ρ(s)dσds = ρ(s)nωns n−1ds. B 0 ∂B(0,s) 0

x x−yDefine ρ(r)(x) = 1 ρ( |r| ), ur(x) = ρ(r)(x) ∗ u = 1 ρ( | r

| )u(y)dy. (Without loss rn rn Ω of generality, we can assume u ∈ L1(Ω) – otherwise we consider near every point s.t. u is integrable.)

Now we have x − y

ur(y) =1

ρ( | |

)u(x)dx rn Ω r

x − y=

1 ρ( | |

)u(x)dx rrn

B(y,r) r x − y

=1

ρ( | |

)u(x)dσds rrn

0 ∂B(y,s) r1 s

= ρ( )u(x)dσds rrn

0 ∂B(y,s) r1 s

= ρ( )nωns n−1 u(y)ds rn

0 r rnωnu(y) s

= ρ( )s n−1ds rn

0 r

= nωnu(y)

1

rρ(t)r n−1tn−1dt rn

0 1

= nωnu(y) ρ(t)tn−1dt 0

= u(y).

But ρ ∈ C∞ ⇒ ur ∈ C∞, so u ∈ C∞. Thus MVP u is harmonic by last lecture. ⇒

Weak Solution

1

For the function Γ(x), we have (in distributional sense ) ΔΓ(x) = δ0(x), i.e. for ϕ ∈ Cc

2(Rn),

Γ(x)Δϕ(x)dx = ϕ(0) = ϕδ(0). Rn

More generally, ΔΓ(x − y) = δy(x).Proof: Choose R large so that Suppϕ ⊂ B(0, R). Choose ρ small. From Green’sformula we get

∂ϕ ∂ΓΓΔϕdx = )ds.

Rn−B(0,ρ) ∂Bρ

(Γ ∂ν

− u∂ν

As ρ → 0, we get

ΓΔϕdx ΓΔϕdx, Rn−B(0,ρ)

→ Rn

∂ϕ cΓ(ρ)

∂ν ds ≤

ρn−2 ρn−1 → 0,

∂Bρ

∂Γ 1 1 u udσ u(0),−

∂Bρ ∂ν

= nωn ρn−1 →

∂Bρ

which give what we claimed.

Application: G(x, y) = G(y, x)

G(x, y) − G(y, x) = (G(x, z)δ(y − z) − G(y, z)δ(x − z))dz Ω

= (G(x, z)ΔzΓ(y − z) − G(y, z)ΔzΓ(x − z))dz Ω

∂ ∂ = (G(x, z) Γ(y − z) − G(y, z) Γ(x − z))dz

∂νz∂Ω ∂νz

= 0.

Weyl’s Lemma: Regularity of weakly harmonic functions

Theorem 2 Suppose u ∈ L1 0(Ω) satisfies u(x)Δϕ(x)dx = 0 for ∀ϕ ∈ Cc

2(Ω). ThenΩ u is harmonic in Ω.

Proof: Without loss of generality, we can assume u ∈ L1(Ω). Again we take

x − yur(x) = ρ(r)(x) ∗ u =

1 ρ( | |

)u(y)dy. rn

Ω r

2

Claim 1. f (y − x)Δg(x)dx = Δy f (y − x)g(x)dx, ∀f, g :Ω Ω

Δy f (y − x)g(x)dx = Δy f (x)g(y − z)dz Ω Ω

= f (z)Δy g(y − z)dz Ω

= f (y − x)Δg(x)dx. Ω

Claim 2. ur (x)Δϕ(x)dx = Ω u(x)Δϕr (x)dx :Ω

x − yur (x)Δϕ(x)dx =

1( ρ(

| |)u(y)Δϕ(x)dy)dx

Ω Ω r n

Ω r x − y

= ( 1

ρ( | |

)u(y)Δϕ(x)dxdy rn Ω Ω r

x − y= u(y)(

1 ρ( | |

)Δϕ(x)dxdy Ω Ω rn r

= u(y)Δy ( r

1 n ρ(

|x − y|)ϕ(x)dx)dy Ω Ω r

= u(y)Δy ϕr (y)dy. Ω

Claim 3. ur (x) is harmonic. In fact, for any ϕ ∈ Cc

2(Ω), Δϕr (y) ∈ Cc 2(Ω), so by the assumption we have

u(y)Δy ϕr (y)dy = 0. Ω

Thus by claim 2, Ω ur (x)Δϕ(x)dx = 0 for any ϕ ∈ Cc 2(Ω).

But ur (x) ∈ C∞(Ω), thus

ur (x)Δϕ(x)dx = Δur (x)ϕ(x)dx. Ω Ω

So we get

Δur (x)ϕ(x)dx = 0, ∀ϕ ∈ Cc 2(Ω),

Ω

which implies Δur (x) = 0, i.e. ur (x) is harmonic.Claim 4. ur uniquely bounded and equicontinuous on any Ω ⊂⊂ Ω.

In fact, ur = ρ(r) ∗ u implies

ur L1 L1 uL1 uL1 , ≤ ρ(r) ≤

3

sosup sup ur L1 .

Ω⊂⊂Ω |Dk ur | ≤ C ·

Ω | | ≤ Cu

Since ur harmonic, we get

1 ur (y) = ur (x)dx,

ωnRnB(y,R)

which implies 1

L1 .|ur (y)| ≤ωnRn u

Claim 5. u is smooth. In fact, by ArzelaAscoli theorem, there is some subsequence ri → 0, i → ∞ s.t.

uri → v ∈ C∞ on Ω ⊂ Ω. But uri = ρ(ri) ∗ u → u in L1 as ri → 0, so u = v on Ω. Thus u is smooth on Ω.

Now since u is smooth, we have

0 = uΔϕ = ϕΔu, ∀ϕ, Ω Ω

so Δu = 0, i.e. u is harmonic.

4

Lecture 4

Removable Singularity Theorem

Theorem 1 Let u be harmonic in Ω \ x0, if

o( x − x02−n) , n > 2,| |

u(x) = o(ln x − x0 ) , n = 2 | |

as x x0, then u extends to a harmonic function in Ω.→

Proof: Without loss of generality, we can assume Ω = B(0, 2), then u|∂B(0,1) is continuous. Thus by Poisson Integral formula, ∃v ∈ C(B(0, 1)) ∩C∞(B(0, 1)) to be harmonic function with boundary condition v = u on ∂B(0, 1).

Choose > 0 and δ > 0 small, consider

ω(x) = u(x) − v(x) − ( x 2−n − 1) , n > 2,| |u(x) − v(x) + log x ) , n = 2,| |

then ω(x) is harmonic on B1(0) \BBδ(0), and ω(x) = 0 on ∂B1(0). On ∂Bδ(0), − x 2−n is the dominate term, thus ω ≤ 0 on ∂Bδ(0) for δ small enough. | |Now by maximum principle, ω ≤ 0 on B1(0) \Bδ(0), i.e.

u(x) ≤ v(x) + ( x 2−n − 1),| |

Thus by letting → 0, we get

u(x) ≤ v(x), ∀x ∈ B1(0) \Bδ(0).

This is true for any δ small, so it is true for ∀x ∈ B1(0) \ 0.By reverting u and v, we can get

v(x) ≤ u(x), ∀x ∈ B1(0) \ 0,

thus v(x) = u(x), ∀x ∈ B1(0) \ 0. Now we can define u(0) = v(0), and extend u to be a harmonic function on B(0, 1),

thus a harmonic function on Ω = B(0, 2).

Example This gives an example of Dirichlet problem that is NOT solvable: Take Ω = B(0, 1) \ 0, then ∂Ω = ∂B(0, 1) ∪ 0. Consider the Dirichlet problem ⎧ ⎨ Δu = 0 , in Ω,

u = 0 , on ∂B(0, 1),⎩ u = 1 , at 0

1

If this is solvable, then the solution u can be extend to a bounded harmonic function on B(0, 1). Now by MVP, u(0) = 0, which is a contradiction.

Laplacian in general coordinate systems

Theorem 2 Let gij be the metric component of a coordinate system, then

1 ∂k (gkj ∂j uΔu = det(grs)).

det(grs)

Proof: Take any ϕ ∈ C∞, we have 0

ϕΔu det(gij )dy = ϕΔudx

= < ϕ, u > dx

= gij ∂iϕ∂j u det(gij )dy

= ϕ∂i(gij ∂j u det(gij ))dy ∂i(gij ∂j u

det(gij ))= ϕ det(gij ) dy. det(gij )

Thus the formula follows.

Laplacian in spherical coordinates (r, ω) 2Now g = dr2 + r gSn−1 , so

1 0 1 0(gij ) = = (gij ) =

0 r−2gSn−1

.2 ij0 r gSn−1 ⇒

so det(gij ) = r2(n−1)det(gSn−1 ) = r n−1 det(gSn−1 ),

thus

2

1Δu = ∂1(g 1j ∂ju r n−1 det(gSn−1 ))

rn−1 det(gSn−1 )1

+ ∂k(gkj ∂j u r n−1 det(gSn−1 ))rn−1 det(gSn−1 )

k>1

1 1 ij= rn−1 ∂r(∂ru r n−1) +

det(gSn−1 ) ∂k(r−2 g

Sn−1 ∂j u det(gSn−1 ))k>1

1 = ∂r(∂ru r n−1) + r−2ΔSn−1 u

rn−1

1 =

rn−1 (urr r n−1 + ur(n − 1)r n−2) + r−2ΔSn−1 u

= urr + (n − 1) ur +

1ΔSn−1 u.

2r r

If u(r, θ) = f (r)B(θ) is variables separated, then

f (r)Δu(r, θ) = (frr + (n − 1)

fr )Bθ + ΔSn−1 B(θ).2r r

Proposition 1 Let B(θ) be a homogeneous harmonic polynomial of degree k restricted to Sn−1, then ΔSn−1 B(θ) = −k(k + n − 2)B(θ).

Remark 1 Let Pk be the set of homogeneous polynomials of degree k on Rn , Hk be the set of harmonic homogeneous polynomials of degree k on Rn, then

2k = k ⊕ r k−2.P H P

It’s not hard to prove

dimPk =(k + n − 1)!

,k!(n − 1)!

so

dimHk =(k + n − 1)! (k + n − 3)!

. k!(n − 1)!

−(k − 2)!(n − 1)!

= (2k + n − 2)(k + n − 3)! k!(n − 2)!

For such a B(θ) ∈ Hk, we have

frΔ(f (r)B(θ)) = (frr + n − 1

fr − k(k + n − 2) 2 )B(θ).

r r

For the solution of the equation

frfrr +

n − 1 fr − k(k + n − 2) = 0,

2r r

let f = rp, then fr = p rp−1, frr = p(p − 1)rp−2, we get

0 = p(p − 1)rp−2 + p(n − 2)rp−2 − k(k + n − 2)rp−2 = (p − k)(p + k + n − 2)rp−2 .

3

Thus p = k or p = −k − n + 2. For p = k, we get u(r, θ) = rk B(θ), where B(θ) ∈ Hk , thus u is just the homogeneous

k harmonic polynomial on Rn . For those p = −k − n + 2, if k = 0, then p = 2 − n and B(θ) = constant, thus

u = c · r2−n, which is the fundamental solution. if k > 0, then p < 2 − n, note that B(θ) is defined on the compact set Sn−1, thus B is bounded, so u grows faster than the fundamental solution near the origin.

From above we get a degree gap of harmonic function:

· · · · · · , −n, −(n − 1), −(n − 2), , 0, 1, 2, · · · · · ·

Notice that we have to have the gap in view of our removable singularity theorem.

Homogeneous expansions

Theorem 3 Any harmonic function in B(0, 1) can be expressed as an infinite sum

∞

u(x) = pk (x), pk ∈ Hk . k=0

Proof: Take the Taylor expansion of u, u = pk , where pk ∈ Pk , we have

0 = Δu = Δpk,

but Δpk ∈ Pk−2, thus Δpk = 0 for all k, i.e. pk ∈ Hk .

4

Lecture 5

Last time: In spherical coordinates, u(x) = U(r, θ),

21Δu(r, θ) = Urr +

n − 1 Ur + ΔSn−1 U.

r r

If U(r, θ) = f(r)B(θ), then

1Δu = (frr +

n − 1 fr )B(θ) +

2 fΔSn−1 B. r r

Proposition 1 Eigenvalues of ΔSn−1 are −k(k + n − 2), where k ≥ 0, so λ1(Sn−1) = n − 1.

Let ΔSn−1 Bk (θ) = −k(k + n − 2)Bk (θ), then

frΔ(f(r)Bk(θ)) = (frr + n − 1

fr − k(k + n − 2) 2 )B(θ).

r r

From this we get that harmonic functions which has form rpBk must satisfies p = k or −k − n + 2, thus we get a gap for the values of p:

· · · · · · , −n, −(n − 1), −(n − 2), , 0, 1, 2, · · · · · ·

The terms before the gap are harmonic functions blowing faster than or as fast as the Green’s function, and the terms after the gap correspond to homogenous harmonic polynomials. The gap comes from the removable singularity theorem.

Laplacian in inverted coordinates: Kelvin transform I yFirst we define the inverted transform T : Rn −0 to be T y = = x.−0 −→ Rn |y|2

Note that T = T −1 . Let’s find the component of the metric tensor of y. i ∂Now the metric of coordinates x is gEuc = (δij ). Let ek =

∂yk . Suppose T∗ek = ∂ ∂xi aik ∂xi , then aik =

∂yk , and

(T ∗gEuc)kl = T ∗δij (ek , el) = δij (T∗ek , T∗el) ∂ ∂ = δij ( amk

∂xm , arl )∂xr

= amk aml

Because ∂xm ∂ m 2y y

amk = = ∂yk ( |

y

y|2 ) = δkm

|y

k

|4

m

,∂yk |y|2 −

1

so we have gkl = (T ∗gEuc)kl = amkaml

m k

( δkm 2y ym

)( δlm 2ylym

= |y|2 − |y|4 |y|2 − |y|4 )

m l l k l k 2δkl 2y yk + 2y y 4y y |y|

= |y|4 − |y|6 +

|y|8

δkl = . |y|4

Thus 1 kl det(gkl) =

|y|4n = |y|−4n , g = |y|4δkl.

So by the formula in last lecture, we get

4Δu(T y) = |y|2n∂i(δij |y| |y|−2n∂j u)

= y 2n∂i(|y|4−2n∂iu)| |2n 2= |y|4

|y|4−2n∂i∂iu + 2(2 − n) y < u, y > | |2= |y| uii + 2(2 − n) y uiyi.| |

On the other hand, we have ∂2 ∂2 n+2 y y 2−n u) = |y|n+2(( y 2−n)u + 2|y 2−n y 2−n uii)| |

∂Y i2 (| | ∂yi2 | | | · u + | |

2 4= 2(2 − n) y uiy i + uii|y| ,| |

so we get Δu(Ty) = |y|n+2Δ( y 2−nu).| |

yDefinition 1 The Kelvin Transform of u is defined to be Ku(y) = |y|2−nu( 2 ). y| |

Corollary 1 If u is harmonic on Rn − 0, then Ku is harmonic on Rn − 0.

Now let’s look at the degree gap again:

· · · · · · , −n, −(n − 1), −(n − 2), , 0, 1, 2, · · · · · ·

As we have known, those terms after “” correspond to homogeneous harmonic polynomials uk = rkBk(θ). Apply the Kelvin’s transform to this harmonic functions, we get

Kuk(y) = |y|2−n uk( y

2 ) = r 2−n r−kBk(θ) = r 2−n−kBk(θ). y| |

Thus those terms before “”, i.e. those blow up faster than the Green’s near origin, are just the Kelvin transform of the homogeneous harmonic polynomials.

2

Also we can know that if we apply Kelvin’s transform to homogeneous harmonic 1 1 1polynomials, then near ∞, we get harmonic functions which decays like

rn−2 , rn−1 , rn , 1and there is no harmonic function which decay like

rn−3 near ∞.

Harmonic at ∞

Definition 2 Suppose u is harmonic on Rn \ K, where K is a compact set, then we say u is harmonic at ∞ if its Kelvin transform Ku is harmonic at the origin.

Remark 1 1) We have to remove a compact set K, otherwise if u is harmonic on Rn

and harmonic at ∞, then u is bounded in Rn, thus u is constant, which we have no interest.

x2) u is harmonic at ∞ = Ku is harmonic at origin = ⇒ Ku(x) = |x|2−nu(⇒ |x|2 ) ≤ CC = ⇒ u(y) ≤ | |n−2 , i.e. u decay at least as fast as the Green’s function Γ.

y

Theorem 1 Suppose n > 2, u is harmonic in Rn \ K, where K is compact subset. If limx→∞ u(x) = 0, then u is harmonic at ∞.

Proof: Ku(x) = |x|2−nu( x 2 ) = o( x 2−n) as x → 0, thus Ku has harmonic extension

x| | | |to 0 by Removable Singularity Theorem.

CCorollary 2 If u harmonic on Rn \ K, and limx→∞ u = 0, then u(x) ≤ | |n−2 . x

This corollary tells us that harmonic function which decays at ∞, must decay at least as fast as the Green’s function: another “gap”.

Now we turn to the “n = 2” case.

Theorem 2 Suppose n = 2, u is harmonic in Rn \ K, where K is compact subset. If limx→∞

u(x) log |x| = 0, then u is harmonic at ∞.

Proof: Just the same as last theorem.

Corollary 3 n = 2. If u harmonic on Rn \ K, and limx→∞ u(x) = 0, then u has a

limit at ∞. log |x|

Kelvin II: Poission integral formula proof. Poission Integral Formula: If u is harmonic on Rn, then

u(x) = P (x, y)u(y)dσy, ∂B

3

2 1where P (x, y) = 1−|x|n nωn

. So x−y| |

Ku(x) = |x|2−n u( x

x 2 ) =

∂B |x|2−nP (

|x |2 , y)u(y)dσy.

x| |

xClaim: |x|2−nP ( | |2 , y) = −P (x, y).x

yIn fact, by using the formula | |y| − |y|x| = x x y , we get | |x| − | | |

1 1

2 , y) = x1 − |x|2

x|x|2−nP ( |xx |

| |2−n

| |x|2 n nωn− y|1

1 = |x|2−n

1 − |x|2

1 y y x n nωnx n| | | |y| − | | |

x 2 1 =

| | − 1 .y y x n nωn| |y| − | | |

But |y| = 1, so we get |x|2−nP (

x 2 , y) = −P (x, y).

x| |

By this claim, we have

Ku(x) = − P (x, y)u(y)dσy, ∀|x| > 1. ∂B

So Ku(x) is harmonic on |x| > 1. But Ku is analytic, thus ΔKu is also analytic, so ΔKu ≡ 0 on Rn − 0, which

means that Ku is harmonic.

4

Lecture 6

Weak maximum principle for linear elliptic operators

Now we consider the more general differential operators

L = aij (x)Dij + bi(x)Di + c(x),

i.e., for any C2 function u,

Lu = aij (x) ∂2u(x)

+ bi(x) ∂u(x)

+ c(x)u(x),∂xi∂xj ∂xi

where aij , bi, c are bounded functions.

Definition 1 Suppose L is like above. 1. If ∃λ(x) > 0 s.t. (aij (x)) > λ(x)I, then L is elliptic. 2. If ∃λ(x) > λ0 > 0 s.t. (aij (x)) > λ(x)I, then L is strictly elliptic. 3. If ∃∞ > Λ > λ0 > 0 s.t. ΛI > (aij (x)) > λ0I, then L is uniformly elliptic.

Theorem 1 Suppose L is elliptic in bounded domain Ω, u ∈ C2(Ω) ∩ C0(Ω), Lu ≥0, c(x) ≡ 0 in Ω, then

sup u = sup u. Ω ∂Ω

If Lu ≤ 0 instead, then inf u = inf u. Ω ∂Ω

Proof: Assume x0 ∈ Ω s.t. u(x0) = supΩ u, then (Dij u(x0)) ≤ 0, Diu(x0) = 0, so we get

Lu(x0) = aij Dij u(x0) ≤ 0.

If Lu > 0, then we have already get a contradiction. So the theorem is true for this simple case.

Now we turn to the general case Lu ≥ 0. Without loss of generality, we can assume rxa11 > 0. Let v = e

1 for some constant r, then

1 rxvi = re rx δ1i, vii = r 2 e 1 δ1i, and vij = 0, ∀i = j.

Thus 2 rx rx1 11 rxLv = a 11 r e

1 + b1 re = (a r 2 + b1 r)e

1 .

Since a11 > 0, we can choose r > 0 large enough such that Lv > 0, then for any > 0, we have

L(u + v) = Lu + Lv > 0.

1

So by the result of the simple case, we get

sup(u + v) = sup(u + v). Ω ∂Ω

Now we let > 0, we get sup u = sup u. Ω ∂Ω

For the second part, the proof is just the same.

To generalize the theorem, we define +u+ = maxu, 0, u− = u − u , Ω+ = x u(x) > 0. |

Theorem 2 With the same assumption as above, and suppose Lu ≥ 0, c ≤ 0, then

sup u ≤ sup u + . Ω ∂Ω

If Lu ≤ 0, c(x) ≤ 0 instead, then

inf u ≥ inf u−. Ω ∂Ω

In particular, if Lu = 0, c(x) ≤ 0 , then

sup |u| = sup u .| |Ω ∂Ω

Proof: Let L0u = aij Dij u + biDiu, then in Ω+ we have L0u ≥ −c(x)u ≥ 0. Thus by the previous theorem, we have

sup u = sup u. Ω+ ∂Ω+

So +sup u = sup u + = sup u = sup u = sup u ≤ sup u + .

Ω Ω Ω+ Ω+ ∂Ω+ ∂Ω

Uniqueness of solutions to Dirichlet problem

Corollary 1 (Uniqueness) Suppose L elliptic, c(x) ≤ 0, u, v ∈ C2(Ω) ∩ C0(Ω), and

Lu = Lv , in Ω, u = v , on ∂Ω,

then u = v in Ω. (Comparison theorem)If

Lu ≥ Lv , in Ω, u ≤ v , on ∂Ω,

then u ≤ v in Ω.

2

A Priori C0 estimates for solutions to Lu = f , c ≤ 0.

Theorem 3 Suppose L is strictly elliptic, c(x) ≤ 0, u ∈ C2(Ω) ∩ C0(Ω), where Ω isbounded domain.If Lu ≥ f , then there exists constant C = C(λ, Ω) s.t.

sup u ≤ sup u + + C sup f − . Ω ∂Ω Ω

| |

If Lu = f , then sup u ≤ sup u + C sup f .| | | | | |Ω ∂Ω Ω

Proof: Let L0 = aij Dij + biDi, then

rx1 11L0e = (a 2 + b1 r) > δ > 1r

for r large enough. Let rd rxv = sup u + + (e − e

1 ) sup |f − ,

∂Ω Ω |

where d > x1 for ∀x ∈ Ω. Then

Lv = L0v + cv ≤ L0v ≤ −δ sup f − ≤ − sup f − . Ω | |

Ω | |

∴ L(v − u) ≤ − sup f − − f ≤ 0, in Ω. Ω | |

But v ≥ u on ∂Ω by definition. Thus the last corollary tells us v ≥ u in Ω, i.e.

sup u ≤ sup u + + C sup f − . Ω ∂Ω Ω

| |

If Lu = f , replacing u by −u and f by −f , we thus get the second result.

Strong maximum principle

First we introduce the Hopf’s lemma.

Lemma 1 Suppose L is uniformly elliptic, c = 0, Lu ≥ 0 in Ω. Let x0 ∈ ∂Ω be such that (i) u is continuous at x0;

(ii) u(x0) > u(x), ∀x ∈ Ω; (iii) ∂Ω satisfies an interior sphere condition at x0.

Then the outer normal derivative of u at x0, if exists, satisfies

∂u (x0) > 0.

∂ν

If c(x) ≤ 0, then it holds for u(x0) ≥ 0. If u(x0) = 0, then it holds for any c(x).

3

|

Proof: Let B(y, R) be the interior sphere, i.e. B(y, R) ⊂ Ω and x0 ∈ ∂B(y, R). Define v(x) = e−αr2 − e−αR2

, where r = x − y . Then| |

Lv = aij Dij v + bi(−α(x i − y i)e−αr2 )

i= aij (−αδij e−αr2 + α2(x i − y i)e−αr2

) + bi(−α(x − y i e−αr2 ))

i= e−αr2 (α2 aij (x i − y i)(xj − yj ) − αa ii − αbi(x − y i))

2> e−αr2 (α2λ0r − αΛ − α sup b r)| | ·

Take A = BR(y) \Bρ(y), 0 < ρ < R, then for α large enough, Lv > 0 in A. The assumption (ii) tells us u(x) < u(x0) in Ω, in particular this holds on ∂B(y, ρ),

so there is some δ > 0 s.t. u(x) − u(x0) < −δ < 0 on ∂Bρ(y). Choose > 0 s.t. u(x) − u(x0) + v ≤ 0 on ∂Bρ(y). Since v = 0 on ∂BR(y), we automatically have u(x) − u(x0) + v ≤ 0 on ∂BR(y). Also we have known

L(u − u(x0) + v) = Lu + Lv > 0,

thus by the comparison theorem, we get

u − u(x0) + v ≤ 0, in A.

So ∂u ∂v

(x0) = v(R) > 0. ∂ν

(x0) ≥ −∂ν

For u(x0) = 0, just look at L − c(x).

Now we give the Strong Maximum Principle.

Theorem 4 Suppose L is uniformly elliptic, c = 0, Lu ≥ 0 in Ω, u ∈ C2(Ω) ∩ C0(Ω). If u achieves its maximum in the interior, then u is constant.

If Lu ≤ 0 and u achieves its minimum in the interior, then u is constant. If c ≤ 0, then u cannot achieve a nonnegative maximum in the interior unless u is

constant.

Proof: Assume u is not constant, and achieves maximum M at x0 in the interior. Let Ω− = x ∈ Ω u(x) < M. By definition we know Ω− ⊂ Ω, and ∂Ω− ∩ Ω = ∅

since u is not constant. Let x1 ∈ Ω− be s.t. x1 is closer to ∂Ω− than ∂Ω, and B(x1, R) be the largest ball in

Ω− centered at x1. Then u(y) = M for some y ∈ ∂B(x1, R). By Hopf’s lemma, we get

∂u (y) > 0.

∂ν This is a contradiction, since y is a maximum of u and so Du(y) should be 0.

4

Lecture 7

Quasilinear equations (minimal surface equation)

For any f : Rn −→ R, the graph of f is (x, f(x)) ⊂ Rn+1 . The tangents of the graph is (0, · · · , 1, 0, · · · , 0, fi), where 1 is on the ith slot. So the

n =normal vector is (−f, 1), and the unit normal vector is √1+

1 |f |

(−f, 1).2

The second fundamental form is a map (x) : TGx −→ TGx, (x)(ei) = ei n. (Since < n >= 1 = ⇒ X < n >= 0 = ⇒ 2 < X n >= 0 = ⇒ X n, n, n, n ∈ TG.) We compute:

∂ 1 n = (−f, 1))ei ∂xi ( 1 + |f |2

∂ 1 1 ∂)(−f, 1) +

1 + |f ∂xi ((−f, 1))= ∂xi (

1 + |f |2 |2

1 2fj fji 1 =

|2)3/2 (−f, 1) +

1 + |f−

2 (1 + |f 2 (f1i, · · · , fni, 0)

|= aij ej

where flflifj 1

aij = fij(1 + |f 2)3/2 −

1 + |f |2|∂ ∂ n = en+1.)(Assuming TxG = (

∂x1 , · · · , ∂xn , 0) and ˆMinimal ⇔ “0 mean curvature”, i.e.

aii = 0

flflifi 1 = = Δf⇒

(1 + |f 2)3/2 1 + |f |2|= flflifi = (1 + |f 2)Δf⇒ |

1 = div( fi) = 0 ⇒

1 + |f |2

1= ∂i( fi) = 0 ⇒

1 + |f |2

In general, the operator L = aij (x, u, Du)Dij u + · · · is called quasilinear. Now we check that the surface is “minimal”, i.e. has minimal area.

1 1Denote T : (x , · · · , xn) −→ (x , · · · , xn, f(x1 , · · · , xn)). Since

n+1 I T∗∂k = akj ∂j , (akj ) =

j=1 f

(n+1)×n

,

1

we get

T ∗gRn+1 (∂k, ∂l) = gRn+1 (T∗∂k, T∗∂l) = (AT A)kl = (I + f f T )kl = δkl + fkfl.

2The matrix I + f f T can be diagonized to diag1 + |f , 1, · · · , 1, so the area of |graph of f is

A(f ) = det(gij )dx = det(I + f f T )dx = (1 + |f 2)1/2dx. Rn Rn Rn

|

Thus

d A(f )h = A(f + th)|t=0 dt

1 =

2(1 + |f 2)−1/2 2 < f, h > dx | · Rn

f = <

1 + |f, h > dx

|2Rn

f = − h · div(

1 + |f |2 )dx.

Rn

Thus Minimal div( f ) = 0. 2

⇐⇒ √1+|f |

Fully nonlinear equations (MongeAmpere equation).

Suppose Ω ⊂ Rn . Now we consider the differential equations like

F [u] = F (x, u, Du, D2 u) = 0,

where F : Ω × R × Rn × S(n) −→ R, and S(n) is the set of all symmetric n × n matrices.

Definition 1 F is elliptic in some subset Γ ⊂ Ω × R × Rn × S(n) if ( ∂F )(γ) > 0,∂rij

∀γ = (x, z, p, r) ∈ Γ. If ∃Λ, λ > 0 such that ΛI > ( ∂F ) > λI for all γ ∈ Γ, then we say F uniformly ∂rij

elliptic. If u ∈ C2(Ω), and F is elliptic on range of x → (x, u, Du, D2u), then we say F is

elliptic with respect to u.

Example: MongeAmpere Equation

F [u] = detD2 u − f (x) = 0.

(Note that Δu = trace(D2u)).

2

We do some computation:

(−1)signσ r1σ(1)r2σ(2) · · ·det rij = rnσ(n),σ∈Sn

∂F Fij (r) = ∂rij = (i, j)−cofactor of r,

1(r−1)ij = Fij (r),det r

Fij (r) = det r · (r−1)ij . So F is elliptic when r is positive definite, and thus F [u] is elliptic if u is strictly

convex. More generally, F [u] = det D2u − f (x, u, Du) = 0 is elliptic for strictly convex

functions. Given F [u], define the linearization of F at a function u to be

F [u] : C2(Ω) → R, d

F [u + th] t=0.h −→ dt

|

d F [u](h) = F [x, u + th, Du + tDh, D2 u + tD2h]|t=0

dt= Fz (u)h + Fpi hi + Frij (u)Dij h

= (Frij (u)Dij + FPi (u)Di + Fz (u))h

= Lh

So our definition of elliptic at u linearization of F at u is an elliptic operator. ⇔

Example: Linearization of MongeAmpere:

F [u] = detD2 u − f (x).

F [u](h) = Frij (D2 u)Dij h

Let λi be eigenvalues of D2u, then eigenvalues of Frij are

λn, λn, λn−1λ2 · · · λ1λ3 · · · · · · , λ1 · · ·Certainly F is not uniformly elliptic.

Elementary Symmetric Function: σk (D2u) = Sum of principal k × k matrix.

σk (λ1, · · · , λn) = λik ,λi1 · · ·<iki1<···

Now for F [u] = detD2u − f (x), F [u](h) = Frij (D2u)Dij h, when is it elliptic?

Theorem 1 If σk > 0, σk−1 > 0, · · · , σ1 > 0, then Frij > 0.

3

Γk = componentof σk > 0. Example: n = 3.

det = λ1λ2λ3,

σ2 = λ1λ2 + λ1λ3 + λ2λ3,

Δ = λ1 + λ2 + λ3.

Γ3 = positive cone.For Γ2, σ2 = 0 is a cone, so σ2 > 0 has two components, Γ+ =2 x2 > 0∩σ1 > 0,

e.v. of Frij on (λ2 + λ3, λ1 + λ3, λ1 + λ2), λ1λ2 + λ1λ3 + λ2λ3 > 0, λ1 + λ2 + λ3 > 0.

Claim: If λ1 ≥ λ2 ≥ λ3, then λ2 > 0. In fact, by λ1λ2 + λ1λ3 + λ2λ3 > 0, we get

λ1(λ2 + λ3) + λ2λ3 > 0, i.e. λ1(λ2 + λ3) > −λ2λ3.

If λ2 ≤ 0,then λ2 + λ3 < 0, thus we get

− λ2 − λ3 < λ1 ≤−λ2λ3

λ2 + λ3

−λ2λ3 = ⇒− λ2 − λ3 <λ2 + λ3

λ2λ3= λ2 + λ3 >⇒λ2 + λ3

= ⇒(λ2 + λ3)2 < λ2λ3

= λ23 < 02 + λ2λ3 + λ2 ⇒

which is a contradiction. So we have λ1, λ2 > 0, thus λ1 + λ2 > 0. If λ1 + λ3 ≤ 0, then λ1λ3 < 0, which contradicts with λ2(λ1 + λ3) + λ1λ3 > 0. Thus

λ1 + λ3 > 0. Also from λ1(λ2 + λ3) + λ2λ3 > 0, we can get λ2 + λ3 > 0 by the same way.

Theorem 2 σ2(D2u) = f (x) is elliptic if f (x) > 0 and Δu ≥ 0. σk (D2u) = f (x) is elliptic if f (x) > 0, D2u ∈ Γ+

k , and σk > 0, σk−1 > 0, · · · , σ1 > 0.

Comparison principle for nonlinear equations.

First we give a maximum principle.

4

0

Theorem 3 Let u, v ∈ C0(Ω) ∩ C2(Ω) satisfy F [u] ≥ F [v] in Ω, u ≤ v on ∂Ω, and (i) F is elliptic along the straight line path tu + (1 − t)v, (ii) Fz ≤ 0. Then u ≤ v in Ω.

Proof: 1 d F [u] − F [v] = F [tu + (1 − t)v]dt

dt0 1 d = (tD2 u + (1 − t)D2 v) + Fpi (Diu − Div) + Fz (u − v)dt Frij dt0 1 1 1

= ( Frij dt)D2 ij (u − v) + ( Fpi dt)Di(u − v) + ( Fz dt)(u − v)

0 0 0

= L(u − v)> 0,

but u ≤ v on ∂Ω. Since elliptic on path, we get aij > 0 and c ≤ 0, thus u ≤ v in Ω.

Corollary 1 Suppose u, v ∈ C0(Ω) ∩ C2(Ω) satisfy F [u] = F [v] in Ω, and (i),(ii) hold, with u = v on ∂Ω. Then u = v in Ω.

Example: MongeAmpere. detD2u = f (x) > 0, detD2v = f (x), with u, v strictly convex, tu + (1 − t)v is also

strictly convex. So (i) works. For (ii), there is no z dependence. So u = v on ∂Ω implies u = v on Ω.

Similarly for σk . Result also works for Minimal surface.

Theorem 4 Suppose u ∈ C2(Ω), F [u] = 0, and F elliptic with respect to u. Also suppose F is C∞, (e.g. detD2u = f (x) > 0 ∈ C∞). Then u ∈ C∞(Ω).

Proof:Use difference quotients. Fix coordinate vector e1. Let v(x) = u(x + he1), h ∈ R, and ut = tv + (1 − t)u, 0 ≤ t ≤ 1. 1 d

F (x + the1, ut, Dut, Dtu))dt = F (x + he1, v, Dv, D2 v) − F (x, u, Du, D2 u) = 0,

dt 1 1 1 1

Fx1 (·)h + Fz (·)(v − u) + Fpi (·)Di(v − u) + Frij (·)Dij (v − u) = 0. 0 0 0 0

We can write this to be L(v − u) = −f · h.

Thus 1

L( v − u

) = L(Δhu) = −f = Fx1 (x + the1, ut, Dut, D

2 ut)dt. h 0

5

Sou ∈ W 2,p u ∈ W 3,pΔh , ∀p = ⇒ .

(We will prove this later.) By Sobolev embedding, u ∈ C2,α. Then f ∈ Cα = = u ∈ C3,α ⇒ Δhu ∈ C2,α ⇒

Δhu ∈ C3,α= = = ⇒ f ∈ C1,α ⇒ ⇒ u ∈ C4,α. Go on with this procedure, we get C∞ at last.

6

|

Lecture 8

A Priori Estimates for Poisson’s Equation. Recall that N f = Ω Γ(x − y)f (y)dy is called the Newtonian Potential of f .

Proposition 1 Suppose Ω is bounded domain, f ∈ L1(Ω), and ω = N f is the Newtonian potential of f . Then ω ∈ C1(Rn) and

Diω(x) = DiΓ(x − y)f (y)dy, ∀ x ∈ Ω. Ω

Proof: Γ = C| x| 2−n = x 1−n, therefore ⇒ | DΓ| ≤ C| |

v(x) = DiΓ(x − y)f (y)dy Ω

is well defined. (| v(x) DiΓ dy ≤ C f L∞ .)| ≤ f L∞ Ω | |

Define η(t) to be C∞ function with properties: (1) η(t) = 0 for t < ; (2) η(t) = 1 for t > 2; (3) 0 ≤ η(t) ≤ 1; (4) Dη| ≤ 2 . Define ω(x) to be

ω(x) = Γ(x − y)η( x − y )f (y)dy. | |Ω

Then ω(x) ∈ C1 and ω(x) −→ ω uniformly.

v(x) − Diω(x) = (DiΓ(x − y) − Di(Γ(x − y)η( x − y )))f (y)dy| |Ω

= Di((1 − η( x − y ))Γ(x − y))f (y)dy | |Ω

= Di((1 − η( x − y ))Γ(x − y))f (y)dy | ||x−y|≤2

So 2

v(x) − Diω(x)| ≤ sup | f | |x−y|≤2

( | Γ(x − y) + DiΓ(x − y) )dy | | | |

2 ≤ sup | f | ( |z|≤2

| Γ(z) dz + | DiΓ(z) dz)|| |z|≤2

2 C C ≤ sup f ( dz + dz)| | |z|≤2 | z| n−2 |z|≤2 | z| n−1

2 ≤ C sup | f | ( r≤

rdr + dr) r≤

sup | f | .≤ C · Now we have ω ω and Diω → v uniformly on compact subsets as → 0, thus →ω ∈ C1(Rn) and Diω = v.

1

Theorem 1 Let u ∈ C2(Ω), f ∈ L∞(Ω) and Δu = f in Ω. Then for any compact subdomain Ω ⊂⊂ Ω,

uC1(Ω ) ≤ C(uC0(Ω) + f L∞(Ω)).

Proof: Let ω be the Newtonian potential of f , i.e. ω(y) = Ω Γ(x − y)f (x)dx. Then from Green’s representation formula,

∂ ∂u v(y) = u(y) − ω(y) = u(x) Γ(x − y) − Γ(x − y) dσx

∂ν ∂Ω ∂νx

is a harmonic function. So

C

y∈Ω Ω x − y n−2 ds ≤ Cf ωC0(Ω) = sup Γ(x − y)f (x)dx| ≤ f L∞(Ω) sup | | L∞(Ω),|

y∈Ω Ω

and

Diω(y) = DiΓ(y − x)f (x)ds Ω

DiΓ(y − x)ds L∞(Ω) ≤ f Ω

C≤ f L∞(Ω) x − y n−1 ds| |

= Cf L∞(Ω)

Thus DiωC0(Ω) ≤ Cf L∞(Ω), and so

ωC1(Ω) ≤ Cf L∞(Ω).

Since v is harmonic, we have

DvC0(Ω ) ≤ CvC0(Ω)

≤ C(uC0(Ω) + ωC0(Ω))

≤ C(uC0(Ω) + f L∞(Ω)).

Thus uC1(Ω vC1(Ω) + ωC1(Ω ) ) ≤

≤ C(uC0(Ω) + f L∞ (Ω)).

More over, one can show that for any 0 ≤ α < 1,

uC1,α(Ω) ≤ C(uC0 + f L∞ ).

This is not true for α = 1.

2

Since we have

x + |ˆy − ξ y − ˆ x − ξ| | ≤ | | |δ

x| +y − ˆ≤ |21

x| + y − ξy − ˆ≤ |2 | |

1= y − ξ y − x|.⇒

2 | | ≤ |

Thus 1

II ≤ C · δ dy |ny − ξ|y−ξ|≥δ | R 1

δ dr ≤ C · δ r

δ(log R − log δ)≤ C · δ(log R + cδα−1) ( − log δ ≤ Cδα−1 for 0 ≤ α < 1)≤ C ·

= Cδ + Cδα ≤ Cδα

= C x − x α .| |

Combine the above results, we get ω ∈ C1,α(Ω). Further more, we get the C1,α estimate

ωC1,α (Ω) ≤ CfL∞ (Ω), 0 ≤ α < 1.

Just as last time, this implies

Corollary 1 Suppose Δu = f, f ∈ L∞(Ω), Ω ⊂⊂ Ω, then for 0 ≤ α < 1,

uC1,α(Ω) ≤ C(uC0(Ω) + fL∞ (Ω)).

Remark 1 Look at the above proof and assume f ∈ Lp(Ω).

Diω(x) − Diω(x) = (DiΓ(x − y) − DiΓ(x − y))f(y)dy Ω

DiΓ(x − y) − DiΓ(x − y)|q dy1/q f(y) pdy≤ | | | 1/p

Ω Ω 1 1 p

+ = 1 = ⇒ q = p − 1

. p q

In the 2nd part of the above proof, we had

1/qx||DDiΓ(ˆII ≤ Ω ||x − ˆ x − y)||q

1)1/qδ(

|nq≤ C ·

y − ξ|y−ξ|≥δ |

δ · (δ−nq+n)1/q≤ C · n

= C · δ1−n+ q .

4

nLet α = 1 − n + np−1 , then p(α − 1 + n) = pn − n, i.e. p = 1−α , we have: p nIf Δu = f, f ∈ Lp(Ω), p = 1−α , Ω

⊂⊂ Ω, then

uC1,α(Ω) ≤ C(fLp(Ω) + uC0(Ω)).

Later we will show uW 2,p(Ω) ≤ C(fLp(Ω) + uC0(Ω)).

So C1,α estimate follows by Sobolev Embedding theorem.

5

| |

Lecture 9

C2 estimates

Proposition 1 Let f be bounded and f ∈ Cα(Ω) (locally), ω(x) = Ω Γ(x − y)f (y)dy be the Newtonian potential of f . Then ω ∈ C2(Ω), Δω = f and

Dij ω(x) = Dij Γ(x − y)(f (y) − f (x))dy − f (x) DiΓ(x − y)νj dσy , Ω0 ∂Ω0

where Ω0 is nice domain (for instance, has C2 boundary) with Ω0 ⊃ Ω, and we just extend f to vanish outside of Ω.

Proof: Define u(x) = Ω0 Dij Γ(x − y)(f (y) − f (x))dy − f (x) DiΓ(x − y)νj dσy , the ∂Ω0

1 αright hand side of above. Since Dij Γ(x − y) ≤ x − y ,|x−y|n , |f (y) − f (x)| ≤ f Cα | |u(x) is well defined.

Take ηε(x) ∈ C∞(R) such that (1) 0 ≤ ηε(x) ≤ 1; (2) ηε(x) = 0 for x ≤ ε; (3) 2ηε(x) = 1 for x ≥ 2ε; (4) |ηε| ≤ . Let ε

vε(x) = DiΓ(x − y)ηε( x − y )f (y)dy. | |Ω

Claim: vε ∈ C1(Ω), and vε → Diω uniformly in Ω. In fact, we have

Dj vε(x) = Dj (DiΓ(x − y)ηε( x − y ))f (y)dy | |Ω

= Dj (DiΓ(x − y)ηε( x − y ))(f (y) − f (x))dy | |Ω0

+ f (x) Dj (DiΓ(x − y)ηε( x − y ))dy | | Ω0

= Dj (DiΓ(x − y)ηε( x − y ))(f (y) − f (x))dy | |Ω0

− f (x) Di(Γ(x − y)ηε( x − y ))νj (y)dsy ∂Ω0

= Dj (DiΓ(x − y)ηε( x − y ))(f (y) − f (x))dy − f (x) DiΓ(x − y)νj (y)dsy .| |Ω0 ∂Ω0

1

| |

| | | | |

Now

u(x) −Dj vε(x) Dj (1 − ηε)DiΓ(f (y) − f (x))dy| | ≤ |x−y|≤2ε

2 Cα (|Dij Γ + DiΓ ) x − y dy ≤ f

|x−y|≤2ε |

ε | | |

|α

1 2 1 ≤ Cf Cα ( |x−y|≤2ε |x − y n−α dy +

|x−y|≤2ε | |n−1−α dy)ε x − y|

2 Cα ((2ε)α + (2ε)1−α)≤ Cf

ε≤ Cf Cα εα .

Thus Dj vε(x) converges to u uniformly on compact subsets as ε → 0. But vε → Diω, so ω ∈ C2(Ω) and u = Dij (Ω).

Now take Ω0 = BR(0) ⊃ Ω. Since ΔΓ = 0 away from 0, the integral formula tells us

Δω(x) = −f (x) DiΓνi(y)dsyx−y =R

f (x) (xi − yi)= νi(y)dy |nnωn |x−y =R x − y| |

f (x) (xi − yi) (xi − yi)= )dy nωn |x−y =R x − y n x − y

f (x) 1 =

nωn Rn−1 x−y =R

dsy | |

= f (x).

Remark 1 In fact, in the proof we only needed Dini continuity, i.e. f (x) − f (y) | | ≤

ϕ(r)ϕ( x − y ) where ∞dr < ∞.0 r| |

C2,α estimates for Poisson equation.

Proposition 2 Consider B1 = BR(x0) ⊂ B2R(x0) = B2, f ∈ Cα(B2R), where 0 < α < 1. Let ω(x) = B2

Γ(x − y)f (y)dy, the Newtonian Potential of f in B2. Then ω ∈ C2,α(BR(x0)) and we have estimate

D2ω0;BR + Rα D2ω α;BR ≤ C(f 0;B2R + Rα f α;B2R ),| | | |

where C = C(n, α) is constant.

2

Proof: Let x1 ∈ B1, in last proposition we have showed

Dij ω(x) = Dij Γ(x − y)(f (y) − f (x))dy − f (x) DiΓ(x − y)νj dσy . B2 ∂B2

Thus

α|Dij ω(x) f Cα(x)C dy + |f (x) DiΓ(x − y) dsy| ≤ | | B2

|DijΓ(x − y)||x − y| | | | ∂B2

1 1 f Cα(x) |n−α dy + C f (x) dsy≤ C| | B2

|x − y| |

∂B2 x − y n−1|

1 |1

f Cα(x) | |n−α dy + C f (x) nωn(2R)n−1≤ C| |B3R(x) x − y

| | Rn−1

f Cα(x)(3R)α + C|f (x)≤ C| | | ≤ C( f (x) + Rα f Cα(x)).| | | |

Take x ∈ B1, then

Dij ω(x) = Dij Γ(x − y)(f (y) − f (x))dy − f (x) DiΓ(x − y)νj dσy , B2 ∂B2

Thus

Dij ω(x) − Dij ω(x) = Dij Γ(x − y)(f (y) − f (x))dy − Dij Γ(x − y)(f (y) − f (x))dy B2 B2 − f (x) DiΓ(x − y)νj dsy + f (x) DiΓ(x − y)νj (y)dsy .

∂B2 ∂B2

Boundary terms are

f (x) (DiΓ(x − y) − DiΓ(x − y))νj dsy (I) ∂B2

(f (x) − f (x)) (DiΓ(x − y))νj dsy (II ) ∂B2

1Solid terms in Bδ (ξ), where δ = x − x , ξ = 2 (x + x), are | |

Dij Γ(x − y)(f (y) − f (x))dy (III) Bδ (ξ)

Dij Γ(x − y)(f (x) − f (y))dy (IV ) Bδ (ξ)

On B2 \ Bδ (ξ), we write Dij Γ(x − y) = (Dij Γ(x − y) − Dij Γ(x − y)) + Dij Γ(x − y), the corresponding solid terms are

(Dij Γ(x − y) − Dij Γ(x − y))(f (y) − f (x))dy (V ) B2 \Bδ (ξ)

(f (x) − f (x)) Dij Γ(x − y)dy (V I) B2\Bδ (ξ)

3

|

| |

Now we begin to estimate these terms separately.

(I) = f (x) (DiΓ(x − y) − DiΓ(x − y))νjdsy ∂B2

≤ |f (x) |DDiΓ(ˆ x dsyx − y)||x − ˆ|∂B2

1 1 ≤ |f (x)|δ · c ∂B2

x − y|n

δ

dsy ≤ c|f (x)|δRn nωn(2R)n−1

1 |ˆ

δ f (x)|δ

R ≤ 2c|f (x) f (x)|(

2R )α≤ c| |

2R ≤ 2c|

δα

≤ c|f (x) .| Rα

(II) = (f (x) − f (x)) (DiΓ(x − y))νjdsy ∂B2

1 ≤ |f |Cα(x)δα c

∂B2 |x − y|n−1 dsy

1 ≤ |f |Cα(x)δα c nωn(2R)n−1

Rn−1

f Cα(x)δα .≤ C| |

(III) = DijΓ(x − y)(f (y) − f (x))dy Bδ (ξ)

dyDijΓ(x − y)||(f (y) − f (x))≤ Bδ (ξ)

| |

c α f Cα(x)dy ≤ c Bδ (ξ) x − y n |x − y| | |

1 f Cα(x) Bδ (ξ) |x − y n−α dy ≤ c| |

|1

f Cα(x)≤ c| |B 3

2δ (x)

|x − y|n−α dy

3α ≤ c|f |Cα (2

)α

≤ c|f |Cα(x)δα .

Similarly,

(I V ) ≤ c f Cα (x)δα .| |

The last two terms are to be continued.

4

Lecture 12

March 30th

, 2004

Remark. Since many of our results rely on the regularity of the Newtonian Potential, and hence

use Proposition 2 of Lecture 9, we will assume througout that the Holder constant α ranges in the

open interval (0, 1).

Review from last time

Regularity Theorem. B ⊆ Rn, a ball, u ∈ C2(B) ∩ C0(B), f ∈ Cα(B), with 0 < α < 1.

Suppose u solves Laplace’s Equation: ∆u = f on B, u = 0 on ∂B. Then u ∈ C2,α(B).

In the interior of B, just use estimates on the Newtonian Potential (NP) and on harmonic

functions. On the boundary of B use translation & inversion maps to map ball to upper half plane

with flat boundary. Then note that the estimates on the NP work upto the boundary and an

inversion map is smooth away from the origin.

Corollary. ϕ ∈ C2,α(B), f ∈ Cα(B), with 0 < α < 1.. Then Poisson’s Equation: ∆u = f on

B, u = ϕ on ∂B, has a unique solution u ∈ C2,α(B).

By the above if we can solve for v such that ∆v = f − ∆ϕ on B, v = 0 on ∂B, then v ∈ C2,α(B).

Let u := v + ϕ ∈ C2,α(B). This u solves our original equation! So we just need to be able to solve

uniquely the above homogeneous equation with a C2(B) ∩ C0(B) solution. Then the Theorem will

guarantee it is actually C2,α(B).

In order to do that, set w := NP(g), where g := f − ∆ϕ ∈ Cα (as f ∈ Cα, ϕ ∈ C2,α). Indeed

w ∈ C2(B) ∩ C0(B) from the elementary properties of the Newtonian Potential. Furthermore

∆w = g. If we could make sure somehow the boundary values would be 0 we would be done

as all assumptions of the Theorem would hold. In order to do that, we need to find a function

1

h ∈ C2(B) ∩ C0(B) solving

∆h = 0 on B,h = −w on ∂B

. And indeed by Poisson’s Integral Formula we

can do this. Letting v := w + h ∈ C2(B) ∩ C0(B) we have indeed a the required solution for the

homogeneous problem.

Solving Poisson’s/Laplace’s equation with regularity upto the boundary on gen-

eral domains

Suppose we are given an (open) domain Ω ⊆ Rn, different than a ball, or equivalently some open

subset in a Riemannian manifold (M,g), and that we would like to develope a similar theory for

the Poisson and Laplace equations on these domains. In other words prove a priori estimates upto

the boundary for these domains.

Localizing to a neighborhood in Rn of a point on the boundary ∂Ω intersected with Ω, we could

map it to a neighborhood of H := x = (x1, . . . , xn) ∈ Rn∣

∣xn ≥ 0. This localization is tantamount

to working with the manifolds local coordinates, and then we must work with ∆g, the Riemannian

Laplacian.

We see that indeed we will be able to extend our theory to these generalized domains once we

show our boundary estimates hold for general elliptic operators.

Constant coefficients operators

Let L0u(x) = AijDiju(x) = f(x) with Aij a constant matrix satisfying 0 < λ|v|2 ≤ Aijvivj ≤

Λ|v|2,∀ 0 6= v ∈ Rn. This two-sided inequality will be referred to as uniform ellipticity.

Theorem. Let u be as above and 0 < α < 1.

I. If u ∈ C2(Ω), f ∈ Cα(Ω), then ∀ Ω′ ⊂⊂ Ω (i.e ”Ω′ precompact in Ω”) there exists C =

C(λ,Λ,Ω′,Ω, n) such that

||u||C2,α(Ω′) ≤ C · (||u||C0(Ω) + ||f ||Cα(Ω)).

2

II. If u ∈ C2(Ω) ∩ C0(Ω ∪ T ), f ∈ Cα(Ω ∪ T ) and u = 0 on T , then ∀ Ω′⊂⊂ Ω there exists

C = C(λ,Λ,Ω′,Ω, n) such that

||u||C2,α(Ω′∪T ′) ≤ C · (||u||C0(Ω∪T ) + ||f ||Cα(Ω∪T )),

where T ′ := Ω′ ∩ T . We assume that T is a flat boundary portion (portion of a hyperplane in Rn)

contained in ∂Ω .

Setup: Let H be an invertible linear transformation represented by multiplication by a constant

matrix Hkl, and let H−1 denote its inverse. Being linear, by rotating if necessary, we may assume

it maps the upper half space to itself, and that the flat boundary portion remains flat. Put

u := u H−1 and y = Hx. Then u : Ω −→ R,

Diu(y) = Dlu(H−1y) · H−1li (summation)

from D applied to u

(

y1...

yn

· [H−1]

)

= u(

[ ylH−1l1 , . . . ,H−1

ln yn ])

. Then

DiDj u(y) = DkDlu(H−1y)H−1kj H−1

li = (H−1)T · D2u(H−1y) · H−1,

⇒ HT · D2u(y) · H = D2u(x).

Plugging this into our elliptic equation we get AlkHilDiDj uHjk = AlkDlku(x) = f(x), or

HilAlkHjkD2u = (HAHT )D2u(y) = f(x) = f(H−1y) =: f(y).

Choosing appropriate H can diagonalize A: HAHT = diag(λ1, . . . , λn). Set

P := Hdiag(λ− 1

21 , . . . , λ

− 12

n ). Then PAPT = I, and in the domain H(Ω), which has flat boundary,

we get the simple Poisson equation ∆u = f ∈ Cα. By the theory we developed earlier in the course

for this equation on such domains, ∀ Ω′ ⊂⊂ H(Ω) we have the interior estimates

||u||C2,α(Ω′) ≤ C · (||u||C0(H(Ω)) + ||f ||Cα(H(Ω))).

Now ||u||C2,α(Ω′) ≤ C · ||u||C2,α(H(Ω′)), ||u||Cα(Ω′) ≤ C · ||u||Cα(H(Ω′)), where we have used for the

last two the identities ||g||C0(H(Ω)) = supy∈H(Ω) |g(y)| = supx∈Ω |g(x)| = ||g||C0(Ω) and

3

||g||Cα(H(Ω)) = supy1 6=y2∈H(Ω)

|g(y1) − g(y2)|

|y1 − y2|α= sup

x1 6=x2∈Ω

|g(x1) − g(x2)|

|Hx1 − Hx2|α

= supx1 6=x2∈Ω

|g(x1) − g(x2)|

|x1 − x2|α

( |x1 − x2|

|Hx1 − Hx2|

)α

≤ ||g||Cα(Ω) · (smallest eigenvalue of H)−1.

Here we use H is a diffeomorphism. The C2,α inequality follows similarly using HT D2u(y)H =

D2u(x). Note that since H,H−1 are both strictly positive, the above inequalities can be shown

to hold in both directions (with different constants). That is to say all norms of u are equivalent

to those of u. This observation combined with the above interior estimates for u gives us interior

estimates for u in Ω′.

As for boundary estimates (part II of the Theorem): we have seen that we can assume wlog that

H maps the upper half plane to itself. Then our above inequalities for equivalence of the norms

extend to the boundary of course, and since our theory (Lecture 11) gives boundary estimates for

u we are done.

Interpolation

Theorem. Let Ω′ ⊂⊂ Ω, u ∈ C2,α(Ω), with 0 < α < 1. For any ǫ > 0, ∃ C(ǫ) such that

|u|Ck,β(Ω′) ≤ C(ǫ) · |u|Co(Ω) + ǫ · |u|C2,α(Ω′).

Note these are the semi-norms not the full norms! Also, as ǫ → 0, C(ǫ) → ∞.

The case k = 1, β = 0. Let x, x′ ∈ Ω′, x′′ ∈ Ω \Ω′, such that all three points lie on a single line

segment parallel to the xi axis and such that Diu(x) = u(x′′)−u(x′)2ǫ

≤2|u|

C0(Ω)

2ǫ. From the fact that

x′′ is not in Ω′ we will get a global C0 norm involved (i.e norm over all Ω instead of just over Ω′).

Now let x ∈ Ω′,

4

∫ x

x

Diiu = Di(x) − Di(x),

from which follows

|Di(x)| ≤ |Di(x)| + |

∫ x

x

Diiu| ≤1

ǫ|u|C0(Ω) + max

x∈segment xx

in xi direction

Diiu · |x − x| ≤1

ǫ|u|C0(Ω) + ǫ · |u|C2(Ω′).

The case k = 2, β = 0. Fix i and look at Diu. Again choose points on a segment in the xl

direction such that Dliu(x) =Diu(x′′) − Diu(x′)

2ǫ≤

2|Du|C0(segment)

2ǫ. Now

|Dliu(x)| ≤ |Dliu(x)| + |Dliu(x) − Dliu(x)| ≤1

ǫ|Du|C0(segment) + |D2u|C2(Ω) · |x − x|α,

which by the first case is

≤1

ǫ·( 1

ǫ′|u|C0(Ω) + ǫ′ · |u|C2(Ω)

)

+ ǫa · |u|C2,α(Ω),

hence

|u|C2(Ω′) ≤ C · |u|C0(Ω) + C ′ · ǫa · |u|C2,α(Ω′).

For the cases to follow let x, y ∈ Ω′ and denote by x a point (to be chosen later) on the line

segment xy.

The case k = 0, β ∈ (0, 1]. We note that we can bound |u|C0,β(Ω′) in a simple manner since

u(x) − u(y)

|x − y|β≤

Du(x) · |x − y|

|x − y|β≤ |u|C1 · |x − y|1−β ≤ |u|C1 · ǫ1−β , if |x − y| ≤ ǫ,

2|u|C0

ǫβ, if |x − y| > ǫ.

or in other words

|u|C0,β(Ω′) ≤2|u|C0

ǫβ· |u|C0(Ω) + ǫ1−β · |u|C1(Ω′).

5

The case k = 1, β ∈ (0, 1]. We again note the dichotomy (x as above)

Diu(x) − Diu(y)

|x − y|β≤

DDiu(x) · |x − y|

|x − y|β≤ |u|C2 · |x − y|1−β ≤ |u|C2 · ǫ1−β , if |x − y| ≤ ǫ,

2|u|C1

ǫβ, if |x − y| > ǫ.

or

|u|C1,β(Ω′) ≤2|u|C1

ǫβ· |u|C1(Ω) + ǫ1−β · |u|C2(Ω). ≤ C · |u|C0(Ω) + C ′ · |u|C2(Ω′)

where in the last inequality we used one of the previous cases.

The case k = 2, β ∈ (0, α). Once again

Diju(x) − Diju(y)

|x − y|β=

Diju(x) − Diju(y)

|x − y|α· |x − y|α−β ≤

|u|C2,α(Ω′)ǫα−β , if |x − y| ≤ ǫ,

2|u|C2(Ω)

ǫβ, if |x − y| > ǫ.

or

|u|C2,β(Ω′) ≤2

ǫβ· |u|C2(Ω′) + ǫα−β · |u|C2,α(Ω′). ≤ C · |u|C0(Ω) + C ′ · |u|C2(α)Ω

′

where in the last inequality we used one of the previous cases.

Remark. The Interpolation technique works also for Ω′ ⊂⊂ Ω with flat boundary involved: we get

inequalities with the flat boundary portion included, by the theory developed in Lecture 11.

6

Lecture 21

May 4th, 2004

Higher Elliptic Regularity

We may obtain using our previous result yet higher regularity. This is a key point in regularity

theory, and is quite beautiful. The previous theorem guaranteed that under certain conditions

on the coefficients any W 1,2(Ω) solution is in fact W 2,2(Ω′) for any Ω′⊂⊂ Ω. Naturally we would

expect that if we had assumed more regularity on the coefficients we would get that any W 2,2(Ω)

solution is in fact W 3,2(Ω′). If this could work then we could thus start from a W 1,2 solution, a

weak solution, and get arbitrary regularity if we are willing to shrink more and more the domain.

Alternatively we could get just enough regularity so that our embedding theorems ensure us the

solution is Holder , then C1,α, C2,α and by repeating this process any desired regularity! (Here we

use the Corollary of Lecture 18) This is really something. We started from a weak solution which

need not be a function and using our theory we are able to show it behaves well and is in fact

smooth! We will make this discussion precise in the sequel.

Let us see how this interpolation process works.

We assume as before u ∈ W 1,2(Ω) is a weak solution of Lu = f (though we know this implies

more regularity in the interior, we won’t use it now). Then

∀ v ∈ C10(Ω) −

∫

Ω

aijDjuDiv +

∫

Ω

(biDiu+ cu)v =

∫

Ω

fv,

and the idea is now to look at a smaller space of test functions v and see what that gives. In a

sense the extra regularity we find will come for free. Take in particular v ∈ C20(Ω). That means

v = Dkw where w ∈ C10(Ω), and

∫

Ω

aijDjuDiDkw =

∫

Ω

(−biDiu− c+ f)Dkw.

1

Since w is twice continuously differentiable we may interchange derivatives above and integration

by parts yields

−

∫

Ω

Dk(aijDju)Diw =

∫

Ω

(−biDiu− c+ f)Dkw,

and

−

∫

Ω

aijDkDjuDiw =

∫

Ω

DkaijDjuDiw +

∫

Ω

(−biDiu− c+ f)Dkw,

and further

∫

Ω

aijDj(Dku)Diw =

∫

Ω

[−Di(DkaijDju) + Dk(b

iDiu− c+ f)] · w =:

∫

Ω

g · w,

which gives that Dku is a weak solution of the second order equation

L(Dku) = g

since this holds ∀ w ∈ C10(Ω).

Now we note that L = aij is strictly elliptic, and that if

• aij ∈ C1,1(Ω)

• bi, c ∈ C0,1(Ω)

• u ∈W 2,2(Ω)

then we will have g ∈ L2(Ω), and the Theorem of the previous lecture applies and we have Dku ∈

W 2,2(Ω′) ∀ Ω′⊂⊂ Ω, i.e u ∈ W 3,2(Ω′) as we wished to show. Indeed we get this extra regularity

seemingly for free and we may continue this for higher derivatives.

Let us see what kind of a priori estimates we get on the higher norms. From the Theorem we

have

||Dku||W 2,2(Ω′) ≤ c(

||Dku||W 1,2(Ω) + ||g||L2(Ω)

)

.

≤ c(

||u||W 2,2(Ω) + ||u||W 1,2(Ω) + ||u||W 2,2(Ω) + ||f ||L2(Ω)

)

2

where the last three terms come from the definition of g. We now shrink from Ω to Ω′ and from Ω′

to Ω′′ so that terms on the lhs are evaluated on Ω′′ and the ones on the rhs on Ω′. But then those

terms on the rhs can be evaluated by Ω terms using our theorem once again! We get altogether

then

||Dku||W 2,2(Ω′′) ≤ c(

||u||W 1,2(Ω) + ||f ||L2(Ω)

)

.

We state this as the following

Theorem. Let u ∈W1,20 (Ω) be a weak solution of Lu = f in Ω, and assume

• L strictly elliptic with (aij) > γ · I, γ > 0

• aij ∈ Ck,1(Ω)

• bi, c ∈ Ck−1,1(Ω)

• f ∈W k,2(Ω).

Then for any precompact set Ω′⊂⊂ Ω, u ∈W k+2,2(Ω′) and

||u||W 2,2(Ω′) ≤ C(||aij ||Ck,1(Ω), ||b||Ck−1,1(Ω), ||c||Ck−1,1(Ω), λ,Ω′,Ω, k, n) ·

(

||u||W 1,2(Ω) + ||f ||Wk,2(Ω)

)

.

What we just did is the analogue in regularity theory of the technique we used in the Holder part

of the course. As there, we want to differentiate the original equation but the lack of regularity

hinders us from doing do directly. We then take difference quotients and get bounds which now

allow us to differentiate and get all higher estimates. This shows us how special solutions of such

partial differetial equations are among general functions in those Sobolev Spaces.

Corollary. Let u ∈W 1,20 (Ω) be a weak solution of Lu = f in Ω, and assume


• f, aij , bi, c ∈ C∞(Ω)

Then on the whole domain, u ∈ C∞(Ω).

3

Proof. For all k ∈ N, f ∈ W k,2(Ω) ⇒ Ω′ ⊂⊂ Ω u ∈ W k+2,2(Ω′). By the Sobolev Embedding

(Corollary to Lecture 18) then u ∈ Cm(Ω′),m < k+2− n2 , hence u ∈ C∞(Ω′). Apply this reasoning

for Ω′ a ball around each point p ∈ Ω! to get u ∈ C∞(Ω). Smoothness is indeed a notion defined

pointwise.

Global Regularity (upto the boundary)

Up until now our regularity results were for the space W 1,20 (Ω), i.e functions which vanish on ∂Ω.

We now study W 1,2(Ω).

Theorem. Suppose u ∈W 1,2(Ω) is a (weak) solution of Lu = f and assume


• aij ∈ C0,1(Ω)

• bi, c ∈ L∞(Ω)

• f ∈ L2(Ω)

• Ω has C2 boundary.

• ∃ ϕ ∈W 2,2(Ω) such that u− ϕ ∈W1,20 (Ω).

Then u ∈W 2,2(Ω) = W 2,2(Ω) on all of Ω with

||u||W 2,2(Ω′) ≤ C(||aij ||C0,1(Ω), ||b||C0(Ω), ||c||C0(Ω), λ,Ω′,Ω, ∂Ω, n)·

·(

||u||W 1,2(Ω) + ||f ||L2(Ω) + ||ϕ||W 2,2(Ω)

)

.

Note that even for the 0 boundary values case this theorem gives a stronger conclusion: regularity

upto the boundary with a uniform estimate on all of Ω. The price is the assumption that the

boundary is regular enough.

4

Proof. We reduce to the zero boundary case in the usual manner. Suppose we could prove the

Theorem for all zero boundary Dirichlet Problems. Then for the problem L(u−ϕ) = f ′ := f −Lϕ

on Ω, u − ϕ = 0 on ∂Ω (this is precisely the assumption u − ϕ ∈ W1,20 (Ω)) we would have the

desired estimates

||u− ϕ||W 2,2(Ω′) ≤ C ·(

||u− ϕ||W 1,2(Ω) + ||f − Lϕ||L2(Ω)

)

⇒

||u||W 2,2(Ω′) ≤ C ·(

||u||W 1,2(Ω) + ||f ||L2(Ω) + ||ϕ||W 2,2(Ω)

)

,

since ϕ ∈W 2,2(Ω) and L is of second order.

So we assume indeed u ∈W1,20 (Ω). We now take a neighborhood containing a boundary portion

and map it through a C2 diffeomorphism ψ (i.e ψ−1 exists and is C2) onto Rn with the boundary

portion mapping into the hyperplane xn = 0.

We pull back everything onto the flat boundary situation using (ψ−1)⋆– the original equation is

∀ v ∈ C10(Ω) −

∫

Ω

aijDjuDiv +

∫

Ω

(biDiu+ cu)v =

∫

Ω

fv,

and the pulled-backed one

∀ v ψ−1 ∈ C10(ψ(Ω)) −

∫

ψ(Ω)

Jac(ψ−1)aij ψ−1(ψ−1)⋆Dju(ψ−1)⋆Div

+

∫

ψ(Ω)

Jac(ψ−1)(bi ψ−1(ψ−1)⋆Diu+ c ψ−1)v ψ−1 =

∫

Ω

fv.

As we can assume ψ−1 preserves the given orientation of Rn and it is a diffeomorphism then

Jac(ψ−1) > 0 and therefore we still have a strictly elliptic equation Lu = f and the C2 of ψ−1 guar-

antees the f ∈ L2(Ω) and that aij is still Lipschitz, and bi and c still bounded (e.g. (ψ−1)⋆(biDiu) =

bi ψ−1 · (ψ−1)⋆Diu = bi ψ

−1 · Di(ψ−1)⋆u = bi ψ

−1 · Di(u ψ−1) = bi ψ−1 · Dku · Di(ψ

−1)k

(summation over k)).

Now we note that the difference quotients proof from last time still works for ∆hl u for each

of the directions l = 1, . . . , n − 1 tangent to the boundary. So by applying that Theorem we

5

get Dlu ∈ W 1,2(ψ(Ω′)) and hence Dij u ∈ L2(ψ(Ω′)) except for i = j = n. Since ψ is a C2

diffeomorphism, the same holds for u.

So in order to finish the proof we go back to the proof. We have W 2,2 except possibly in the

boundary, so may write the equation

Lu = aijDiju+ DiaijDju+ biDiu+ c · u = f,

a.e. All terms are in L2 except annDnnu. But then isolating it on one side of the equation we see

it must be as well, so it can not blow up at the boundary.

So now indeed we are done: we cover all of Ω with a finite number of small ball cover the

boundary portion and another Ω′ covering the rest of the interior and we have the desired estimate

on each of those domains.

We now have higher regularity upto the boundary:

Corollary. Let u ∈ W 1,2(Ω) be a weak solution of Lu = f in Ω, and u = ϕ on ∂Ω (i.e

u− ϕ ∈W1,20 (Ω)) and assume



• bi, c ∈ Ck−1,1(Ω)

• f ∈W k,2(Ω).

• ∂Ω is Ck+2.

Then u ∈W k+2,2(Ω) uniformly on the whole domain and

||u||Wk+2,2(Ω)

≤ C(||aij ||Ck,1(Ω), ||b||Ck−1,1(Ω), ||c||Ck−1,1(Ω), λ, ∂Ω,Ω′,Ω, k, n) ·(

||u||W 1,2(Ω) + ||f ||Wk,2(Ω) + ||ϕ||Wk+2,2(Ω)

)

.

If k = ∞ then u ∈ C∞(Ω).

6

The only difference from the compactly supported case is that we need now to have at our

disposal a modified Sobolev Embedding: W k+2,2 ⊆ Cm(Ω) instead of the one we proved with

Wk+2,20 ⊆ Cm(Ω). This is indeed the case as one can show by modifying the latter’s proof under

the assumption of smooth enough boundary.

Improvement of our estimate

Assume u ∈W 1,20 (Ω), Lu = f ∈ L2(Ω), aij , bi, c ∈ L∞(Ω). Then

||u||W 2,2(Ω) ≤ c ·(

||u||L2(Ω) + ||f ||L2(Ω)

)

.

Proof. During the proof which involved the W 1,2(Ω) norm on the rhs we arrived at the inequality

λ||Du||L2(Ω) ≤

∫

Ω

aijDiuDjϕ =

∫

Ω

(−biDiu− cu+ f)ϕ

for all test functions ϕ ∈ C10(Ω) but in fact also for all test functions in the completion W 1,2

0 (Ω)! In

particular we can take ϕ = u ! But actually for ϕ = u we can get this directly just from the strict

ellipticity without having to go through the difference quotients process (just true for this special

choice of v !). In particular we needn’t assume more than L∞(Ω) regularity on the aij now! We

continue then and get

λ||Du||L2(Ω) ≤

∫

Ω

aijDiuDju =

∫

Ω

(−biDiu− cu+ f)u

= −

∫

Ω

(−biu(Diu) +

∫

Ω

(−cu2 + fu)

≤1

2ǫ

∫

Ω

|Du|2 +

n∑

i=1

1

2

1

ǫ

∫

Ω

|biu|2 +

∫

Ω

(−cu2 + fu).

For ǫ < λ one can move the first term to the lhs to conclude that (thanks to strict ellipticity we

can now divide out by λ and get a uniform bound!)

7

||Du||L2(Ω) ≤1

2||u||L2(Ω) + ||f ||L2(Ω).

Now we can plug this in to our original estimate to get the desired improvement

||u||W 2,2(Ω) ≤ c ·(

||u||W 2,1(Ω) + ||f ||L2(Ω)

)

.

= c ·(

||u||L2(Ω) + ||Du||L2(Ω) + ||f ||L2(Ω)

)

.

≤ c′ ·(

||u||L2(Ω) + ||f ||L2(Ω)

)

.

Similarly this improvement applies to u ∈W 1,2(Ω) though it will not apply up to the boundary;

we will have

||u||W 1,2(Ω′) ≤ c(

||u||L2(Ω) + ||f ||L2(Ω)

)

by taking ϕ = η ·u with η = 1 on Ω′ and applying the above argument. If we want an estimate on all

of Ω we need to add a term ||ϕ||W 2,2(Ω) to the rhs by applying the above result to u−ϕ ∈W1,20 (Ω)

for f ′ := f − Lϕ.

8

Lecture 22

May 6th, 2004

Define u+ := maxu, 0, u− := minu, 0. For a generalized function u ∈ W 1,2(Ω) we say

u ≤ 0 on ∂Ω if u+ ∈ W1,20 (Ω). Similarly we say u ≤ v on ∂Ω if u − v ≤ 0 on ∂Ω. Finally define

sup∂Ω

u := infc : u ≤ c on ∂Ω.

Weak L2Maximum Principle

We consider the divergence form equation

Lu := Di(aijDju) + biDiu + cu = f,

with c ≤ 0.

Theorem. Suppose u ∈ W 1,2(Ω). Assume

• c ≤ 0


• ||bi||C0(Ω) ≤ Λ

• f ∈ W k,2(Ω)

Then

If Lu ≥ 0 then supΩ u ≤ sup∂Ω u+.

If Lu ≤ 0 then infΩ u ≥ inf∂Ω u−.

If c = 0 then the above holds with |u| instead of u.

The last conclusion follows from the first two since in that case u and −u each satisfy one

inequality.

1

Proof. From the statement we have that u satisfies an inequality in the weak sense, the integral

inequality

∀ v ∈ W1,20 (Ω) −

∫

Ω

aijDjuDiv +

∫

Ω

(biDiu + cu)v ≥ 0

or

∫

Ω

aijDjuDiv ≤

∫

Ω

biDiuv +

∫

Ω

cuv.

Now restrict to v such that u · v ≥ 0. Since c ≤ 0

∫

Ω

aijDjuDiv ≤

∫

Ω

biDiuv ≤ Λ

∫

Ω

v|Du|.

If supΩ u > sup∂Ω u+ then choose k ∈ R such that sup∂Ω u+ ≤ k < supΩ u. Now pick a specific v,

v := (u − k)+. This v is 0 everywhere except where u exceed k, and in particular where it exceeds

the supremum of the boundary values. Indeed we have v ∈ W1,20 (Ω) as well as

Dv =

Du for u > k (there v > 0)0 for u ≤ k (there v = 0)

.

And so

∫

Ω

aijDjvDiv ≤ Λ

∫

Γ

v|Dv|,

where Γ := suppDv ⊆ suppv. Now by strict ellipticity the lhs majorizes λ∫

Ω|Dv|2 hence

λ||Dv||2L2(Ω) = λ

∫

Ω

|Dv|2 ≤ Λ

∫

Γ

v|Dv| ≤ Λ||v||L2(Γ)||Dv||L2(Ω)

by the Holder Inequality (HI) (for p = q = 2) and therefore

||Dv||L2(Ω) ≤ c(λ,Λ) · ||v||L2(Γ) = c ·(

∫

Γ

v2)

12

≤ c ·(

∫

Γ

(v2)n

n−2

n−2n

∫

Γ

1n2

2n

)12

= c · Vol(Γ)1n ||v||

L2n

n−2 (Γ)

2

once again by the HI for p = nn−2 , q = n

2 . On the other hand by the Sobolev Embedding

||v||L

2nn−2 (Ω)

≤ C||Dv||L2(Ω) and so over all

||v||L

2nn−2 (Ω)

≤ C||Dv||L2(Ω) ≤ C||v||L2(Ω)c · Vol(Γ)1n ||v||

L2n

n−2 (Ω)

and therefore Vol(Γ)1n ≥ C where the constant is independent of k ! (note v ∈ L2(Ω)). Let therefore

k → supΩ

u. Then we see u must still attain its maximum on a set of positive measure! But then

Dv = Du = 0 there! Which in turn contradicts this previous bound on the volume of Γ = supp(Dv).

So we conclude that there exists no k ∈ [sup∂Ω

u+, supΩ

u), in other words sup∂Ω

u+ ≥ supΩ

u. The

second case of the Theorem follows now since if Lu ≤ 0 then L(−u) ≥ 0 and the first case

applies.

Corollary. Let L be strictly elliptic with c ≤ 0. Assume u ∈ W1,20 (Ω) satisfies Lu = 0 on Ω.

Then u = 0 on Ω.

An a priori Estimate

We improve slightly on the aesthetics of the higher regularity proved in the previous lecture for

the case c ≤ 0.

Theorem. Let u ∈ W1,20 (Ω) ∩ W k+2,2(Ω) be a weak solution of Lu = f in Ω, and assume



• bi, c ∈ Ck−1,1(Ω) (for k = 0, C−1,1 := C0 = L∞)

• f ∈ W k,2(Ω)

• ∂Ω is Ck+2

3

Then

||u||W k+2,2(Ω) ≤ c · ||Lu||W k,2(Ω).

Note that the assumption u ∈ W k+2,2(Ω) is superfluous once u ∈ W1,20 (Ω) in light of our previous

results.

Also note that this is exactly analogous to what we did in our Holder theory study; there we

proved Lu = f ∈ Ck,α(Ω), c ≤ 0 implies ||u||Ck+2,α(Ω) ≤ c||f ||Ck,α(Ω).

Proof. Case k = 0. We want to prove ||u||W 2,2(Ω) ≤ c · ||Lu||W 2,2(Ω). and we already know that

||u||W 2,2(Ω) ≤ c ·(

||u||L2(Ω) + ||Lu||W 2,2(Ω)

)

,

so we now try to demonstrate ||u||L2(Ω) ≤ c||Lu||W 2,2(Ω) for all u ∈ W 2,2(Ω)∩W1,20 (Ω). If not, pick

a sequence um ⊆ W 2,2(Ω)∩W1,20 (Ω) with ||um||L2(Ω) = 1, ||Lum||W 2,2(Ω) −−−−→

m→∞ 0 and hence by

what we know

||um||W 2,2(Ω) ≤ c.

Since W 2,2(Ω) is a Hilbert space exists a subsequence which converges weakly to u ∈ W 2,2(Ω)

(note Alouglou’s Theorem applies as we have separability and every Hilbert space is a reflexive

Banach space). Since W 2,2(Ω) → L2(Ω) is a compact embedding we actually have um → u ∈ L2(Ω)

(i.e strongly). But now ||Lum||L2(Ω) → 0, hence Lu = 0 weakly. Since c ≤ 0 this implies by our

previous work u = 0 ! In contradiction with ||um||L2(Ω) = 1 as um → u in L2(Ω) so ||u||L2(Ω) = 1

allora . . .

Corollary. Let Ω ⊆ Rn be a bounded domain with Ck+2 boundary. Then the map

∆ : W k+2,2(Ω) ∩ W1,20 (Ω) −→ W k,2(Ω)

is an isomorphism.

4

Proof. Injective: By the previous Corollary if L(u1 − u2) = 0 on Ω and u1 − u2 ∈ W1,20 (Ω) then

u1 − u2 = 0. This actually applies also to any two such functions in W 1,2(Ω) with equal boundary

values.

Surjective: Let f ∈ W k,2(Ω). We can find a solution Lu = f with u in W2,20 (Ω) by Riesz

Representation Theorem and our regularity theory. So ∆−1 exists and by our above Theorem

satisfies

||∆−1f ||W k+2,2(Ω) ≤ C · ||f ||W k,2(Ω).

So ∆−1 is continuous. From the definition of ∆ we see that

||∆u||W k,2(Ω) ≤ ||u||W k+2,2(Ω)

(note no constant on rhs ) we see also ∆ itself is a continuous map between those spaces (wrt to

their topologies).

Corollary. For appropriate L (see above Theorems) with c ≤ 0

L : W k+2,2(Ω) ∩ W1,20 (Ω) −→ W k,2(Ω)

is an isomorphism.

Proof. Injective: Exactly as above.

Surjective: We employ the Continuity Method (CM) which will work out exactly as in the

Schauder case. Consider the family of equations

Ltu := (1 − t)Du + tLu = f.

Recall that the CM will provide for the surjectivity of L based on the surjectivity of ∆ (proved

above) once we can prove

||u||W k+2,2(Ω) ≤ c · ||Ltu||W k,2(Ω)

5

with c independent of t. And this is indeed the case since each of the Lt satisfies the assumptions

of the previous Theorem.

Negative Sobolev Spaces

What happens for the k = −1 case? Where does ∆ map to? ∆u is not defined as a function,

though it is as a distribution: given v ∈ W1,20 (Ω) one can define

∆u(v) := −

∫

Ω

∇u · ∇v

which realizes ∆u as a linear functional on W1,20 (Ω), in other words

∆ : W1,20 (Ω) −→ (W 1,2

0 (Ω))⋆.

The motivation for this definition lies in the fact that when we look at the equation −

∫

Ω

∇u ·∇v =

∫

Ω

∆uv we actually mean

∫

Ω

v · (∆udx) and ∆udx gives a distribution under the identification of

distributions with measures.

Recall the inner product as we defined it in W1,20 (Ω) is

(u, v) = +

∫

Ω

∇u · ∇v.

By the Riesz Representation Theorem given any element F ∈ (W 1,20 (Ω))⋆ there exists a unique

u ∈ W1,20 (Ω) such that F (v) = (u, v), so

F (v) = (u, v) = +

∫

Ω

∇u · ∇v = (−∆u)(v),

as distributions. Therefore ∆ is surjective. Injectivity follows from the definition of ∆. Continuity

of the inverse is also provided for by the Riesz Representation Theorem

||u||W 1,20 (Ω) = ||−∆u||(W 1,2

0 (Ω))⋆ .

6

We conclude from this short discussion that ∆ : W1,20 (Ω) −→ (W 1,2

0 (Ω))⋆ =: W−1,2(Ω) is an

isomorphism of Hilbert Spaces. This is a natural extension to our previous results, and adopting

this notation they all extend now to the case k = −1.

7

Lecture 23

May 11th, 2004

LpTheory

Take f any measurable function on a domain Ω ⊆ Rn and define the distribution function of f

µf (t) := |x ∈ Ω : |f(x)| > t|. We use alternatively | · | and λ(·) to denote the Lebesgue measure.

Proposition. Assume f ∈ Lp(Ω) for some p > 0.

I) µf (t) ≤ t−p

∫

Ω

|f |pdx.

II)

∫

Ω

|f |pdx = p

∫ ∞

0

tp−1µf(t)dt.

In order for the second equation to make sense we need the distribution function to be measurable

and indeed it is as f itself is.

Proof. First

∫

Ω

|f |pdx ≥

∫

f>t

|f |pdx ≥ tpλ(x : f(x) > t) = tpµf (t).

Second, assume first p = 1. By Fubini’s Theorem one can interchange order of integration in

∫

Ω

|f | =

∫

Ω

∫ |f(x)|

0

dtdx =

∫ ∞

0

∫

Ω

Ix∈Ω: f(x)>tdxdt =

∫ ∞

0

µf (t)dt.

For general p

µfp(t) = |x : fp(x) > t| = |x : f(x) >p√

t| = µf (p√

t) =

and so

1

p

∫ ∞

0

tp−1µf (t)dt =

∫ ∞

0

µfp(tp)d(tp) =

∫

Ω

|f |pdx.

Marcinkiewicz Interpolation Theorem. Let 1 ≤ q < r < ∞ and let T : Lq(Ω) ∩ Lr(Ω) −→

Lq(Ω) ∩ Lr(Ω) be a linear map. Suppose there exist constants T1, T2 such that

∀ f ∈ Lq(Ω) ∩ Lr(Ω) µTf (t) ≤(T1||f ||Lq(Ω)

t

)q

, µTf (t) ≤(T2||f ||Lr(Ω)

t

)r

, ∀ t > 0.

Then for any exponent in between q < p < r, T can be extended to a map Lp(Ω) −→ Lp(Ω) for all

f ∈ Lq(Ω) ∩ Lp(Ω). And moreover,

||Tf ||Lp(Ω) ≤[ p

q − p(2T1)

q +p

r − p(2T2)

r]

1p

||f ||Lp(Ω).

Otherwise stated: weak (q, q) & weak (r, r) =⇒ strong (p, p) p ∈ (q, r), though not for the

endpoints, the constants blow-up there (we say an operator is strong (p1, p2) if it maps functions

in Lp1 to functions in Lp2 . We say it is weak (p1, p2) if its domain is in Lp1 and its distribution

function satisfies the first inequality in the assumptions above with q replaced by p2).

Proof. Take f ∈ Lq(Ω) ∩ Lr(Ω), and let s > 0. Let

f1 :=

f(x) |f(x)| > s

0 |f(x)| ≤ s

f2 :=

0 |f(x)| > s

f(x) |f(x)| ≤ s

indeed one notices that f = f1 + f2. The trick will be to let this splitting of f vary by letting s

itself vary. So |Tf | ≤ |Tf1| + |Tf2|. If Tf(x) > t at some point x ∈ Ω then either Tf1 > t/2 or

Tf2 > t/2. This translates into

2

µTf(t) ≤ µTf1(t/2) + µTf2

(t/2)

≤( T1

t/2

)q∫

Ω

|f1|q +

( T2

t/2

)r∫

Ω

|f2|r.

We choose the smaller exponent q for the terms where f is large (f1) and larger one r for where f

is small (f2), intuitively. This will make sense in a moment when it will be clear how this guarantees

that our two integrals — with different integration domains — are finite. By the Proposition we

have

∫

Ω

|Tf |pdx = p

∫ ∞

0

tp−1µTf (t)dt.

and once we substitute in the above inequality we get

∫

Ω

|Tf |pdx ≤ p

∫ ∞

0

tp−1[( T1

t/2

)q∫

Ω

|f1|q +

( T2

t/2

)r∫

Ω

|f2|r]

dt

= p(2T1)q

∫ ∞

0

(

∫

|f |>s

|f1|q)

tp−1−qdt + p(2T2)r

∫ ∞

0

(

∫

|f |≤s

|f2|r)

tp−1−rdt.

We chose s > 0 arbitrary in the above construction of fi. In particular we may let it vary. This

is a neat trick. We set s = t to get

p(2T1)q

∫ ∞

0

(

∫

|f |>s

|f |q)

sp−1−qds + p(2T2)r

∫ ∞

0

(

∫

|f |≤s

|f |r)

sp−1−rds

= p(2T1)q

∫

Ω

|f |qdx

∫ |f |

0

sp−1−qds + p(2T2)r

∫

Ω

|f |rdx

∫ ∞

|f |

sp−1−rds

= (2T1)q p

q − p

∫

Ω

|f |p + (2T2)r p

r − p

∫

Ω

|f |p.

Altogether

∫

Ω

|Tf |pdx ≤[ p

q − p(2T1)

q +p

r − p(2T2)

r]

·

∫

Ω

|f |p.

3

Remark. In Gilbarg-Trudinger, p.229, a different constant is achieved which is slightly stronger

than ours (as can be seen using the AM-GM Inequality). This is done by introducing an additional

constant A, letting t = As and later choosing A appropriately.

Back to the Newtonian Potential

We defined the Newtonian Potential of f

ω ≡ Nf :=

∫

Ω

Γ(x − y)f(y)dy =1

n(2 − n)ωn

∫

Ω

1

|x − y|n−2dy.

Claim. (Young’s Inequality) N : Lp(Ω) −→ Lp(Ω). Moreover continuously so– ∃ C such

that ||Nf ||Lp(Ω) ≤ C||f ||Lp(Ω).

Remark. For p = 2 we proved in the past much more: ∆(Nf) = f ∈ L2(Ω) ⇒ Nf ∈ W 2,2(Ω).

Also our previous estimates on the Newtonian Potential can actually be made to extend our Claim

to W 1,p(Ω) regularity. These estimates can not give though W 2,p(Ω) estimates (see the beginning

of the next Lecture).

Proof.

ω : = Γ ⋆ f =

∫

Ω

Γ(x − y)f(y)dy

=

∫

Ω

f(y)Γ(x − y)1p Γ(x − y)1−

1p dy

≤

∫

Ω

|f(y)pΓ(x − y)|dy

1p

∫

Ω

|Γ(x − y)|dy1− 1

p

≤ C ·

∫

Ω


1p .

4

since Γ(x−y) ∼1

|x − y|n−1and therefore is integrable over R

n. Therefore we have an upper bound

on ωp which we can integrate

∫

Ω

ωpdx ≤

∫

Ω

Cp

∫

Ω


dx

= Cp

∫

Ω

∫

Ω

|f(y)|p|Γ(x − y)|dxdy

= Cp

∫

Ω

|f(y)|p(

∫

Ω

|Γ(x − y)|dx)

dy

≤ C

∫

Ω

|f(y)|pdy.

where we applied Fubini’s Theorem.

Theorem. Let f ∈ Lp(Ω) for some 1 < p < ∞ and let ω = Nf be the Newtonian Potential of

f . Then ω ∈ W 2,p(Ω) and ∆w = f a.e. and

||D2w||Lp(Ω) ≤ c(n, p,Ω) · ||f ||Lp(Ω).

For p = 2 we have even

∫

Rn|D2ω|2 =

∫

Ω

f2.

Proof. We prove just for p = 2, leaving the hard work for the next and last lecture. First we assume

f ∈ C∞0 (Rn). From long time ago: f ∈ C∞

0 (Rn) ⇒ ω ∈ C∞(Rn) and ∆ω = f (Holder Theory for

the Newtonian Potential).

Let B := BR a ball containing suppf ⇒

∫

BR

(Dω)2 =

∫

BR

f2 =

∫

Ω

f2 (1)

We embark now on our main computation

5

∫

BR

|D2ω|2 =

∫

BR

DijωDijω (summation) = −

∫

BR

Dj(Dijω)Diω +

∫

∂BR

DijωDiωνjdθ

= −

∫

BR

Di(Djjω)Diω +

∫

∂BR

DijωDiωνjdθ

= −

∫

BR

Di(∆ω)Diω +

∫

∂BR

∂

∂νDωDωdθ

=

∫

BR

(∆ω)2 −

∫

∂BR

∆ω ·∂

∂νωdθ +

∫

∂BR

∂

∂νDω · Dωdθ

=

∫

BR

(∆ω)2 +

∫

∂BR

∂

∂νDω · Dωdθ.

The last equality results from our assumption that f vanishes on ∂B, i.e has compact support

inside Ω. Now since f is smooth

Diω(x) =

∫

Ω

DiΓ(x − y)f(y)dy ≤C

Rn−1,

Dijω(x) =

∫

Ω

DijΓ(x − y)f(y)dy ≤C

Rn.

Therefore as we let R −→ ∞, the second term - which is integrated only over the sphere of radius

R in Rn – tends to 0. Then we have in the limit the desired result (after substituting (1) for the

rhs).

Now if f ∈ L2(Ω), approximate it by functions fm ∈ C∞0 (Rn) (possible by the density argument

used in the past: C∞0 (Ω) = L2(Ω)) such that fm −−−−→L2(Ω) f . From the Claim above ||Nf ||Lp(Ω) ≤

C||f ||Lp(Ω), hence ||N(fi − fj)||Lp(Ω) ≤ C||fi − fj ||Lp(Ω), from which ωm ≡ Nfm −−−−→L2(Ω) Nf ≡ ω.

Now ∆ωj = fj and by the C∞0 (Ω) case applied to the Dirichlet Problem ∆(ωi − ωj) = fi − fj

∫

Rn|D2(ωi − ωj)|

2 =

∫

Ω

|fi − fj |2.

6

As the rhs tends to 0 for i, j large we have that D2ωm converges in L2(Ω), i.e ωm converges in

W 2,2(Ω). Since we already know its limit is ω ∈ L2(Ω) we conclude that in fact ω ∈ W 2,2(Ω) !

7

Lecture 13

April 1st, 2004

Now we would like to extend our estimates to general domains.

Proposition. Let Ω be a C2,α domain and u ∈ C2,α(Ω), with 0 < α < 1. Given ǫ > 0,∃ c = c(ǫ,Ω)

s.t. for k = 0, 1, β ∈ (0, 1) and k = 2, β < α

||u||Ck,β (Ω) ≤ c · ||u||C0(Ω) + ǫ · |u|C2,α(Ω).

Note this is a global estimate, i.e. upto the boundary.

We will use the remark from last time concerning domains with a portion of a hyperplane on

the boundary which will provide us with a needed estimate. We choose a C2,α injective function

Ψ which maps B(x, r) ∩ Ω (x ∈ ∂Ω) in such a manner as to map B(x, r) ∩ ∂Ω onto a portion of

a hyperplane. Its inverse S := Ψ−1 is also C2,α. Set u := u Ψ−1, the pulled-back function, and

T ′ := Ψ(Br/2(x) ∩ ∂Ω). For the domain in the image we have estimates as just mentioned

|u|Ck,β(Ψ(Br/2(x))∩T ) ≤ c(ǫ) · |u|C0(Ψ(Br(x))) + ǫ · |u|C2,α(Ψ(Br/2(x))∩T ). (1)

Let S = (S(1), . . . , S(n)). Now calculate

Diu = Dlu · (S(l))i ≡ Dlu · Di(S(l)) summation over l understood

DjDiu = DkDlu(S(k))j(S(l))i + DluDj(S(l))i

|u|C0(Br(x)) = |u|C0(Ψ(Br(x))).

Now

|u|Cα(Ψ(Br(x))) = supy1 6=y2

|u(y1) − u(y2)|

|y1 − y2|α= sup

x1 6=x2

|u(x1) − u(x2)|

|Ψ(x1) − Ψ(x2)|α

≤ supx1 6=x2

|u|Cα(Br(x))|x1 − x2|

α

|Ψ(x1) − Ψ(x2)|α.

1

Now since Ψ,Ψ−1 are C2,α they are in particular Lipschitz (C0,1), i.e ∃K > 0 s.t.

|Ψ(x1) − Ψ(x2)| ≤ K|x1 − x2|, |Ψ−1(y1) − Ψ−1(y2)| ≤ K|y1 − y2|,

or

1/K · |x1 − x2| ≤ |Ψ(x1) − Ψ(x2)| ≤ K · |x1 − x2|.

When plugged-in to our previous computation this yields

≤ |u|Cα(Br(x)) · K−α.

Analogously we get

|u|Cα(Br(x)) ≤ |u|Cα(Br(x)) · Kα.

We also have analogously for the first derivatives

|Du|C0 ≤ |Du|C0 · K1

and

|Du|C0 ≤ |Du|C0 · K−11

where K1 depends on |Ψ|C1 .

And for the second derivatives ∃K2 depending on |Ψ|C2 (i.e depending on the domain Ω !) with

|D2u| ≤ K1 · |D2u| + K2|Du| ≤ K ′||u||C2 .

Similarly

|D2u| ≤ K ′′||u||C2

Finally for C2,α norms, can again show norms on both sides are equivalent using K,K ′,K ′′. So

we conclude - for each point x ∈ ∂Ω we have a ball and the estimate (1) holds there. The above

long discussion yields that we have furthermore the desired estimate concerning the norms therein:

||u||Ck,β(Br/2(x)∩Ω) ≤ c · ||u||C0(Br(x)∩Ω) + ǫ · |u|C2,α(Br/2(x)∩Ω).

2

Since Ω is compact we can cover ∂Ω by finitely many balls. On the interior of each we have this

(desired) estimate. To complete the proof we therefore just need to make sure this estimate also

holds in the interior of Ω. We take a set Ω′ ⊂⊂ Ω and a number D > 0 s.t.

A) if x, y ∈ Ω and d(x, y) ≤ D then either

i) both x, y are both contained in one of the small ball covering the boundary of Ω.

There we have the desired estimate already.

or

ii) both x, y are in Ω′. Then we have the desired estimate as well from our interior

estimate from the previous lecture for the semi-norms, which extends to give the des-

ired estimate for the norms.

B) if d(x, y) > D we have|D2u(x) − D2u(y)|

|x − y|α≤ 1/Dα · 2 · |D2u|C0(Ω). And so also in this case

we get the desired estimate.

We now try to extend our results to the case of non-constant coefficients , which was in fact our

original goal.

Theorem. Let u ∈ C2,α(Ω), L = aij(x)Dij + bi(x)Di + c(x) (summation understood over double

indices), and suppose Lu = f ∈ Cα(Ω). Assume furthermore that L has coefficients in Cα(Ω) and

is uniformly elliptic, i.e

•1

Λ· δij ≤ aij(x) ≤ Λ · δij

• ||aij(x)||Cα(Ω), ||bi(x)||Cα(Ω), ||c(x)||Cα(Ω) < Λ.

Then ∀ Ω′⊂⊂ Ω, ∃ c = c(Λ, n,Ω′,Ω) such that

||u||C2,α(Ω′) ≤ c(

||u||C0(Ω) + ||f ||Cα(Ω)

)

.

Thanks to the interpolation proposition we really only need to bound the C2,α semi-norm with

the above rhs since all the other semi-norms contained in the C2,α norm are bounded above by it

together with the C0 semi-norm which is part of the rhs already.

3

We will try to make use of the Holder constant of the coefficients to relate our situation to the

constant coefficients case. The idea is that locally the coefficients are almost constant, and the

degree to which this almost true is good enough for us (continuous wouldn’t be good enough).

In this spirit, rewrite Lu = f as

(aij(x) − aij(x0) + aij(x0))uij(x) + bi(x)ui(x) + c(x)u(x) = f,

or

aij(x0)uij(x) = −(aij(x) − aij(x0))uij(x) − bi(x)ui(x) − c(x)u(x) + f.

Calling the lhs L0u and the rhs F (x) we thus define a constant coefficient uniformly elliptic

operator and a function.

Let Ω′⊂⊂ Ω with x0 ∈ Ω′. Take B(x0, µ · D) ∈ Ω′, B(x0,D) ∈ Ω with µ > 0 small and denote

d := µ ·D. As mentioned above we only need to prove an estimate for the |D2u|Cα(Ω′). We observe

that for any y0 ∈ Ω′, and for d small enough

D2u(x0) − D2u(y0)

|x0 − y0|α≤

c · (||u||C0(B(x0, d2 )) + ||F ||Cα(B(x0, d

2 ))), if y0 ∈ B(x0,d

2)

2|u|C2(Ω′)

dα+ ||F ||C0(Ω′)), if |x0 − y0| ≥

d

2.

since when d is small enough, we can think of having a a uniform elliptic equation with almost

constant coefficients in B(x0,d2 ) and then apply our previous results. We therefore get

|D2u|Cα(Ω′) ≤ c′ · (||u||C0(Ω′) +1

dα|u|C2(Ω′) + sup

x0∈Ω′

||F ||Cα(B(x0, d2 ))). (2)

We first estimate the last term which is of a local nature. Observe that for two Holder functions

k, l ∈ Cα(Ω), |k · l|Cα(Ω) ≤ |k|Cα(Ω) · |l|C0(Ω) + |k|C0(Ω) · |l|Cα(Ω) ≤ C(Ω) · |k|Cα(Ω)|l|Cα(Ω), while for

the norms themselves we see from the first inequality that ||k · l||Cα(Ω) ≤ ||k||Cα(Ω) · ||l||Cα(Ω). So

||F ||Cα(B(x0, d2 ))) ≤ ||aij(x) − aij(x0)||Cα(B(x0, d

2 )) · ||uij ||Cα(B(x0, d2 ))+

+||bi||Cα(B(x0, d2 )) · ||ui||Cα(B(x0, d

2 )) + ||c||Cα(B(x0, d2 ))||u||Cα(B(x0, d

2 )) + ||f ||Cα(B(x0, d2 )).

4

Remember that the Cα norm includes the C0 and Cα semi-norms. We have

||aij(x)−aij(x0)||Cα(B(x0, d2 )) = sup |aij(x)−aij(x0)|+ |aij(x)−aij(x0)|Cα ≤ c · ||aij ||Cα(Ω′)|x−x0|

α

=⇒

||F ||Cα(B(x0, d2 )) ≤ c ·Λdα(||D2u||Cα(B(x0, d

2 )) + ||Du||Cα(B(x0, d2 )) + ||u||Cα(B(x0, d

2 ))) + ||f ||Cα(B(x0, d2 ))

and using interpolation for the 2nd and 3rd terms we find

≤ c · Λdα(|u|C2,α(B(x0, d2 )) + |u|C0(Ω′)) + ||f ||Cα(B(x0, d

2 )). (3)

We now come back to the 2nd term of (2).

We now let x0 range over all points of Ω′. For each x0 we will find different d,D such that (2)

above holds.

It so happens that this inequality is useless unless we can not control from below the term

involving d−α. The problem is that as x0 → ∂Ω, d = d(x0) → 0. To overcome this we assume

we chose Ω′ such that Ω′⊂⊂ Ω′′ ⊂⊂ Ω and such that dist(∂Ω′, ∂Ω′′) ≥ δ0 > 0. Then for problematic

points x0 → ∂Ω and y0 6∈ B(x0,d2 ) we can replace the term

2|u|C2(Ω′)

dα by2|u|

C2(Ω′)

(d+δ0)α . Therefore we

have overcome the problem by means of introducting a constant c′′ depending on Ω′ (and Ω) and

we can replace (2) by

|D2u|Cα(Ω′) ≤ c′ · (||u||C0(Ω′) + c′′|u|C2(Ω′′) + supx0∈Ω′

||F ||Cα(B(x0, d2 ))). (4)

The 2nd term can be bounded above using our interpolation theory:

|u|C2(Ω′′) ≤ c(ǫ1) · ||u||C0(Ω) + ǫ1 · |u|C2,α(Ω′′) (5)

with ǫ1 > 0 of our choice. And so we have all in all

|u|C2,α(Ω′) ≤ c′·(

||u||C0(Ω)+c(ǫ1)·||u||C0(Ω)+ǫ1·|u|C2,α(Ω′′)+cΛdα(|u|C2,α(Ω′)+|u|C0(Ω′))+||f ||Cα(Ω′)

)

.

Now choose d small enough such that cΛdα <1

4and ǫ1 <

1

4·|u|C2,α(Ω′)

|u|C2,α(Ω′′).

5

Remark. This Theorem gives still just a Schauder type interior estimate. Next time we will try

to extend it to the boundary.

6

Lecture 14

April 6,th 2004

Extending interior Schauder estimates to flat boundary part

Theorem. u ∈ C2,α(Ω∩T ), Lu = f, u = 0 on T , with 0 < α < 1. Assume coefficients are bounded

in C2,α(Ω∩T ) as well as uniformly elliptic. Then ∀Ω′ ∩T ′ ⊂⊂ Ω∩T, ∃ c = c(Λ, n,Ω′,Ω, T ′, T ) such

that

||u||C2,α(Ω′∩T ′) ≤ c(

||u||C0(Ω∩T ) + ||f ||Cα(Ω∩T )

)

.

Proof. As in the last remark we see that our proof consisted of perturbing the equation at any

x0 ∈ Ω′ and relying on our constant coefficients estimates and interpolation methods. Both of these

hold upto the flat boundary from our previous work.

Global Schauder estimates

Theorem. Let Ω be a C2,α domain and u ∈ C2,α(Ω)⋆ with 0 < α < 1. Let L be uniformly elliptic

with Cα(Ω) bounds on coefficients . Let

Lu = f, f ∈ Cα(Ω),u = ϕ on ∂Ω.

Then ∃ c = c(Ω,Λ, n) such that

||u||C2,α(Ω) ≤ c(

||u||C0(Ω) + ||f ||Cα(Ω) + ||ϕ||C2,α(∂Ω)

)

.

⋆ We note that Gilbarg-Trudinger intend by this notation locally Holder while we will take it

henceforth to mean globally Holder in the sense that we assume supx0 6=y0∈ΩD2

u(x0)−D2u(y0)

|x0−y0|αis

finite.

1

Here we let ||ϕ||C2,α(∂Ω) := infϕ:Ω→R

||ϕ||C2,α(Ω).

Proof. It is enough to prove for the case of zero boundary values: if we can solve the Dirichlet

problem

Lv = f − Lϕ =: f ′ ∈ Cα on Ω,

v = 0 on ∂Ω.

we can also solve our original one by setting v + ϕ solves the original equation. And if we have

the above announced estimates for v then by the triangle inequality (for the relevant norms) and

the uniform ellipticity (which gives ||Lϕ||Cα(Ω) ≤ c · ||ϕ||C2,α(Ω)) the same estimates will hold for

u, possibly with a different constant.

So indeed we may assume ϕ = 0.

By definition of a C2,α domain ∃Ψ,Ψ−1 ∈ C2,α(Rn → Rn) mapping each small portion of the

boundary of Ω, say B(x0, R) ∩ ∂Ω for x0 ∈ ∂Ω to flat boundary. We set as in computations

in the past u := u Ψ−1 and then Du = Du Ψ−1′, D2u = D2u · Ψ−1′ + Du · D2Ψ−1. These

computations convince us once more that the relevant norms on a, b, c and a, b, c are equivalent

using Ψ,Ψ−1 ∈ C2,α (e.g we find ||b||Cα(Ω) ≤ ||b||Cα(Ω)(|Ψ|C1,α(+)|Ψ|C2,α(Ω) ≤ C · Λ).

We have for the flat boundary

||u||C2,α(Ψ(B(x0, 12 R)∩Ω)) ≤ c(

||u||C0(Ψ(B(x0,R)∩Ω)) + ||f ||Cα(Ψ(B(x0,R)∩Ω))

)

.

Now by our above work we know this holds also for u in B(x0, R) ∩ Ω

||u||C2,α(B(x0, 12 R)∩Ω) ≤ c(

||u||C0(B(x0,R)∩Ω) + ||f ||Cα(B(x0,R)∩Ω)

)

.

Now we patch up the estimates over a countable cover of ∂Ω by small balls B(xi,12Ri). ∂Ω being

compact we may choose a finite subcover say after relabeling B(xi,12Ri)

Ni=1. Finally we adjoin

to these estimates an interior estimate for some Ω′ such that Ω \ ∪Ni=1B(xi,

12Ri)⊂⊂ Ω′ ⊂⊂ Ω. And

having this we are done by analysing the different cases that might arise in a similar fashion to

previous proofs.

2

Banach Spaces

Let V be a vector space equipped with a norm || · || : V → R i.e i) ||x|| ≥ 0 with equality

⇔ x = 0; ii) ||αx|| = |α|||x||; iii) ∆− inequality. With a norm we have a metric d(x, y) := ||x−y||

and we can talk about topology induced from it, convergence etc.

Cauchy sequence: xi such that d(xn, xm) −→N→∞ 0, ∀m,n ≥ N.

Banach space: a normed space complete wrt the norm metric ⇔ every Cauchy sequence con-

verges (wrt the norm metric) in V (limit in V ).

We mention in passing a few examples.

• The Bolzano-Weierstrass theorem showing (Rn, | · |) is complete carries over to show

finite dimensional normed spaces are Banach.

• (C0(Ω), || · ||L1) is incomplete, so is not Banach;

• On the other handwhile (C0(Ω), || · ||C0(Ω)) and in general (Ck,α(Ω), || · ||Ck,α) are Banach,

as can be demonstrated using the Arzela-Ascoli theorem [cf. Peterson, Riemannian Geometry,

Chapter 10].

• Sobolev spaces are yet another example.

Contraction Mapping Theorem. Let B a Banach space and T : B → B a contraction mapping

(wrt to the norm metric). Then T has a unique fixed point.

Proof. Here the assumption translates into ||Tx − Ty|| ≤ θ · ||x − y|| for θ ∈ [0, 1). The idea is to

look at the sequence xn := T nx0 and show it is Cauchy using the ∆-inquality . Let x ∈ V be its

limit; we see that

Tx = T lim xn = lim Txn (by continuity of T!) = limxn+1 = x.

As for uniqueness, if x, y are two fixed points,

3

||x − y|| = ||Tx − Ty|| ≤ θ||x − y|| ⇒ ||x − y|| = 0

and by the norm properties x = y.

4

Lecture 15

April 8th

, 2004

The Continuity Method

Let T : B1 → B2 be linear between two Banach spaces. T is bounded if

||T || = supx∈B1

||Tx||B2

||x||B1

< ∞ ⇔ ||Tx||B2≤ c · ||x||B1

for some c > 0.

Continuity Method Theorem. Let B be a Banach space , V a normed space, L0, L1 : B → V

bounded linear operators. Assume ∃c such that Lt := (1 − t)L0 + tL1 satisfies

||x||B ≤ c · ||Ltx||V , ∀t ∈ [0, 1]. (∗)

Then – L0 is onto ⇔ L1 is.

Proof. Assume Ls is onto for some s ∈ [0, 1]; by (∗) Ls is also 1-to-1 ⇒ L−1s exists. For t ∈ [0, 1], y ∈

V solving Ltx = y is equivalent to solving Ls(x) = y + (Ls − Lt)x = y + (t − s)L0x + (t − s)L1x.

By linearity now x = L−1s y + (t − s)Ls

−1 (L0 − L1)x.

Define a linear map T : B → B, Tx = L−1s y + (t − s)Ls

−1 (L0 − L1)x. One has ||Tx1 −

Tx2||B = ||(t − s)Ls−1 (L0 − L1)(x1 − x2)||. (∗) now gives us a bound on Ls

−1: since Ls is onto

∀x ∈ B, ∃y ∈ B such that Lsy = x and so

||Ls−1x||B ≤ c · ||Ls Ls

−1x||V

||Ls−1x||B ≤ c · ||x||V ⇒ ||Ls

−1|| ≤ c.

As an application we see that

||Tx1 − Tx2||B ≤ (t − s)c · (||L0|| + ||L1||)||x1 − x2||,

1

and for t close enough to s (precisely for t ∈ [s− 1c(||L0||+||L1||)

, s+ 1c(||L0||+||L1||)

]) we therefore have

a contraction mapping! Therefore T has a fixed point by the previous theorem which essentially

means that we can solve Ltx = y for any fixed y or that Lt is onto. Repeating this c(||L0||+ ||L1||)

many times we cover all t ∈ [0, 1].

Remark. Note as in the beginning of the proof that once such operators are onto they are in fact

invertible as long as (∗) holds.

Elliptic uniqueness

Let us summarize the properties we have establised for uniformly elliptic equations. Let Ω be a

bounded domain in Rn. Let L = aij(x)Dij + bi(x)Di + c(x) be uniformly elliptic, i.e

1

Λ· δij ≤ aij(x) ≤ Λ · δij

and assume c(x) ≤ 0.

Let u ∈ C2(Ω) ∩ C0(Ω) be a solution of Lu = f ∈ Cα(Ω) with 0 < α < 1. Then we have the

following a priori estimates –

A. supΩ

|u| ≤ c(γ,Λ,Ω, n) · (sup∂Ω

|u| + supΩ

|f |).

B. Under the additional assumptions

• in the case L has α − Holder continuous coefficients with Holder constant Λ,

• Ω has C2,α boundary

• u ∈ C2,α(Ω), f ∈ Cα(Ω),

we had the global Schauder estimate

||u||C2,α(Ω) ≤ c(γ,Λ,Ω, n)(

||u||C0(Ω) + ||f ||Cα(Ω)

)

.

C. Under the assumptions of B, when c(x) ≤ 0

2

||u||C2,α(Ω) ≤ c(sup∂Ω

|u| + supΩ

|f |).

D. The above applies to the Dirichlet problem

Lu = f on Ω, u = ϕ on ∂Ω

and in particular when ϕ = 0 we get very simply

||u||C2,α(Ω) ≤ c · ||Lu||Cα(Ω).

Theorem. Let Ω be a C2,α domain, L uniformly elliptic with Cα(Ω) coefficients and (x) ≤ 0.

Look at all u ∈ C2,α(Ω) and assume f ∈ Cα(Ω). Then the Dirichlet problem Lu = f on Ω, u =

ϕ on ∂Ω has a unique solution u ∈ C2,α(Ω) provided that the Dirichlet problem for ∆ is solvable

∀f ∈ Cα(Ω), ∀ϕ ∈ C2,α(Ω)!

Proof. Connect L and ∆ via a segment: [0, 1] → Lt := (1 − t)L + t∆. Since those operators are

all linear it is enough to prove for ϕ = 0 as we have seen previously. C2,α(Ω) is a Banach space

(Lecture 14), and so is its subspace B(Ω) := u ∈ C2,α(Ω), u = 0 on ∂Ω. As a matter of fact Lt is

a bounded operator B(Ω) → Cα(Ω) by the assumptions on the coefficients of L. And, by uniformly

elliptic we see from D above

||u||C2,α(Ω) = ||u||C2,α(B(Ω)) ≤ c · ||Ltu||Cα(Ω),

with c independent of t (depends just on L). Note Cα(Ω) is a Banach space and in particular a

vector space. The Continuity Method thus applies.

Strangely enough, we are now back to solving Dirichlet’s problem for ∆ in domains.

Our methods so far were good for providing solution in balls, spherically symmetric domains.

In other words we were able to solve (in C2,α(B(0, R))!) ∆u = f ∈ Cα(Ω) on B(0, R), u =

ϕ on ∂B(0, R) using the Poisson Integral Formula and estimates for the Newtonian Potential. We

used conformal mappings (inversion) to get indeed C2,α upto the boundary. We conclude therefore

that

3

Corollary. We can solve the Dirichlet Problem for any L satisfying the assumptions of the

Theorem in balls.

Perron’s Method gives a solution in quite general domains but we will not go into its details as

later on our regularity theory (weak solutions, Sobolev spaces etc.) will give us those answers.

Elliptic C2,α regularity

Let B :=ball, T :=some connected boundary portion.

Theorem. Let L be uniformly elliptic with Cα coefficients and assume c(x) ≤ 0. Let u ∈

C2(Ω) ∩ C0(Ω) be a solution of the Dirichlet problem Lu = f ∈ Cα(B) in B, u = ϕ ∈ C0(∂B) ∩

C2,α(T ) on ∂B has a unique solution u ∈ C2,α(B ∪ T ) ∩ C0(B).

We know by the previous theorem that if ϕ ∈ C2,α(∂B) (and not just on T ) then unique solvability

would be equivalent to the unique solvability of ∆ on B which we have! Therefore this Theorem is

a slight generalization.

Proof. As was just outlined the crucial problem lies in the (possible) absence of regularity of ϕ

on part of the boundary. So we approximate ϕ by a sequence ϕk ⊂ C3(B) such that both

||ϕk − ϕ||C0(B) −→ 0 and ||ϕk − ϕ||C2,α(B) −→ 0. Solve Luk = f, in B, uk = ϕkon ∂B.

Now L(ui − uj) = 0, in B, ui − uj = ϕi − ϕj on ∂B. And by A above (as c(x) ≤ 0)

||ui − uj ||C0(B) ≤ C sup∂B |ϕi − ϕj |. So we conclude our solutions uk form a Cauchy sequence

wrt the C0 norm, i.e in the Banach space C0(B). Therefore we know ∃u ∈ C0(B) with ui −→C0(B)u

(not just subconvergence!) and furthermore this u satisfies u = ϕ on pB.

Now we shift our look to the C2,α situation; by our interior estimates we have for any B′ ⊂⊂ B

||ui −uj ||C2,α(B′) ≤ c(||ui −uj ||C0(B) + ||0||Cα(B)).. That is our sequence is also a Cauchy sequence

in the Banach space C2,α(B′) ⇒ converges in C2,α(B′) (in particular limit is C2,α regular). This

limit must equal the limit u|′B we obtained through the C0 norm. We do this for any B′ ⊂⊂ B ⇒

get convergence in C2,α(B) ⇒ u satisfies Lu = f on B and has the desired C2,α regularity on B.

4

We now turn to the boundary portion: ∀x0 ∈ T and ρ > 0 such that B(x0, ρ) ∩ ∂B ⊆ T

we have the usual boundary Schauder estimates (for smooth enough functions) which give us

||ui −uj ||C2,α(B(x0,ρ)∩B) ≤ c ·(

||ui − uj ||C0(B) + ||ϕi −ϕj ||C2,α(B(x0,ρ)∩B)

)

. This means that in fact

ui −−−−−−−−−−−→C2,α(B(x0,ρ)∩B) u and in particular u ∈ C2,α at x0. ∀x0 ∈ T .

5

Lecture 16

April 13th

, 2004

Elliptic regularity

Hitherto we have always assumed our solutions already lie in the appropriate Ck,α space and

then showed estimates on their norms in those spaces. Now we will avoid this a priori assumption

and show that they do hold a posteriori. This is important for the consistency of our discussion.

Precisely what we would like to show is —

A priori regularity. Let u ∈ C2(Ω) be a solution of Lu = f and assume 0 < α < 1. We

do not assume c(x) ≤ 0 but we do assume all the other assumptions on L in the previous Theorem

hold. If f ∈ Cα(Ω) then u ∈ C2,α(Ω)

• Here we mean the Cα norm is locally bounded, i.e for every point exists a neighborhood where

the Cα-norm is bounded. Had we written Cα(Ω) we would mean a global bound on supx,y

|f(x) − f(y)|

|x − y|α

(as in the footnote if Lecture 14).

• This result will allow us to assume in previous theorems only C2 regularity on (candidate)

solutions instead of assuming C2,α regularity.

Proof. Let u be a solution as above. Since the Theorem is local in nature we take any point in Ω

and look at a ball B centered there contained in Ω. We then consider the Dirichlet problem

L0v = f ′ on B,

v = u on ∂B.

where L0 := L − c(x) and f ′(x) := f(x) − c(x) · u(x). This Dirichlet problem is on a ball, with

”c ≤ 0”, uniform elliptic and with coefficients in Cα. Therefore we have uniqueness and existence

of a solution v in C2,α(B) ∩ C0(B). But u satisfies Lu = f or equivalently L0u = f ′ on all of Ω so

1

in particular on B. By uniqueness on B therefore we have u|B = v, and so u is C2,α smooth there.

As this is for any point and all balls we have u ∈ C2,α(Ω).

It is insightful to note at this point that these results are optimal under the above assumptions.

Indeed need C2 smoothness (or atleast C1,1) in order to define 2nd derivatives wrt L! If one takes

u in a larger function space, i.e weaker regularity of u, and defines Lu = f in a weak sense then

need more regularity on coefficients of L! Under the assumption of Cα continuity on the coefficients

indeed we are in an optimal situation.

Higher a priori regularity. Let u ∈ C2(Ω) be a solution of Lu = f and 0 < α < 1. We

do not assume c(x) ≤ 0 but we assume uniformly elliptic and that all coefficients are in Ck,α. If

f ∈ Ck,α then u ∈ Ck+2,α. If f ∈ C∞ then u ∈ C∞.

Proof. k = 0 was the previous Theorem.

The case k = 1. The proof relies in an elegant way on our previous results with the combination

of the new idea of using difference quotients. We would like to differentiate the u three times and

prove we get a Cα function. Differentiating the equation Lu = f once would serve our purpose but

it can not be done naively as it would involve 3 derivatives of u and we only know that u has two.

To circumvent this hurdle we will take two derivatives of the difference quotients of u, which we

define by (let e1, . . . , en denote the unit vectors in Rn)

∆hu :=u(x + h · el) − u(x)

h=: .

uh(x) − u(x)

h.

Namely we look at

∆hLu =Lu(x + h · el) − Lu(x)

h=

f(x + h · el) − f(x)

h= ∆huf.

Note ∆hv(x) −→h→0Dlv(x) if v ∈ C1 (which we don’t know a priori in our case yet).

Expanding our equation in full gives

2

1

h

[

(aij(x + h · el) − aij(x) + aij(x))Dijuh − aij(x)Diju(x)

+bi(x + h · el)Diu(x + h · el) − bi(x)Diu(x) + c(x + h · el)u(x + h · el) − c(x)u(x)]

= ∆haijDijuh − aijDij∆

hu + ∆hbiDiuh + biDi∆

hu + ∆hc · uh + c · ∆hu = ∆hf.

or succintly

L∆hu = f ′ := ∆hf − ∆haij · Dijuh − ∆hbi · Diu

h − ∆hc · uh

where uh := u(x + h · e1).

We now analyse the regularity of the terms. f ∈ C1,α so so is ∆hf , but not (bounded) uniformly

wrt h (i.e C1,α norm of ∆hf may go to ∞ as h decreases). On the otherhand ∆hf ∈ Cα(Ω)

uniformly wrt h (∀h > 0): ∆huf = f(x+h·el)−f(x)h

= Dlf(x) for some x in the interval, and rhs

has a uniform Cα bound as f ∈ C1,α on all Ω! (needed as x can be arbitrary).

For the same reason ∆haij ,∆hbi,∆hc ∈ Cα(Ω). By the k = 0 case we know u ∈ C2,α(Ω) and not

just in C2(Ω). ⇔ Dijuh ∈ Cα(Ω) uniformly.

Remark. We take a moment to describe what we mean by uniformity. We say a function

gh = g(h, ·) : Ω → R is uniformly bounded in Cαwrt h when ∀Ω′⊂⊂ Ω exists c(Ω) such that

|gh|Cα(Ω′) ≤ c(Ω). Note this definition goes along with our local definition of a function being in

Cα(Ω) (and not in Cα(Ω)!).

Putting the above facts together we now see that both sides of the equation L∆hu = f ′ are in

Cα(Ω). And they are also in Cα(Ω′) with rhs uniformly so with constant c(Ω′).

By the interior Schauder estimate, ∀Ω′′ ⊂⊂ Ω′ and for each h

||∆hu||C2,α(Ω′′) ≤ c(γ,Λ,Ω′′) ·(

||∆hu||C0,Ω′(+)||f

′||Cα,Ω′ (

)) ≤ c(γ,Λ,Ω′′,Ω′,Ω, ||u||C1(Ω),

which is independent of h! If we assume the Claim below taking the limit h → 0 we get Dlu ∈

C2,α(Ω′′),∀l = 1, . . . , n u ∈ C3,α(Ω′′). ∀Ω′′⊂⊂ Ω′ ⊂⊂ Ω ⇔ u ∈ C3,α(Ω).

3

Claim. ||∆hg||Cα(A) ≤ c independently of h ⇔ Dlg ∈ Cα(A).

First we we show g ∈ C0,1(A). This is tantamount to the existence of limh→0 ∆hg(x) (since if it

exists it equals Dluγ(x) - that’s how we define the first l-directional derivative at x). Now ∆hgh>0

is family of uniformly bounded (in C0(A)) and equicontinuous functions (from the uniform Holder

constant). So by the Arzela-Ascoli Theorem exists a sequence ∆hig∞i=1 converging to some

w ∈ Cα(A) in the Cβ(A) norm for any β < α. But as we remarked above w necessarily equals Dlg

by definition.

Second, we show g ∈ C1(A) (i.e such that derivative is continuous not just bounded) and actually

∈ C1,α(A):

c ≥ ||∆hg||Cα(A) ≥ limh→0

∆hg(x) − ∆hg(y)

|x − y|α=

Dlg(x) − Dlg(y)

|x − y|α= |Dlg|Cα(A)

where we used that c is independent of h.

The case k ≥ 2. Let k = 2. By the k = 1 case we can legitimately take 3 derivatives as

u ∈ C3,α(Ω). One has

L(Dlu) = f ′ := Dlf − Dlaij · Diju − Dlbi · Diu − Dlc · u

with Dlu, f ′ ∈ C1,α(Ω). So again by the k = 1 case we have now Dlu ∈ C3,α(Ω), hence u ∈ C4,α(Ω).

The instances k ≥ 3 are in the same spirit.

Boundary regularity

Let Ω be a C2,α domain, i.e whose boundary is locally the graph of a C2,α function. Let L be

uniformly elliptic with Cα coefficients and c ≤ 0.

Theorem. Let f ∈ Cα(Ω), ϕ ∈ C2,α(∂Ω), u ∈ C2(Ω)∩C0(Ω) satisfyingLu = f on Ω,

u = ϕ on ∂∂Ω.

with 0 < α < 1. Then u ∈ C2,α(Ω).

4

Proof. Our previous results give u ∈ C2,α(Ω) and we seek to extend it to those points in ∂Ω. Note

that even though u = ϕ on ∂Ω and ϕ is C2,α there this does not give the same property for u. It

just gives that u is C2,α in directions tangent to ∂Ω, but not in directions leading to the boundary.

The question is local: restrict attention to B(x0, R) ∩ Ω for each x0 ∈ ∂Ω. We choose a C2,α

homeomorphism Ψ1 : Rn → R

n sending B(x0, R) ∩ ∂Ω to a portion of a (flat) hyperplane and

∂B(x0, R) ∩ Ω to the boundary of half a disc. We then choose another C2,α homeomorphism

Ψ2 : Rn → R

n sending the whole half disc into a disc (= a ball). Therefore Ψ2 Ψ1 maps our

original boundary portion into a portion of the boundary of a ball.

Similarly to previous computations of this sort we define the induced operator L on the induced

domain Ψ2 Ψ1(B(x0, R) ∩Ω) and define the induced functions u, ϕ, f and we get a new Dirichlet

problem with all norms of our original objects equivalent to those of our induced ones. Note that

still c := c Ψ1−1 Ψ2

−1 ≤ 0, therefore by our theory exists a unique solution v ∈ C2,α(Ψ2

Ψ1(B(x0, R)∩Ω) ∪ Ψ2 Ψ1(B(x0, R)∩∂Ω))∩C0(Ψ2 Ψ1(B(x0, R)∩ Ω)) for the induced Dirichlet

problem . Now our u also solves it. So by uniqueness u = v and u has C2,α regularity as the

induced boundary portion, and by pulling back through C2,α diffeomorphisms we get that so does

u.

Remark. The assumption c ≤ 0 is not necessary although modifying the proof is non-trivial

without this assumption (exercise). We needed it in order to be able to use our existence result.

But since we already assume a solution exists we may use some of our previous results which do

not need c ≤ 0 and which secure C2,α regularity upto the boundary.

5

Lecture 17

April 15th

, 2004

Higher boundary regularity

We extend our results to include the boundary.

Higher a priori regularity upto the boundary. Let u ∈ C2(Ω) ∩ C0(Ω) be a solution of

Lu = f on Ω,

u = ϕ on ∂Ω.

Assume uniformly elliptic and that all coefficients are in Ck,α(Ω) with 0 < α < 1 and that Ω is a

Ck+2,α domain. If f ∈ Ck,α(Ω) and ϕ ∈ Ck+2,α(∂Ω) then u ∈ Ck+2,α.

Proof. For k = 0 our previous results apply unchanged (the case c 6≤ 0 can be handled if one

believes the Remark above).

k = 1, the crucial case, we use once again difference quotients. As usual, localize to B+ :=

B(x0, R) ∩ Ω, x0 ∈ ∂Ω. Then flatten the boundary with the help of a C3,α diffeomorphism Ψ.

Assume the flat portion lies on the xn = 0 hyperplane. We get

L∆hu = ∆hf − ∆haij · Dij uh − ∆hbi · Diu

h − ∆hc · uh.

We know the rhs is uniformly Cα(Ψ(B+)) bounded, while the lhs is only so for the directions

l = 1, . . . , n − 1, the tangent directions on Ψ(∂B+), since the equation u = ϕ holds there and may

be differentiated in those directions (and ϕ has 3 derivatives).

We therefore use Schauder estimates for ∆hu which give it is uniformly bounded in C2,α(Ψ(B+′)),

∀B+′⊂⊂ B+ similarly to the higher regularity Theorem for the interior. This is so since the estimates

used there hold, in fact, upto the boundary. We get therefore Dlu ∈ C2,α(Ψ(B+)), l = 1, . . . , n− 1.

1

We treat the remaining derivative. We have DiDluC1,α(B+), i = 1, . . . , n, l = 1, . . . , n − 1 ⇔

Dnlu = Dl(Dnu) (mixed derivatives commute as u ∈ C2!). So all we have to show now is Dnnu ∈

C1,α(B+).

From Lu = f we find

Dnnu =1

ann

(

f − (L − ann)u)

.

From previous calculations of the form of a we see that choosing Ψ appropriately we may diagonalize

it. Then uniformly elliptic gives1

ann> γ > 0. The rhs is C1,α(B+) ⇔ so is lhs ⇔ Du ∈

C2,α(Ψ(B+)) ⇔ u is C3,α near x0.

The cases k ≥ 2 are handled as in the interior Theorem.

This wraps up our discussion on Holder spaces/norms.

Hilbert spaces

Let V be a vector space over the field R. Let (·, ·) be a map V × V → R such that

i) (x, y) = (y, x)

ii) (α1x1 + α2x, y) = α1(x1, y) + α2(x2, y), ∀αi ∈ R

iii) (x, x) > 0, ∀x 6= 0

Let ||x|| := (x, x)12 . One can then demonstrate

||(x, y)|| ≤ ||x|| · ||y|| Schwarz inequality

||x + y|| ≤ ||x|| + ||y|| triangle inequality

The 2nd affirms that || · || defines a norm.

If || · || is complete(

V, (·, ·))

is a Hilbert space.

2

Let F : V → R be linear, i.e a linear functional on V . If sup06=x∈V

|F (x)|

||x||=: ||F ||V ⋆ < ∞, F is

bounded. Here V ⋆ = bounded linear functional on V . Similary for a Hilbert space H define

similarly H⋆.

We give the statement of the main theorem regarding Hilbert spaces . Like the Continuity

Method it will serve us as a strong tool for us to attack abstract questions, a tool from Functional

Analysis.

Riesz Representation Theorem. Let H be a Hilbert space , F ∈ H⋆. Then ∃ ! f ∈ H such

that

i) F (x) = (f, x), ∀x ∈ H

ii) ||F ||H⋆ = ||f ||H

In particular ⇔ H = H⋆.

Sobolev Spaces

Motivation

If ∆u = f, u ∈ C2(Ω) then ∀ϕ ∈ C10(Ω) ϕ∆u = ϕf and

−

∫

Ω

∇ϕ∇u =

∫

Ω

∆u · ϕ =

∫

Ω

f · ϕ.

This observation lies at the heart of weak formulations of the Laplace equation.

Define an inner product on C10(Ω) :=compactly supported functions in C1(Ω)

(ϕ1, ϕ2) :=

∫

Ω

∇ϕ1∇ϕ2.

(

C10(Ω), (·, ·)

)

is not complete: a sequence of functions may degenerate to a function which is not

everywhere differentiable though continuous. Denote by W1,20 (Ω) the completion of C1

0(Ω) wrt

3

this norm. It is nice to note that (·, ·) extends to an inner product on W1,20 (Ω): represent any two

elements there as limits of sequences of elements in C10(Ω) and take the limit of the inner products

of those, which are well defined. Hence W1,20 (Ω) is a Hilbert space.

At this stage we do not yet know how W1,20 (Ω) looks like. Maye its elements are not even

functions.

We continue with the motivation for defining those spaces. Let F (ϕ) := −∫

Ωf ·ϕ, ∀ϕ ∈ C1

0(Ω).

F = F (f,Ω) extends to a linear functional on W1,20 (Ω).

Claim. F is bounded.

||F || = supϕ∈W

1,20

(Ω)=C10(Ω)

ϕ 6=0

|F (ϕ)|

||ϕ||W

1,20

= supϕ∈C1

0(Ω)

ϕ 6=0

|F (ϕ)|

||ϕ||W

1,20

since C10(Ω) is dense in its completion W

1,20 (Ω).

|F (ϕ)|

||ϕ||W

1,20

=|∫

Ωϕ · f |

(

∫

Ω|∇ϕ|2

)12

≤

(

∫

Ωϕ2

)12

·(

∫

Ωf2

)12

(

∫

Ω|∇ϕ|2

)12

.

Using the Poincare inequality

∫

Ω

ϕ2 ≤ c(Ω) ·

∫

Ω

|∇ϕ|2 we find a bound depending on Ω, f but not

on ϕ.

Hence by the Riesz Representation Theorem exists u ∈ W1,20 (Ω), though we do not know it is a

function or even if so whether it has any regularity, such that

F (ϕ) = (u, ϕ)

by def.// ∖

by def.∖

−

∫

Ω

f · ϕ

∫

Ω

∇u∇ϕ.

4

We do not know if u ∈ C10(Ω), just that u ∈ W

1,20 (Ω). We have a weak formulation of

∆u = f on Ω,

u = 0 on ∂Ω.

for any f ∈ L2(Ω)! Our plan is now: if f has certain regularity , u has that regularity +2.

The philosophy is instead of classically solving the ∆-equation with an exact explicit solution like

Poisson’s Integral Formula etc. we just enlarge our function spaces. Then the existence of a solution

in the enlarged space becomes trivial (following Riesz). The work comes down to showing that the

solution actually lies back in our original space of functions! That is regularity theory in a nutshell.

We will focus on that in the sequel.

Weak derivatives

For u, vi ∈ L1loc

(Ω) say ”vi = Diu” if

∫

Ω

vϕ = −

∫

Ω

u · Diϕ, ∀ϕ ∈ C10(Ω).

If such v exists ∀i = 1, . . . , n then u is weakly differentiable in Ω with ∇u =weak (v1, . . . , vn).

If each Dju satisfies the above conditions we say u is twice weakly differentiable. We will omit

the quotations marks in what follows.

The derivative does not exist pointwise in general. But by the Lesbegues Theorem it does exist

pointwise almost everywhere (a.e).

Definition

We are now in a position to define Sobolev spaces. Let ||u||Lp(Ω) :=(

∫

Ω|u|p

)1p

. Define

Lp(Ω) := equivalence classes of measurable functions such that || · ||Lp(Ω) < ∞

where f ∼ g if f = g a.e.

5

Define

W k(Ω) := k-times weakly differentiable functions ∩ L1loc(Ω) ⊆ L1

loc(Ω),

Similarly define the Sobolev spaces

W k,p(Ω) ≡ Lk,p(Ω) = u ∈ W k(Ω),Dαu ∈ Lp(Ω)all multi-

indices α, |α| ≤ k⊆ L1

loc(Ω),

equipped with the norm

|| · ||W k(p)Ω :=

∑

|α|≤k

∫

Ω

|Dα · |p

1p

(still need to prove it is a norm!). An equivalent norm is given by

∑

|α|≤k

∫

Ω

||Dα · ||L0(Ω).

Lp(Ω) is a Banach space ! (Riesz-Fischer Theorem). Also W k,p(Ω) = Lk,p(Ω) are.

Claim. C∞(Ω) ∩ W k,p(Ω) is dense in W k,p(Ω). i.e. we could have defined W k,p(Ω) as the

completion of C∞(Ω) wrt || · ||W k(p)Ω.

Given u ∈ W k,p(Ω) mollify it to

uh(x) :=

∫

Rn

1

hnρ( |x − y|

h

)

u(y)dy,

with ρ a smooth bump function on R with mass 1 and support in [− 12 , 1

2 ]. Now u ∈ C∞(Ω)∩W k,p(Ω)

and uh → u in the W k,p(Ω) norm.

We now define Sobolev spaces of compactly supported objects

Wk,p0 (Ω) := completion of Ck

0 (Ω) wrt || · ||W k,p(Ω).

6

Note functions in Ck0 (Ω) vanish on ∂Ω so in a sense W

1,p0 (Ω) (respectively W

k,p0 (Ω)) can be thought

of as (weak) functions which vanish on ∂Ω (whose first k − 1 derivatives vanish on ∂Ω).

Equivalence of norms

For ϕ ∈ W1,20 (Ω) we defined two norms. One using the inner product

∫

Ω

∇ϕ1 · ∇ϕ2 on C10(Ω)

which gave us the norm

||ϕ|| =

∫

Ω

|∇ϕ|2

12

and another norm

||ϕ||′ =

∑

|α|≤1

∫

Ω

|Dαϕ|2

12

=

∫

Ω

|ϕ|2 +n

∑

i=1

|Diϕ|2

12

≤ ||ϕ||L2(Ω) + ||∇ϕ||L2(Ω).

These norms are indeed equivalent since we are assuming compact support! The Poincare in-

equality shows || · ||′ ≤ (1 + c(Ω)) · || · ||. This inequality fails grossly for non-compactly supported

functions, e.g the constant function. Since || · || ≤ || · ||′ the norms are equivalent.

Remark. in both of the above norms we define first the norms of functions which are also in

C10(Ω) and then we extend the norm to the completion by means of norms of limits of sequences

whose elements are all in C10(Ω) (those are dense).

7

Lecture 18

April 22nd, 2004

Embedding Theorems for Sobolev spaces

Sobolev Embedding Theorem. Let Ω a bounded domain in Rn, and 1 ≤ p < ∞.

W1,p0 (Ω) ⊆

Lnp

n−p (Ω), p < n

C0,α(Ω), α = 1 − np, p > n,

i.e in particular ⊆ C0(Ω).

Furthermore, those embeddings are continuous in the following sense: there exists C(n, p,Ω) such

that for u ∈ W1,p0 (Ω)

||u||L

npn−p (Ω)

≤ C · ||∇u||Lp(Ω), ∀p < n

supΩ

|u| ≤ C ′ · Vol(Ω)1n− 1

p · ||Du||Lp(Ω), ∀p > n.

We start with a function whose derivative and itself belong to Lp. The above theorem gives us

more regularity for the function – it belongs to Lp· nn−p – based on its regular derivative.

Proof. C10(Ω) is dense in W

1,p0 (Ω). We prove first for u ∈ C1

0(Ω) and will later justify why the proof

actually extends to the larger space.

Case p = 1. fix an index i ∈ 1, . . . , n and observe

u(x) =

∫ xi

−∞

Diu(x1, . . . , t, . . . , xn)dt.

From which

1

|u(x)| ≤

∫ xi

−∞

|Diu|(x1, . . . , t, . . . , xn)dt

≤

∫ ∞

−∞

|Diu|(x1, . . . , t, . . . , xn)dt. (1)

Write this down for each i, take a product of the terms and take the n − 1th root of the result to

yield altogether

|u(x)|n

n−1 ≤

n∏

i=1

(

∫ ∞

−∞

|Diu|dxi

)1

n−1

.

Quick Reminder. Holder’s inequality (HI) tells us

1

p+

1

q= 1 ⇒

∫

u · v ≤(

∫

up)

1p

·(

∫

vp)

1q

,

or more generally

1

p1+ . . . +

1

pk

= 1 ⇒

∫

u1 · · · uk ≤(

∫

up1

1

)1

p1· · ·(

∫

upk

k

)1

pk.

Coming back to our inequality, we integrate over the x1 axis and subsequently apply the Holder

inequality with k = n − 1, pi = n − 1 –

∫ ∞

−∞

|u(x)|n

n−1 dx1 ≤

∫ ∞

−∞

n∏

i=1

(

∫ ∞

−∞

|Diu|dxi

)1

n−1

dx1.

=(

∫ ∞

−∞

|D1u|dx1

)1

n−1

·

∫ ∞

−∞

n∏

i=2

(

∫ ∞

−∞

|Diu|dxi

)1

n−1

dx1.

≤Holder’s Ineq.

(

∫ ∞

−∞

|D1u|dx1

)1

n−1

·

n∏

i=2

(

∫ ∞

−∞

[

∫ ∞

−∞

|Diu|dxi

]

n−1n−1

dx1

)1

n−1

.

=(

∫ ∞

−∞

|D1u|dx1

)1

n−1

·(

∫ ∞

−∞

∫ ∞

−∞

|D2u|dx2dx1

)1

n−1

·

·

n∏

i=3

[

∫ ∞

−∞

(

∫ ∞

−∞

|Diu|dxi

)

dx1

]1

n−1

.

2

Now courageously continuing with this confusing calculation, we integrate over the x2 axis. This

is the reason we singled out the second terms from the n−2 others ones; if we would have integrated

now over the xj axis we would have choosen a term involving integration over that axis. And indeed

now the middle term is a constant wrt this operation, that is only the other two terms appear in

this integral, hence –

∫ ∞

−∞

∫ ∞

−∞

|u(x)|n

n−1 dx1dx2

≤(

∫ ∞

−∞

∫ ∞

−∞

|D2u|dx2dx1

)1

n−1

·

(

∫ ∞

−∞

(

∫ ∞

−∞

|D1u|dx1

)1

n−1

·

n∏

i=3

[

∫ ∞

−∞

|Diu|dxi

]

dx2

)1

n−1

.

and using the Holder Inequality the second term transforms, and we have

=(

∫ ∞

−∞

∫ ∞

−∞

|D2u|dx2dx1

)1

n−1

·(

∫ ∞

−∞

[

∫ ∞

−∞

|D1u|dx1

]

n−1n−1

dx2

)1

n−1

·

·(

n∏

i=3

∫ ∞

−∞

∫ ∞

−∞

∫ ∞

−∞

|Diu|dxidx1dx2

)1

n−1

.

In the same vein, we now isolate among the n terms the only term involving integration over the

x3 axis, integrate over that axis and then once again apply the Holder Inequality for the remaining

n − 1 terms (at each stage we always have n − 1 terms except from the isolated one; the Holder

Inequality allows us to lift the 1n−1

exponent and let another new dxi come in to the integral of

those n − 1 terms).

Finally, therefore, we will arrive at

∫ ∞

−∞

· · ·

∫ ∞

−∞

|u(x)|n

n−1 dx1 · · · dxn ≤n∏

j=1

(

∫ ∞

−∞

· · ·

∫ ∞

−∞

|Dju|dx1 · · · dxn

)1

n−1

.

In other words if we restrict to Ω

(

||u||L

nn−1 (Ω)

)n

n−1

≤

n∏

j=1

(

∫

Ω

|Dju|dx)

1n−1

.

3

or still

||u||L

nn−1 (Ω)

≤

n∏

j=1

(

∫

Ω

|Dju|dx)

1n

≤1

n·

n∑

j=1

.

∫

Ω

|Dju|dx ≤1

n·

n∑

j=1

.

∫

Ω

|Du|dx =

∫

Ω

|Du|dx

= ||∇u||L1(Ω).

This concludes the p = 1 < n case. Let us remark that of course we neglected at the last steps

to seek the best possible Sobolev constant and contented ourselves with the constant 1:

||u||L

n·1n−1 (Ω)

≤ 1 · ||∇u||L1(Ω).

In fact the best possible Sobolev constant c is achieved for Ω = B(0, r), u = IB(0,r) (IA is the

characteristic function on the set A, evaluating to 1 on A and 0 otherwise); believing that, we

compute

Vol(B(0, r))n

n−1 = c ·

∫

B(0,r)

|DIB(0,r)|dx = c ·

∫

B(0,r)

|δ∂B(0,r)|dx = c · Area(S(r)),

i.e

(ωnrn)n

n−1 = c · nωnrn−1 ⇒ c =1

n n√

ωn

.

Case 1 < p < ∞. A little trick will make our previous work apply to this case as well. Let γ > 1

be a constant to be specified. We have by our previous case

|| |u|γ ||L

nn−1 (Ω)

≤

∫

Ω

∣

∣D|u|γ∣

∣dx ≤ γ

∫

Ω

|u|γ−1 · |Du|dx.

Let q be such that1

p+

1

q= 1. One has using the Holder Inequality

(

∫

Ω

|u|γ·n

n−1 dx)

nn−1

≤(

∫

Ω

|u|(γ−1)qdx)

1q

·(

∫

Ω

|Du|pdx)

1p

.

We have q = pp−1 . Choose γ = n−1

n−p· p in order to have (γ − 1)q = n

n−1 · γ. Hence

4

(

∫

Ω

|u|(n−1n−p

·p)· nn−1 dx

)n

n−1

≤(

∫

Ω

|u|(n−1n−p

·p)−1)·( p

p−1 )dx)

p−1q

·(

∫

Ω

|Du|pdx)

1p

,

or succintly

||u||L

npn−p (Ω)

=

∫

Ω

|u|np

n−p

n−1n

− p−1p

≤n − 1

n − p· p||∇u||Lp(Ω).

This deals with the case p < n indeed. We remark that characteristic functions no longer give

the best Sobolev constants in the case 1 < p < n.

Remark. The above proof holds and is valid for u ∈ C10(Ω)! We did not prove for distributional

coefficient. If u is only in W1,p0 (Ω), take a sequence um ⊆ C1

0(Ω) such that um → u in the

W1,p0 (Ω)-norm. This means that also

||ui − uj ||L

npn−p (Ω)

≤ c · ||Dui − duj ||Lp(Ω) → 0.

um is thus a Cauchy sequence in Lnp

n−p (Ω). Lnp

n−p (Ω) is a Banach space d’apres Riesz-Fischer, i.e

u′ := limum ∈ Lnp

n−p (Ω); ⇔ u′ = u is in that space too. Now

||um||L

npn−p (Ω)

≤ c · ||Dum||Lp(Ω) → 0.

↓ ↓

||u||L

npn−p (Ω)

≤ c · ||Du||Lp(Ω) → 0.

So we our Theorem applies equally well to functions in the larger space. In this last line we needed

also mention that Dum → Du, but this is true since Dum is a Cauchy sequence from same

computation as above. Its limit lies in Lp again as this is a Banach space and so indeed Dum → Du

in Lp and hence also ||Dum||Lp(Ω) → ||Du||Lp(Ω).

Case p > n. We postpone the proof of this case to state a Corollary.

5

Corollary. By iterating, ∀k ≥ 2 holds

Wk,p0 (Ω) ⊆

Lnp

n−k·p (Ω), kp < n

Cm, 0 ≤ m ≤ k − np.

Proof. For instance, if k = 2, u ∈ W 2,p ⇒ u,Du ∈ W 1,p. By the k = 1 case above we have

u,Du ∈ Lnp

n−p . That means Du ∈ W 1,np

n−p ⇒ (by k = 1 case once again) u ∈ W 1,p′

where

p′ =n · ( np

n−p)

n − ( npn−p

)=

n2p

n2 − np − np=

np

n − 2p.

This proof repeated carries over ∀k ∈ N.

Now for the second inclusion, the promised postponed. We will need the following lemma en

passant.

Lemma. Let Ω be a bounded domain, B :=Ball ⊆ Ω, u ∈ W 1,1. Then for all x ∈ Ω

∣

∣u(x) −1

Vol(B)

∫

B

udx∣

∣ ≡∣

∣u(x)− 6

∫

B

udx∣

∣ ≤ c ·

∫

B

|Du(y)|

|x − y|n−1dy.

Proof. By our density theorem C10(Ω) is dense in W

1,p0 (Ω) and thus work with u in the former.

Take x, y ∈ Ω. Let ω := y−x|y−x|

,

u(x) − u(y) =

∫ |x−y|

0

Dru(x + rω)dr.

Integrating over some ball B

Vol(B) · u(x) −

∫

B

u(y) =

∫

B

(

∫ |x−y|

0

Dru(x + rω)dr)

dy.

Put

6

v(x) =

Dru(x), x ∈ Ω0, x 6∈ Ω.

Take now a particular ball B(x,R) ⊆ Ω to get

∣

∣u(x)− 6

∫

B

u(y)dy∣

∣ ≤1

Vol(B)

∫

|x−y|<2R

(

∫ ∞

0

|v(x + rω)|dr)

dy.

Switch order of integration, and change coordinates to spherical ones

=1

Vol(B)

∫ ∞

0

(

∫ 2R

0

(

∫

Sn−1(1)

|v(x + rω)|ρn−1dωSn−1(1)

)

dρ)

dr

after rescaling, where (ρ, ω) are the spherical coordinates, i.e ω are coordinates on the unit sphere.

Now

=(2R)n

nVol(B)

∫ ∞

0

(

∫

Sn−1(1)

|v(x + rω)|dωSn−1(1)

)

dr

=(2R)n

nVol(B)

∫ ∞

0

(

∫

Sn−1(1)

|v(x + rω)|

rn−1rn−1dωSn−1(1)

)

dr.

Set z := x + rω, → r = |rω| = |x − z|, rn−1dωSn−1(1)dr = dz,

=(2R)n

nVol(B)

∫

B

|v(z)|

|x − z|n−1dz,

and as B(x,R) ⊆ Ω ⇒

≤(2R)n

nVol(B)

∫

Ω

|Dru(z)|

|x − z|n−1dz.

Claim.

∫

BR

|x − y|1−n|Du(y)|dy ≤ CR1−np ||Du||Lp(BR), ∀p > n.

for BR := B(x0, R) ⊆ Rn.

7

Proof. By the Holder inequality, ∀q such that 1q

+ 1p

= 1

∫

BR

|x − y|1−n|Du(y)|dy ≤

∫

BR

|x − y|(1−n)qdy

1q

· ||Du||Lp(BR)

≤ supx∈Ω

∫

BR

|x − y|(1−n)qdy

1q

· ||Du||Lp(BR)

= c ·

∫

BR

|x0 − y|(1−n)qdy

1q

· ||Du||Lp(BR)

= c ·

∫ R

0

r(1−n)qrn−1dr

1q

· ||Du||Lp(BR)

= c ·

∫ R

0

rn−11−p dr

1q

· ||Du||Lp(BR)

= c ·(n − 1

1 − p+ 1)

R

(

n−11−p

+1

)/

q

· ||Du||Lp(BR)

= C(n, p)Rp−n

p · ||Du||Lp(BR)

as(n − 1

1 − p+ 1)

·1

q=(n − 1

1 − p+ 1)

·p

1 − p= −

n − 1

p+

p − 1

p=

p − n

p.

Now we can finally, combining those last two results conclude the second inclusion in our Theorem

as well as the estimate therein. First, using the triangle inequality together with the first lemma

we have

|u(x) − u(y)| ≤∣

∣u(x)− 6

∫

B

udx∣

∣+ 6

∫

B

udy − u(y)∣

∣ ≤ 2c ·

∫

B

|Du(y)|

|x − y|n−1dy

which in turn is

≤ c(n, p)|x − y|1−np ||Du||Lp(B)

once we choose a ball B = B(x, |x − y|) and apply the Claim. Since this is for any x, y ∈ Ω, and

u ∈ W 1,p(Ω) then u ∈ C1− np (Ω), if p > n.

Second and finally, we have as well

8

|u(x)| ≤∣

∣u(x)− 6

∫

B

udx∣

∣ ≤ 2c ·

∫

Ω

|Du(y)|

|x − y|n−1dy ≤ c(n, p) · diam(Ω)1−

np ||Du||Lp(Ω)

= c′(n, p) · Vol(Ω)1n− 1

p ||Du||Lp(Ω)

which gives the desired sup norm. Indeed for k ≥ 2 the Corollary follows by iterating: we get first

Holder regularity of u, then we have Du is W1,p0 (Ω) so we apply the first Theorem to it and get Du

is Holder and so on.

9

Lecture 19

April 27th, 2004

We give a slightly different proof of

Theorem. Let Ω a bounded domain in Rn, and 1 ≤ p < ∞.

W1,p0 (Ω) ⊆ C0,α(Ω), α = 1 −

n

p, p > n,

and ∃C(n, p,Ω) such that for u ∈ W1,p0 (Ω)

||u||C0,α(Ω) ≤ C · ||u||W 1,p(Ω), ∀p > n,

in other words

supΩ

|u| + |u|C0,α(Ω) ≤ C ·

||u||Lp(Ω) + ||∇u||Lp(Ω)

, ∀p > n.

Note the inequality is stronger than the one we stated in the previous lecture.

Proof. We take u ∈ C10(Ω) as before, wlog (density argument). Extend u to R

n trivially, i.e set

u = 0 on Rn \ Ω. Let x, y ∈ Ω and σ = |x − y| and let p be the point

x + y

2. Put B = B(p, σ) and

take z ∈ B. By the Fundamental Theorem of Calculus

u(x) − u(z) =

∫ 1

0

d

dtu(x + (1 − t)z)dt

=

∫ 1

0

∇u(x + t(z − x)) · (z − x) dt.

Integrating over z ∈ B

1

∣

∣

∫

B

u(z)dz − Vol(B)u(x)∣

∣ ≤

∫

B

∫ 1

0

|∇u(x + t(z − x))| · |z − x|dtdz

≤ 2σ

∫

B

∫ 1

0

|∇u(x + t(z − x))|dtdz

= 2σ

∫ 1

0

(

∫

B

|∇u(x + t(z − x))|dz)

dt.

Change variables to

z := x + t(z − x), → dz = tndz.

For z ∈ B(x, σ) ⇒ z ∈ B(x, tσ) =: B. In the new variable we have now

∣

∣

∫

B

udz − Vol(B)u(x)∣

∣ ≤ 2σ

∫ 1

0

t−n(

∫

B

|∇u(z)|dz)

dt.

By the Holder Inequality for q such that1

p+

1

q= 1

∫

B

|∇u(z)|dz)

dt ≤

∫

B

1q

1q ·

∫

B

|∇u(w)|pdw

1p

= Vol(B(tσ))1q ||∇u||Lp(B)

≤ Vol(B(tσ))1q ||∇u||Lp(Ω)

= ω1qn t

nq σ

nq ||∇u||Lp(Ω) ⇒

∣

∣

∫

B

udz − Vol(B)u(x)∣

∣ ≤ 2σ1+ nq ω

1qn

(

∫ 1

0

t−n · tnq dt

)

||∇u||Lp(Ω).

Divide now throughout by Vol(B) = ωnσn

∣

∣ 6

∫

B

u(z)dz − u(x)∣

∣ ≤ σ1+ nq−nω

1q−1

n

(

∫ 1

0

t−n(1− 1q)dt

)

||∇u||Lp(Ω)

= σ1−np ω

− 1p

n

(

∫ 1

0

t−n( 1p)dt

)

||∇u||Lp(Ω)

2

and the integral evaluates to[ t−

np+1

1 − np

]

∣

∣

∣

∣

∣

1

0

which is finite iff p > n. We thus conclude

∣

∣ 6

∫

B

u(z)dz − u(x)∣

∣ ≤ c(n, p) · σ1−np ||∇u||Lp(Ω).

We repeat the above computation with x replaced by y and use the triangle inequality, which gives

us

∣

∣u(x) − u(y)∣

∣ ≤ 2c(n, p) · |x − y|1−np ||∇u||Lp(Ω)

and subsequently

|u(x) − u(y)|

|x − y|1−np

≤ 2c(n, p) · ||∇u||Lp(Ω).

And concluding

||u||Cα(Ω) = ||u||L∞(Ω) + supx6=y∈Ω

|u(x) − u(y)|

|x − y|1−np

≤ C(n, p,Ω) · ||∇u||Lp(Ω).

since both C0 and L∞ norms coincide, being just supΩ, and finally because by our above computa-

tions we can also bound the L∞ norm in terms of the Lp norm of Du

|u(x)| ≤ 2c(n, p,diam(Ω)) · ||∇u||Lp(Ω)

so ||u||L∞(Ω) is bounded by the same rhs .

Compactness Theorems

Lemma. Let Ω be a bounded domain in Rn, and 1 ≤ p < ∞. Let S be a bounded set in Lp(Ω).

In other words,

∀ u ∈ S, ||u||Lp(Ω) ≤ MS .

3

Suppose ∀ ǫ > 0, ∃ δ > 0 such that

∀ u ∈ S, ∀ |z| < δ

∫

Ω

|u(y + z) − u(y)|pdy < ǫ.

Then S is precompact in Lp(Ω) (denoted S ⊂⊂ Lp(Ω)), i.e every sequence of functions in S has

convergent subsequence (”subconverges”), or equivalently S is compact.

This is an Arzela-Ascoli type theorem: bounded equicontinuous family is precompact. We just

have to show somehow that the integral equicontinuity-type condition implies equicontinuity.

Proof. Mollify u as done previously in the course

uh =

∫

Rn

ρh(x − y)u(y)dy, ρh(z) =1

hnρ(

|z|

h).

Set Sh := uh, u ∈ S.

We compute

uh =

∫

Rn

ρh(x − y)u(y)dy =

∫

Rn

ρh(x − y)|u(y)|dy

=

∫

Rn

ρ1q

h ρ1p

h |u(y)|dy

≤

∫

Rn

ρh

1q ·

ρh|u(y)|1p dy

≤ ||u||Lp(Ω).

Now

uh(x + z) − uh(x) =

∫

Rn

[

ρh(x + z − y) − ρh(x − y)]

u(y)dy

=

∫

Rn

[

ρh(x − y)[

u(y + z) − u(y)]

dy

and the same estimate as above yields

uh(x + z) − uh(x) ≤ 1 ·

∫

Ω

|u(y + z) − u(y)|p

1p ≤ ǫ

1p .

4

Now by our assumption for δ > 0 small enough and |z| < δ we will attain any desired ǫ on the

rhs. Note∫

Rn ρh = 1 is fixed for all h by our choice of ρ. Hence by definition we see that Sh is an

equicontinuous family, and bounded wrt the Lp(Ω) norm as inside S, hence by the Arzela-Ascoli

theorem Sh is precompact in the space Lp(Ω).

Now limh→0 Sh → S as we have seen in previous lectures. So as the above estimates are inde-

pendent of h, S is precompact itself in Lp(Ω).

Theorem (Kodrachov) Let Ω be bounded in Rn.

(I) p < n : W1,p0 (Ω) ⊆ Lq(Ω) ∀ 1 ≤ q <

np

n − p.

(II) p > n : W1,p0 (Ω) ⊆ C0,α(Ω) ∀ 0 < α < 1 −

n

p.

and moreover W1,p0 (Ω) is compactly embedded in each of the rhss.

We have then a curious situation– W1,p0 (Ω) ⊆ L

np

n−p ⊆ Lq for 1 ≤ q < npn−p

but the first

inclusion is only continuous! Only for q stricly smaller than npn−p

is it compact... And similarly for

the case p > n.

For the sake of clarity: we say B1 ⊆ B2 is compactly embedded if for every bounded set S in B1,

i(S) ⊆ B2 is precompact, where i : B1 → B2 is the inclusion map.

Proof. Case q = 1. By the density argument we mentioned repeatedly we assume wlog S ⊆ C10(Ω)

and that MS = 1. Let u ∈ S. Then ||u||Lp(Ω) ≤ 1, ||Du||Lp(Ω) ≤ 1. Hence ||u||L1(Ω) =∫

Ω|u(x)| ≤

∫

Ω1

1q

∫

Ω|u|p

1p ≤ Vol(Ω)

1q · 1, in other words S is also bounded in L1. Once we show the

condition of the Lemma holds then we will have precompactness in L1(Ω). And indeed

u(y + z) − u(y) =

∫ 1

0

du

dt(y + tz)dt =

∫ 1

0

∇u(y + tz) · zdt ⇒

∫

Ω

|u(y + z) − u(y)|dy ≤ |z|Vol(Ω)1q ||∇u||Lp(Ω) ≤ c|z|.

5

Case 1 < q < npn−p

. We try to find some estimates for the Lq(Ω) norm using the indispensible

Holder Inequality. Naturally we will be able to take care of boundedness of all such q together if

we allude to the fact that the Λnp

n−p (Ω) is bounded, indeed the Lp norms are increasing in p– first

choose λ such that qλ + q(1 − λ)n − p

np= 1

∫

|u|q =

∫

|u|qλ · |u|q(1−λ) ≤

∫

(

|u|qλ)

1qλ

qλ

·

∫

(

|u|q(1−λ))

np

n−p1

q(λ−1)

q(1−λ)( n−p

np)

⇒

||u||Lq(Ω) ≤ ||u||λL1(Ω) · ||u||1−λ

Lnp

n−p (Ω)

≤ ||u||λL1(Ω) · c · ||∇u||1−λLp(Ω)

≤ ||u||λL1(Ω) · c · 1

≤ c(n, p,Vol(Ω)),

where we applied our Theorem from the previous lecture. Now note that we are done using the

q = 1 case: S is bounded in Lq(Ω) and hence a subsequence converges in Lq(Ω), but then by the

above inequality it will also converge in Lq(Ω)!

Case p > n. By the Theorem of the previous lecture W1,p0 (Ω) ⊆ C0,α(Ω) continuously. But

now C0,α(Ω) ⊆ C0,β(Ω) compactly for any 0 ≤ β < α as mentioned in one of the previous lec-

tures.

Remark. Replacing W1,p0 (Ω) by W 1,p(Ω) (the completion of C1(Ω) wrt the W 1,p norm) in the

above embedding theorems require that the domain be Lipschitz, i.e ∂Ω is of class C0,1 (this is a

local requirement).

6

Lecture 20

April 29th, 2004

Difference Quotients and Sobolev spaces

Define

∆hi u :=

u(x+ h · el) − u(x)

h, h 6= 0.

Lemma. Let Ω be a bounded domain in Rn, and u ∈ W 1,p(Ω), for some 1 ≤ p < ∞. Then for

any Ω′ ⊂⊂ Ω such that dist(Ω′, ∂Ω) > h holds

||∆hi u||Lp(Ω′) ≤ ||Diu||Lp(Ω).

Proof.

|∆hi u| =

∣

∣

u(x+ h · el) − u(x)

h

∣

∣ ≤1

h

∫ h

0

∣

∣Diu(x1, . . . , xi + ζ, . . . , xn)∣

∣dζ

≤1

h

∫ h

0

1q1q

·

∫ h

0

∣

∣Diu(x1, . . . , xi + ζ, . . . , xn)∣

∣

pdζ

1p ⇒

|∆hi u|

p ≤ hp

q−p ·

∫ h

0

∣

∣Diu(x1, . . . , xi + ζ, . . . , xn)∣

∣

pdζ

=1

h·

∫ h

0

∣

∣Diu(x1, . . . , xi + ζ, . . . , xn)∣

∣

pdζ ⇒

∫

Ω′

|∆hi u|

p ≤1

h·

∫

Ω′

∫ h

0

∣

∣Di|pdζdx =

1

h·

∫ h

0

∫ ′

Ω

∣

∣Di|pdxdζ

=1

h

∫ h

0

||Diu||Lp(Ω′) = ||Diu||Lp(Ω′) ≤ ||Diu||Lp(Ω),

1

where we applied Fubini’s Theorem in order to switch order of integration.

Conversely we have

Lemma. Let u ∈ Lp(Ω) for some 1 ≤ p < ∞ and suppose ∆hi u ∈ Lp(Ω′) with ||∆h

i u||Lp(Ω′) ≤ K

for all Ω′ ⊂⊂ Ω and 0 < |h| < dist(Ω′,Ω). Then the weak derivative satisfies ||Diu||Lp(Ω) ≤ K.

Consequently if this holds for all i = 1, . . . , n then u ∈W 1,p(Ω).

Proof. We will make use of

Alouglou’s Theorem. A bounded sequence in a separable, reflexive Banach space has a weakly

convergent subsequence.

A topological space is called separable if it contains a countable dense set.

A Banach space is called reflexive if (B⋆)⋆ = B.

A sequence xn in a Banach space is said to converge weakly to x when limn→∞

F (xn) → F (x) for

all linear functionals F ∈ B⋆. This is sometimes denoted limn→∞

xn x.

Example: Let ℓ2 :=

(a1, a2, . . .) :

∞∑

i=1

a2i <∞

. Consider the sequence xi := (0, . . . , 0, 1, 0, . . .)

⊆ ℓ2. Any bounded linear functional on ℓ2 will be some linear combination of the linear functionals

Fj , defined by Fj(a1, . . .) = aj (each such linear combination corresponds exactly to a point in ℓ2.

That makes sense, indeed by the Riesz Representation Theorem (ℓ2)⋆ = ℓ2 (note ℓ2 is a Hilbert

space not just a Banach space as it has an inner product structure).). For any such F = (a1, . . .),

limi→∞

F (xi) = limi→∞

ai = 0. So xi converges to the 0 vector weakly, though certainly not strongly:

by Fourier Theory each point in ℓ2 corresponds to a periodic function on [0, 1], i.e an element of

L2(S1), and of course limn→∞

exp(n2π√−1z) 6→ 0(z).

2

We come back to the proof. For the Banach space B = Lp(Ω), B⋆ = Lq(Ω) with1

p+

1

q= 1.

This can be seen directly: If F ∈ (Lp(Ω))⋆, then exists f such that F (g) =∫

Ωg · f, ∀g ∈ Lp(Ω),

and this will be bounded iff f ∈ Lq(Ω). So we get an identification F ∈ (Lp(Ω))⋆ ∼= Lq(Ω).

By Alouglou’s Theorem there exists a sequence hm → 0 with ∆hm

i u v ∈ Lp(Ω). In other

words

∫

Ω

ψ · ∆hm

i u→

∫

Ω

ψ · v ∈ Lp(Ω), ∀ ψ ∈ Lq(Ω).

And in particular for any ψ ∈ C10(Ω) (which is dense in Lq(Ω) so will suffice to look at such ψ as

will become clear ahead)

∫

Ω

ψ∆hm

i u =

∫

Ω

ψ1

h(u(x+ h · el) − u(x))dx

=1

h

∫

Ω

ψ(x− hei)u(x)dx −1

h

∫

Ω

ψ(x)u(x)dx

=

∫

Ω

1

h(ψ(x− hei) − ψ(x))u(x)dx

=

∫

Ω

−∆hi ψ(x)u(x)dx

h→0−→

∫

Ω

−Diψ(x)u(x)dx

since ψ is continuously differetiable. Altogether

∫

Ω

ψ · v ∈ Lp(Ω) =

∫

Ω

−Diψ(x)u(x)dx,

which by definition means v is the weak derivative of u in the direction of the xi axis, or simply

the undistinctive notation v = Diu.

We also get the desired estimate, using the Fatou Lemma∫

lim inf ≤ lim inf∫

∫

Ω

|Diu|pdx =

∫

Ω

lim inf |∆hi u|

pdx ≤ lim inf

∫

Ω

|∆hi u|

pdx ≤ Kp,

i.e ||Diu||Lp(Ω) ≤ K.

3

L2Theory

Consider the second order equation in divergence form

Lu ≡ L(u) := Di(aijDju) + biDiu+ c · u = f,

with aij , bi, c ∈ L1(Ω) (integrable coefficients).

We call u ∈W 1,2(Ω) a weak solution of the equation if

∀ v ∈ C10(Ω) −

∫

Ω

aijDjuDiv +

∫

Ω

(biDiu+ cu)v =

∫

Ω

fv.

Elliptic Regularity

Let u ∈W 1,2(Ω) be a weak solution of Lu = f in Ω, and assume


• aij ∈ C0,1(Ω)

• bi, c ∈ L∞(Ω)

• f ∈ L2(Ω)

Then for any Ω′ ⊂⊂ Ω, u ∈W 2,2(Ω′) and

||u||W 2,2(Ω′) ≤ C(||aij ||C0,1(Ω), ||b||C0(Ω), ||c||C0(Ω), λ,Ω′,Ω, n) ·

(

||u||W 1,2(Ω) + ||f ||L2(Ω)

)

.

Note L∞(Ω) stands for bounded functions on Ω while C0(Ω) are functions that are also continuous

(Ω being bounded).

Proof. Start with the definition of u being a solution in the weak sense, ∀ v ∈ C10(Ω):

4

∫

Ω

aijDjuDiv =

∫

Ω

(biDiu+ c− f)v.

and take difference quotients, that is replace v with ∆−hv.

∫

Ω

aijDjuDi(∆−hv) =

∫

Ω

(biDiu+ c− f)(∆−hv).

Taking −h is a technicality that will unravel its reason later on, and we really mean ∆−hk v for

some k ∈ 1, . . . , n and then eventually repeat the computation for all k in that range. This

will become clear as well. Finally our goal will be to use the Chain Rule and move the difference

quotient operator onto u under the integral sign and get uniform bounds on ∆hDu and in this way

get a priori W 2,2(Ω) estimates.

The Chain Rule gives

∆h(aijDju) =

1

h

(

aiju(x+ h · ek)Dju(x+ h · ek) − aij(x) − aij(x+ h · ek) + aij(x+ h · ek)Dju(x))

= aiju(x+ h · ek)∆hDju− ∆haijDju.

And applied to our previous equation, a short calculation verifies that we can ’integrate by part’

wrt ∆h–

∫

Ω

aijDjuDi(∆−hv) =

∫

Ω

∆h(aijDju)Div ⇒

∫

Ω

aiju(x+ h · ek)∆hDjuDiv =

∫

Ω

−∆haijDjuDiv +

∫

Ω

(biDiu+ c− f)(∆−hv) ⇒

∣

∣

∫

Ω

aiju(x+ h · ek)∆hDjuDiv∣

∣ ≤ ||∆haijDju||L2(Ω)||Div||L2(Ω)+

+ ||biDiu+ cu− f ||L2(Ω)||∆−hv||L2(Ω),

5

where we have used the Holder Inequality for p = q = 2. This in turn can be bounded by

≤ ||aij ||C0,1(Ω)||Du||L2(Ω)||Dv||L2(Ω)+

+(

||bi||L∞(Ω)||Du||L2(Ω) + ||c||L∞(Ω)||u||L2(Ω) + ||f ||L2(Ω)

)

||Dv||L2(Ω)

≤ C(||u||W 1,2(Ω) + ||f ||L2(Ω)) · ||Dv||L2(Ω).

where we have used the Holder Inequality for p = 1, q = ∞, i.e a simple bounded integration

argument (e.g ||cu||L2(Ω) =(

∫

c2 · |u|2)

12 ≤

(

sup |c|2∫

O

|u|2)

12 ), and ∆haij → Dka

ij as aijC0,1(Ω).

Take a cut-off function η ∈ C10(Ω), 0 ≤ |η| ≤ 1, η

∣

∣

Ω′= 1. We now choose a special v: v := η2∆hu.

From uniform ellipticity (aijζiζj ≥ λ|ζ|2)

λ

∫

Ω

|ηD∆hu|2 ≤

∫

Ω

η2aij(x+ h · ek)Di∆huDj∆

hu.

Now

Div = 2ηDiη∆hu+ η2Di∆

hu

which we substitute into our previous inequality,

∫

Ω

η2aij(x+ h · ek)Dj∆huDj∆

hu ≤

∫

Ω

aij(x+ h · ek)Dj∆hu · (Div − 2ηDiη∆

hu)

≤ C(||u||W 1,2(Ω) + ||f ||L2(Ω))||Dv||L2(Ω)+

+ C ′||(D∆hu)η||L2(Ω)||Dη∆hu||L2(Ω)

again by the Holder Inequality. Now since η ≤ 1

||Div||L2(Ω) ≤ C ′′(

||Diη∆hu||L2(Ω) + ||D∆hu||L2(Ω)

)

.

Combining all the above and again using η ≤ 1,

6

λ

∫ ′

Ω

|ηD∆hu|2 ≤ C(

||u||W 1,2(Ω) + ||f ||L2(Ω)

)

· C ′′(

||Dη∆hu||L2(Ω′) + ||D∆hu||L2(Ω′)

)

+ C ′||(D∆hu)||L2(Ω′)||Dη∆hu||L2(Ω′)

≤ c(

||u||W 1,2(Ω) + ||f ||L2(Ω) + ||Dη∆hu||L2(Ω′)

)

· ||(D∆hu)||L2(Ω′)

+ c(

||u||W 1,2(Ω) + ||f ||L2(Ω)

)

· ||Dη∆hu||L2(Ω′).

Using the AM-GM Inequality ab =

√

1

ǫa2 · ǫb2 ≤

1

2

(1

ǫa2+ǫb2

)

for the first term and the inequality

(a+ b)c ≤ 12 (a+ b+ c)2 for the second

λ

∫

Ω′

|ηD∆hu|2 ≤1

ǫc2

(

||u||W 1,2(Ω) + ||f ||L2(Ω) + ||Dη∆hu||L2(Ω′)

)2+ ǫ||(D∆hu)||2L2(Ω′)

+ c(

||u||W 1,2(Ω) + ||f ||L2(Ω) + ||Dη∆hu||L2(Ω′)

)2.

Choose any 0 < ǫ < λ/2. Then subtract the second term on the first line of the rhs from the

lhs to get

||ηD∆hu||2L2(Ω′) ≤ c(

||u||W 1,2(Ω) + ||f ||L2(Ω) + ||Dη∆hu||L2(Ω′)

)2⇒

||ηD∆hu||L2(Ω′) ≤ c(

||u||W 1,2(Ω) + ||f ||L2(Ω) + ||Dη∆hu||L2(Ω′)

)

≤c(

||u||W 1,2(Ω) + ||f ||L2(Ω) + supΩ

|Dη| · ||∆hu||L2(Ω′)

)

≤c(

||u||W 1,2(Ω) + ||f ||L2(Ω)

)

·(

1 + supΩ

|Dη|)

,

since ||∆hu||L2(Ω) ≤ ||Du||L2(Ω) ≤ ||u||W 1,2(Ω) ≤ ||u||W 1,2(Ω) + ||f ||L2(Ω) where we have applied the

first Lemma to u ∈W 1,2(Ω). Now we are done as we can choose η such that first η∣

∣

Ω′= 1 (for the

lhs !) and second |Dη| ≤ dist(Ω′, ∂Ω) (for the rhs ) and so

||1 · D∆hu||2L2(Ω′) ≤ c(

||u||W 1,2(Ω) + ||f ||L2(Ω)

)

,

7

independently of h. So by our second Lemma the uniform boundedness of the difference quotients

of Du in L2(Ω′) implies Du ∈ W 1,2(Ω′) ⇒ u ∈W 2,2(Ω′) and we have the desired estimate for

its W 2,2(Ω′) norm by the above inequality combined with the Lemma.

Now that u ∈W 2,2(Ω′) then the our original equation holds in the usual sense

Lu = aijDiju+ DiaijDju+ biDiu+ c · u = f,

a.e !

8

Lecture 0 - alove7 2 1. A Harnack inequality implies Cα regularity for 0 < α < 1. 2. A positive...

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