Sobolev Spaces - ustc.edu.cnstaff.ustc.edu.cn/~bjxuan/Sobolev.pdf · Chapter 2 Sobolev Spaces...

Chapter 2

Sobolev Spaces

Sobolev spaces turn out often to be the proper setting in which to

apply ideas of functional analysis to get information concerning par-

tial differential equations. Here, we collect a few basic results about

Sobolev spaces. A general reference to this topic is Adams [1], Gilbarg-

Trudinger [29], or Evans [26].

2.1 Holder spaces

Assume Ω ⊂ Rn is open and 0 < γ ≤ 1.

Definition 2.1.1 (i) If u : Ω → R is bounded and continuous, we

write

‖u‖C(Ω) := supx∈Ω

|u(x)|.

(ii) The γth-Holder semi-norm of u : Ω → R is

[u]C0,γ(Ω) := supx,y∈Ω, x 6=y

|u(x) − u(y)|

|x− y|γ

,

and the γth-Holder norm is

‖u‖C0,γ (Ω) := ‖u‖C(Ω) + [u]C0,γ (Ω).

(iii) The Holder space

Ck,γ(Ω)

11

12 CHAPTER 2. SOBOLEV SPACES

consists of all functions u ∈ Ck(Ω) for which the norm

‖u‖Ck,γ (Ω) :=∑

|α|≤k

‖Dαu‖C(Ω) +∑

|α|=k

[Dαu]C0,γ (Ω) (2.1.1)

is finite.

Theorem 2.1.2 The space of functions Ck,γ(Ω) is a Banach space.

Proof. It is easy to verify that ‖ · ‖Ck,γ(Ω) is a norm (Exercise). It

suffices to check the completeness of Ck,γ(Ω).

1. Construction of limit functions.

Let um∞m=1 be a Cauchy sequence in Ck,γ(Ω), i.e.,

‖um − un‖Ck,γ(Ω) → 0, as m,n→ +∞.

Since any Cauchy sequence is a bounded sequence, so ‖um‖Ck,γ (Ω) is

uniformly bounded. Particularly, ‖um‖C(Ω) is uniformly bounded, that

is, um(x) is a uniformly bounded sequence. Furthermore, the bound-

edness of ‖Dum‖C(Ω) or [Dαum]C0,γ (Ω) implies the equicontinuity of the

sequence um(x). The Arzela-Ascoli theorem implies there exists a

subsequence, still denoted by um∞m=1, which converges to a bounded

continuous function u(x) uniformly in Ω. The same argument can be

repeated for any derivative of Dαum of order up to and including k.

That is, there exist functions uα ∈ C(Ω), such that

Dαum → uα, uniformly in Ω, ∀ |α| ≤ k.

2. Dαu = uα.

Deduce by induction, first let α = (1, 0, · · · , 0), since the series

u1(x) ++∞∑

j=1

(uj+1 − uj) = limm→∞

um+1(x) = u(x),

uniformly in Ω and the series

Dαu1(x) +

+∞∑

j=1

(Dαuj+1 −Dαuj) = limm→∞

Dαum+1(x) = uα(x),

2.1. HOLDER SPACES 13

uniformly in Ω. The theory of series implies that u(x) is differentiable

in Ω and Dαu = uα. The induction method shows that Dαu = uα is

valid for all multi-index α such that |α| ≤ k.

3. u ∈ Ck,γ(Ω).

It suffices to show that for any multi-index α such that |α| = k,

there holds [Dαu]C0,γ (Ω) < +∞.

For any x, y ∈ Ω, x 6= y, since Dαum → Dαu uniformly in Ω, there

exists a N large enough, such that:

|DαuN (z) −Dαu(z)| ≤1

2|x− y|γ , ∀ z ∈ Ω.

Calculate:

|Dαu(x) −Dαu(y)| ≤ |Dαu(x) −DαuN (x)| + |DαuN (x) −DαuN (y)|

+|DαuN (y) −Dαu(y)|

≤ |x− y|γ + |x− y|γ‖DαuN‖C0,γ(Ω)

≤ (1 + supm

‖Dαum‖C0,γ (Ω))|x− y|γ ,

Thus the boundedness of ‖Dαum‖C0,γ(Ω) implies [Dαu]C0,γ(Ω) < +∞.

4. um → u in Ck,γ(Ω).

It also suffices to show that for any multi-index α such that |α| = k,

there holds

[Dαum −Dαu]C0,γ (Ω) → 0, as m→ ∞.

In fact, for any x, y ∈ Ω, x 6= y, there holds

|Dαum(x) −Dαu(x) − (Dαum(y) −Dαu(y))|

≤ lim supn→∞

|Dαum(x) −Dαun(x) − (Dαum(y) −Dαun(y))|

≤ |x− y|γ limn→∞

[Dαum −Dαun]C0,γ (Ω).

Since um∞m=1 is a Cauchy sequence in Ck,γ(Ω), [Dαum−Dαu]C0,γ(Ω) ≤

limn→∞

[Dαum −Dαun]C0,γ(Ω) → 0 as m→ 0.


2.2 Sobolev spaces

2.2.1 Lp spaces

The Lp(Ω) space consists of all the function u : Ω → R such that∫

Ω|u|p dx <∞, where 1 ≤ p ≤ ∞; with norm

‖u‖Lp(Ω) =

(

∫

Ω|u|p dx

)1/p, if 1 ≤ p <∞,

esssupΩ

|u|, if p = ∞.

In order to deduce some properties of Lp space, let us start from some

basic inequalities, which will be used latter.

Lemma 2.2.1 (Young inequality) Let 1 < p, q < ∞, 1p + 1

q = 1,

then

ab ≤ap

p+bq

q, a, b > 0.

Proof. Method 1. The mapping x 7→ ex is convex, and consequently,

ab = elog a+log b = e1p

log ap+ 1q

log bq

≤1

pelog ap

+1

qelog bq

=ap

p+bq

q.

Method 2. Let y = xp−1, x ≥ 0, then x = y1

p−1 = yq−1, Consider the

area of the region restricted by y = 0, x = a, and curve y = xp−1 is

I =

∫ a

0xp−1 dx =

ap

p;

and the area of the region restricted by x = 0, y = b, and curve

x = yq−1 is

II =

∫ b

0yq−1 dy =

bq

q;

The area of the rectangle [0, a]×[0, b] is ab, Thus ab ≤ I+II ≤ap

p+bq

q,

and the equality holds if and only if b = ap−1.

It is easy to obtain the following Young inequality with ε (Exercise):

2.2. SOBOLEV SPACES 15

Lemma 2.2.2 (Young inequality with ε) Let 1 < p, q < ∞, 1p +

1q = 1, then

ab ≤ εap

p+

1

εq/p

bq

q, a, b > 0.

Lemma 2.2.3 (Holder inequality) Let 1 ≤ p, q ≤ ∞, 1p + 1

q = 1,

then if u ∈ Lp(Ω), v ∈ Lq(Ω), we have

∫

Ω|uv| dx ≤ ‖u‖Lp(Ω)‖v‖Lq(Ω).

Proof. 1. In the cases where p = ∞ or q = ∞, it is easy, because there

exists a subset Ω′ ⊂ Ω, with |Ω′| = |Ω|, such that supΩ′

|u| = ‖u‖L∞(Ω)

or supΩ′

|v| = ‖v‖L∞(Ω).

2. In the cases where 1 < p, q < ∞. By the homogeneity of the

inequality, we may assume that ‖u‖Lp(Ω) = ‖v‖Lq(Ω) = 1. Then the

Young inequality implies that

∫

Ω|uv| dx ≤

1

p

∫

Ω|u|p dx+

1

q

∫

Ω|u|q dx = 1 = ‖u‖Lp(Ω)‖v‖Lq(Ω).

By induction argument, it is easy to show the following Generalized

Holder inequality (Exercise):

Lemma 2.2.4 (Generalized Holder inequality) Let 1 ≤ p1, · · · , pm ≤

∞, 1p1

+ · · · + 1pm

= 1, then if uk ∈ Lpk(Ω) for k = 1, · · · ,m, we have

∫

Ω|u1 · · · um| dx ≤

mΠ

k=1‖uk‖Lpk (Ω).

Lemma 2.2.5 (Minkowski inequality) Assume 1 ≤ p ≤ ∞, u, v ∈

Lp(Ω), then

‖u+ v‖Lp(Ω) ≤ ‖u‖Lp(Ω) + ‖v‖Lp(Ω).


Proof. From the Holder inequality, we have

‖u+ v‖pLp(Ω) =

∫

Ω|u+ v|p dx

≤

∫

Ω|u+ v|p−1(|u| + |v|) dx

≤(

∫

Ω|u+ v|

(p−1) pp−1 dx

)p−1

p(

(

∫

Ω|u|p dx)1/p + (

∫

Ω|v|p dx)1/p

)

= ‖u+ v‖p−1Lp(Ω)(‖u‖Lp(Ω) + ‖v‖Lp(Ω)).

Lemma 2.2.6 (Interpolation inequality for Lp- norms) Assume 1 ≤

s ≤ r ≤ t ≤ ∞, and 1r = θ

s + (1−θ)t . Suppose also u ∈ Ls(Ω) ∩ Lt(Ω).

Then u ∈ Lr(Ω), and

‖u‖Lr(Ω) ≤ ‖u‖θLs(Ω)‖u‖

(1−θ)Lt(Ω).

Proof. From the Holder inequality, we have

∫

Ω|u|r dx =

∫

Ω|u|θr|u|(1−θ)r dx

≤(

∫

Ω|u|θr s

θr dx)

θrs(

∫

Ω|u|

(1−θ)r t(1−θ)r dx

)

(1−θ)rt .

Theorem 2.2.7 (Completeness of Lp(Ω)) Lp(Ω) is a Banach space

if 1 ≤ p ≤ ∞.

Proof. It is easy to check that ‖·‖Lp(Ω) is a norm (Exercise). It suffices

to verify the completeness of Lp(Ω)

1. Construction of the limit function.

Assume that 1 ≤ p < ∞ and let um be a Cauchy sequence in

Lp(Ω). There is a subsequence umj of um such that

‖umj+1 − umj‖p ≤ 1/2j , j = 1, 2 · · · .


Let vm(x) =m∑

j=1|umj+1(x) − umj (x)|, then

‖vm‖p ≤

m∑

j=1

‖umj+1 − umj‖p ≤

m∑

j=1

1/2j < 1.

Putting v(x) = limm→∞

vm(x), which may be infinite for some x, we obtain

by the Fatou Lemma that∫

Ω|v(x)|p dx ≤ lim inf

m→∞

∫

Ω|vm(x)|p dx ≤ 1.

Hence v(x) < ∞ a.e. in Ω and its absolutely convergence implies the

series

um1 +∞∑

j=1

(umj+1(x) − umj (x)) (2.2.1)

converges to a limit u(x) a.e. in Ω. Let u(x) = 0 whenever it is

undefined as the limit of series (2.2.1). Thus

limj→∞

umj (x) = u(x) a.e. in Ω.

2. u ∈ Lp(Ω) and um → u in Lp(Ω).

For any ε > 0, there exists N such that if m,n ≥ N , then ‖um −

un‖p < ε. Hence by the Fatou Lemma again, we have∫

Ω|um(x) − u(x)|p dx =

∫

Ωlim

j→∞|um(x) − umj (x)|

p dx

≤ lim infj→∞

∫

Ω|um(x) − umj (x)|

p dx

≤ εp.

If m ≥ N , thus u = (u − um) + um ∈ Lp(Ω), and ‖um − u‖p → 0 as

m→ ∞.

3. In the case where p = ∞, let um be a Cauchy sequence in L∞(Ω),

then there exists a subset A ⊂ Ω, with |A| = 0, such that if x ∈ Ω \A,

then for n,m = 1, · · · , there holds

|um(x)| ≤ ‖um‖∞, |um(x) − un(x)| ≤ ‖um − un‖∞.

That is, um(x) is uniformly bounded in Ω \ A, and um(x) is a Cauchy

sequence for any x ∈ Ω\A, thus um(x) converges uniformly on Ω\A to


a bounded function u(x). Set u(x) = 0 if x ∈ A. We have u ∈ L∞(Ω)

and ‖um − u‖∞ → 0 as m→ ∞.

Theorem 2.2.8 (Riesz Representation Theorem for Lp(Ω)) Let

1 < p <∞ and L ∈ [Lp(Ω)]′. Then there exists v ∈ Lp′(Ω),1

p+

1

p′= 1

such that for all u ∈ Lp(Ω), there holds

L(u) =

∫

Ωu(x)v(x) dx.

Moreover, ‖v‖Lp′ (Ω) = ‖L‖[Lp(Ω)]′ , that is,

[Lp(Ω)]′ ≅ Lp′(Ω).

Theorem 2.2.9 (Riesz Representation Theorem for L1(Ω)) Let

L ∈ [L1(Ω)]′. Then there exists v ∈ L∞(Ω) such that for all u ∈ L1(Ω),

there holds

L(u) =

∫

Ωu(x)v(x) dx.

Moreover, ‖v‖L∞(Ω) = ‖L‖[L1(Ω)]′ , that is,

[L1(Ω)]′ ≅ L∞(Ω).

Theorem 2.2.10 (Reflexivity of Lp) Lp(Ω) is reflexive if and only

if 1 < p <∞.

Theorem 2.2.11 C∞0 (Ω) is dense in Lp(Ω) if 1 ≤ p <∞.

Theorem 2.2.12 (Separability of Lp(Ω)) Lp(Ω) is separable 1 ≤

p <∞.

2.2.2 Weak derivatives

Let C∞0 (Ω) denotes the space of infinitely differentiable functions φ :

Ω → R, with compact support in Ω. We will call a function φ belonging

to C∞0 (Ω) a test function.


Motivation for definition of weak derivatives. Assume we are

given a function u ∈ C1(Ω). Then if ϕ ∈ C∞0 (Ω), we see from the

integration by parts that

∫

Ωuφxi dx = −

∫

Ωuxiφdx (i = 1, · · · , n). (2.2.2)

More general now, if k is a positive integer, u ∈ Ck(Ω), and α =

(α1, · · · , αn) is a multi-index of order |α| = α1 + · · · + αn = k, then

∫

Ωu(Dαφ) dx = (−1)|α|

∫

Ω(Dαu)φdx. (2.2.3)

We next examine that formula (2.2.3) is valid for u ∈ Ck(Ω), and

ask whether some variant of it might still be true even if u is not k times

continuously differentiable. Note that the left-hand side of (2.2.3) make

sense if u is only locally summable: the problem is rather that if u is

not Ck, then the expression Dαu on the right-hand side of (2.2.3) has

no obvious meaning. We overcome this difficulty by asking if there

exists a locally summable function v for which formula (2.2.3) is valid,

with v replacing Dαu:

Definition 2.2.13 Suppose u, v ∈ L1loc(Ω), and α is a multi-index. We

say that v is the αth-weak partial derivative of u, written

Dαu = v,

provided that∫

ΩuDαφdx = (−1)|α|

∫

Ωvφ dx (2.2.4)

for all test functions ϕ ∈ C∞0 (Ω).

Lemma 2.2.14 (Uniqueness of weak derivative) The weak partial

derivative of u, if it exists, is uniquely defined up to a set of measure

zero.

Example 2.2.15 Let n = 1, Ω = (0, 2), and

u(x) =

x, if 0 < x ≤ 1,

1, if 1 ≤ x < 2.


Define

v(x) =

1, if 0 < x ≤ 1,

0, if 1 ≤ x < 2.

From the definition 2.2.13, u′ = v. (Exercise)

Example 2.2.16 Let n = 1, Ω = (0, 2), and

u(x) =

x, if 0 < x ≤ 1,

2, if 1 ≤ x < 2.

We assert u′ does not exist in the weak sense. (Exercise)

2.2.3 Definition of Sobolev spaces

Fix 1 ≤ p ≤ ∞ and let k be a nonnegative integer. We define now cer-

tain function spaces, whose members have weak derivatives of various

orders lying in various Lp spaces.

Definition 2.2.17 i) The Sobolev space

W k,p(Ω)

consists of all locally summable functions u : Ω → R such that

for each multi-index α with |α| ≤ k, Dαu exists in the weak sense

and belongs to Lp(Ω).

ii) If u ∈W k,p(Ω), we define its norm to be

‖u‖W k,p(Ω) :=

(∑

|α|≤k

∫

Ω|Dαu|p dx

)1/p, (1 ≤ p <∞),

∑

|α|≤k

ess supΩ

|Dαu|, (p = ∞).

Definition 2.2.18 (Convergence in W k,p(Ω)) i) Let um∞m=1, u ∈

W k,p(Ω). We say um converges (strongly) to u in W k,p(Ω), writ-

ten

um → u in W k,p(Ω),

provided that

limm→∞

‖um − u‖W k,p(Ω) = 0.


ii) We write

um → u in W k,ploc (Ω)

to mean

um → u in W k,p(U),

for each U ⊂⊂ Ω.

Definition 2.2.19 We denote by

W k,p0 (Ω)

the closure of C∞0 (Ω) in W k,p(Ω).

Remark 2.2.20 (i) If p = 2, we usually write

Hk(Ω) = W k,2(Ω), Hk0 (Ω) = W k,2

0 (Ω), (k = 0, 1, · · · ).

The letter H is used, since - as we will see - Hk(Ω) is a Hilbert

space. Note that H0(Ω) = L2(Ω).

(ii) We henceforth identity functions in W k,p(Ω) which agree almost

everywhere.

Example 2.2.21 Let Ω = B(0, 1), the open unit ball in Rn, and

u(x) = |x|−α (x ∈ Ω, x 6= 0), α > 0.

Then u ∈ W 1,p(Ω) if and only if 0 < (α + 1)p < n. In particular

u 6∈W 1,p(Ω) for each p ≥ n. (Exercise)

Example 2.2.22 Let rk∞k=1 be a countable, dense subset of Ω =

B(0, 1) ∈ Rn. Set

u(x) =

∞∑

k=1

1

2k|x− rk|

−α (x ∈ Ω).

Then u ∈ W 1,p(Ω) if and only if (α + 1)p < n. If 0 < (α + 1)p < n,

we see that u belongs to W 1,p(Ω) and yet is unbounded on each open

subset of Ω. (Exercise)

Theorem 2.2.23 (Properties of weak derivatives) Assume u, v ∈

W k,p(Ω), |α| ≤ k. Then


(i) Dαu ∈ W k−|α|,p(Ω) and Dβ(Dαu) = Dα(Dβu) = Dα+βu for all

multi-indices α, β with |α+ β| ≤ k;

(ii) For each λ, µ ∈ R, λu + µv ∈ W k,p(Ω) and Dα(λu + µv) =

λDαu+ µDαv, |α| ≤ k;

(iii) If Ω′ is an open subset of Ω, then u ∈W k,p(Ω′);

(iv) If ξ ∈ C∞0 (Ω), then ξu ∈W k,p(Ω) and

Dα(ξu) =∑

β≤α

(

α

β

)

DβξDα−βu (Leibniz′ formula),

where

(

α

β

)

=α!

β!(α − β)!.

Theorem 2.2.24 (Sobolev spaces as function spaces) For each k

= 0, 1, · · · and 1 ≤ p ≤ ∞, Sobolev space W k,p(Ω) is a Banach space;

W k,p(Ω) is reflexive if and only if 1 < p < ∞; For 1 ≤ p < ∞, the

subspace W k,p∩C∞(Ω) is dense in W k,p(Ω), thus W k,p(Ω) is separable;

Moreover, W k,2(Ω) is a Hilbert space with scalar product

(u, v)W k,2 =∑

α≤k

∫

ΩDαuDαv dx.

Proof. 1. It is easy to check that ‖ · ‖W k,p(Ω) is a norm (Exercise).

Next to show that W k,p(Ω) is complete.

2. Construction of limit functions.

Assume um∞m=1 is a Cauchy sequence in W k,p(Ω). Then for each

|α| ≤ k, Dαum∞m=1 is a Cauchy sequence in Lp(Ω). Since Lp(Ω) is

complete, there exist functions uα ∈ Lp(Ω), such that

Dαum → uα, in Lp(Ω),

for each |α| ≤ k. In particular,

um → u(0,··· ,0) =: u in Lp(Ω).

3. u ∈W k,p(Ω) and Dαu = uα, |α| ≤ k.


In fact, let ψ ∈ C∞0 (Ω), then

∫

ΩuDαψ dx = lim

m→∞

∫

ΩumD

αψ dx

= limm→∞

(−1)|α|∫

Ω(Dαum)ψ dx

=

∫

Ωuαψ dx.

4. um → u in W k,p(Ω).

Since Dαum → uα = Dαu in Lp(Ω) for all |α| ≤ k, it follows that

um → u in W k,p(Ω).

5. Regard W k,p(Ω) as a closed subspace of a Cartesian product space

of spaces Lp(Ω): Let N = N(n, k) =∑

0≤|α|≤k

1 be the number of multi-

indexes α satisfying 0 ≤ |α| ≤ k. For 1 ≤ p ≤ ∞, let LpN =

NΠ

j=1Lp(Ω),

the product norm of u = (u1, · · · , uN ) in LpN being given by

‖u‖LpN

=

(

N∑

j=1‖uj‖

pp

)1/p, if 1 ≤ p <∞,

max1≤j≤N

‖uj‖∞, if p = ∞.

Since a closed subspace of a reflexive (Resp. separable, uniformly con-

vex) Banach space is also reflexive (Resp. separable, uniformly convex),

thus W k,p(Ω) is reflexive if and only if 1 < p <∞; and W k,p(Ω) is sep-

arable if 1 ≤ p <∞. Furthermore,(

C∞(Ω))N

is dense in LpN and thus

in W k,p(Ω) if 1 ≤ p <∞.

6. If p = 2, define the scalar product as

(u, v)W k,2 =∑

|α|≤k

∫

ΩDαuDαv dx.

It is easy to verify W k,2 = Hk is a Hilbert space.

2.2.4 Inequalities

Theorem 2.2.25 (Gagliardo-Nirenberg-Sobolev inequality) Let

1 ≤ p < n, p∗ =np

n− p. There exists a constant C, depending only on


p and n, such that

‖u‖Lp∗ (Rn) ≤ C‖Du‖Lp(Rn), (2.2.5)

for all u ∈ C10 (Rn).

Proof. 1. First assume p = 1.

Since u has compact support, for each i = 1, 2 · · · , n and x ∈ Rn

we have

u(x) =

∫ xi

−∞uxi(x1, · · · , xi−1, yi, xi+1, · · · , xn) dyi,

and so

|u(x)| ≤

∫ +∞

−∞|Du(x1, · · · , xi−1, yi, xi+1, · · · , xn)| dyi, i = 1, 2 · · · , n.

Consequently

|u(x)|n

n−1 ≤nΠ

i=1

(

∫ +∞

−∞|Du(x1, · · · , xi−1, yi, xi+1, · · · , xn)| dyi

)1

n−1.

Integrate this inequality with respect to x1:∫ +∞

−∞|u(x)|

nn−1 dx1

≤

∫ +∞

−∞

nΠ

i=1

(

∫ +∞

−∞|Du(x1, · · · , xi−1, yi, xi+1, · · · , xn)| dyi

)1

n−1dx1

=(

∫ +∞

−∞|Du(y1, x2 · · · , · · · , xn)| dy1

)1

n−1

·

∫ +∞

−∞

nΠ

i=2

(

∫ +∞

−∞|Du(x1, · · · , xi−1, yi, xi+1, · · · , xn)| dyi

)1

n−1dx1

≤(

∫ +∞

−∞|Du| dy1

)1

n−1( n

Πi=2

∫ +∞

−∞

∫ +∞

−∞|Du| dx1dyi

)1

n−1,

(2.2.6)

the last inequality follows from the Generalized Holder inequality.

Now integrate (2.2.6) with respect to x2:

∫ +∞

−∞

∫ +∞

−∞|u(x)|

nn−1 dx1 dx2

≤(

∫ +∞

−∞

∫ +∞

−∞|Du| dx1dy2

)1

n−1

∫ +∞

−∞

nΠ

i=1,i6=2I

1n−1

i dx2,


where

I1 =

∫ +∞

−∞|Du| dy1, Ii =

∫ +∞

−∞

∫ +∞

−∞|Du| dx1dyi, i = 3, · · · , n.

Applying once more the Generalized Holder inequality, we find

∫ +∞

−∞

∫ +∞

−∞|u(x)|

nn−1 dx1 dx2

≤(

∫ +∞

−∞

∫ +∞

−∞|Du| dx1dy2

)1

n−1(

∫ +∞

−∞

∫ +∞

−∞|Du| dy1dx2

)1

n−1

·nΠ

i=3

(

∫ +∞

−∞

∫ +∞

−∞

∫ +∞

−∞|Du| dx1dx2dyi

)1

n−1.

We continue by integrating with respect to x3, · · · , xn, eventually to

find

∫

Rn

|u(x)|n

n−1 dx ≤nΠ

i=1

(

∫ +∞

−∞· · ·

∫ +∞

−∞|Du| dx1 · · · dxn

)1

n−1

=(

∫

Rn

|Du| dx)

nn−1

.

(2.2.7)

This is conclusion for p = 1.

2. Consider now the case that 1 < p < n. We apply (2.2.7) to v := |u|γ ,

where γ > 1 is to be selected. Then

(

∫

Rn

|u(x)|γn

n−1 dx)

n−1n

≤

∫

Rn

|D|u|γ | dx = γ

∫

Rn

|u|γ−1|Du| dx

≤ γ(

∫

Rn

|u|(γ−1) pp−1 dx

)p−1

p(

∫

Rn

|Du|p)

1p .

(2.2.8)

We choose γ so thatγn

n− 1= (γ − 1)

p

p − 1. That is, we get

γ =p(n− 1)

n− p> 1,

in which caseγn

n− 1= (γ − 1)

p

p − 1=

np

n− p= p∗. Thus, (2.2.8)

becomes (2.2.5).


Remark 2.2.26 The exponent p∗ =np

n− pis called critical Sobolev

exponent, because it is the unique exponent q = p∗ to make the following

inequality:

‖u‖Lq(Rn) ≤ C‖Du‖Lp(Rn)

be true for all u ∈ C10(Rn).

Proof. For any u ∈ C10 (Rn) and λ > 0, define the scaled function

uλ(x) = u(λx). Direct calculations show that

‖uλ‖Lq(Rn) = λ−n

q ‖u‖Lq(Rn)

and

‖Duλ‖Lp(Rn) = λ1−np ‖Du‖Lp(Rn).

If the inequality in remark is true for all u ∈ C10 (Rn), replacing u by

uλ, there holds

‖u‖Lq(Rn) ≤ Cλ1−np+ n

q ‖Du‖Lp(Rn).

If 1− np + n

q 6= 0, upon sending λ to either 0 or ∞, we will get C = ∞.

Theorem 2.2.27 (Poincare inequality for W 1,p0 (Ω)) Assume Ω is

a bounded, open subset of Rn, Suppose u ∈ W 1,p0 (Ω) for some 1 ≤ p <

n. Then there exists a constant C, depending only on p, n and Ω, such

that

‖u‖Lq(Ω) ≤ C‖Du‖Lp(Ω), (2.2.9)

for each q ∈ [1, p∗], the constant C depending only on p, q, n and Ω.

Proof. Since u ∈W 1,p0 (Ω), then exists a sequence of functions um∞m=1 ⊂

C∞0 (Ω) converging to u in W 1,p

0 (Ω). We extend each function um(x)

to be 0 on Rn \ Ω and apply Theorem 2.2.25 to discover ‖u‖Lp∗ (Ω) ≤

C‖Du‖Lp(Ω). Since |Ω| <∞, the Holder inequality implies ‖u‖Lq(Ω) ≤

C‖u‖Lp∗(Ω) if 1 ≤ q ≤ p∗.

Remark 2.2.28 In view of Theorem 2.2.27, on W 1,p0 (Ω), the norm

‖Du‖Lp(Ω) is equivalent to ‖u‖W 1,p0 (Ω) if Ω is bounded.


It follows (2.2.11).

2. supRn

|u| ≤ C‖u‖W 1,p(Rn).

Now fix x ∈ Rn. We apply (2.2.11) as follows:

|u(x)| ≤ −

∫

B(x,1)|u(x) − u(y)| dy + −

∫

B(x,1)|u(y)| dy

≤ C

∫

B(x,1)

|Du(y)|

|y − x|n−1dy + C‖u‖Lp(B(x,1))

≤ C‖Du‖Lp(Rn)

(

∫

B(x,1)

dy

|x− y|(n−1) p

p−1

)p−1

p+ C‖u‖Lp(Rn)

≤ C‖u‖W 1,p(Rn).(2.2.12)

The last estimate holds since p > n implies (n− 1) pp−1 < n; so that

∫

B(x,1)

dy

|x− y|(n−1) p

p−1

<∞.

As x ∈ Rn is arbitrary, (2.2.12) implies

supRn

|u| ≤ C‖u‖W 1,p(Rn). (2.2.13)

3. [u]C0, 1−n/p(Rn) ≤ C‖Du‖Lp(Rn).

Choose any two points x, y ∈ Rn and write r := |x − y|. Let

W := B(x, r) ∩B(y, r). Then

|u(x) − u(y)| ≤ −

∫

W|u(x) − u(z)| dz + −

∫

W|u(y) − u(z)| dz. (2.2.14)

On the other hand, (2.2.11) allows us to estimate

−

∫

W|u(x) − u(z)| dz ≤ C −

∫

B(x,r)|u(x) − u(z)| dz

≤ C‖Du‖Lp(Rn)

(

∫

B(x,1)

dz

|x− z|(n−1) p

p−1

)p−1

p

≤ C(

rn−(n−1) pp−1

)p−1

p ‖Du‖Lp(Rn)

= Cr1−n

p ‖Du‖Lp(Rn).(2.2.15)


Similarly, there holds

−

∫

W|u(y) − u(z)| dz ≤ Cr1−

np ‖Du‖Lp(Rn).

Substituting this estimate and (2.2.15) into (2.2.14) yields

|u(x) − u(y)| ≤ Cr1−n

p ‖Du‖Lp(Rn) = C|x− y|1−n

p ‖Du‖Lp(Rn).

Thus

[u]C0, 1−n/p(Rn) = supx 6=y

|u(x) − u(y)|

|x− y|1−n/p

≤ C‖Du‖Lp(Rn).

This inequality and (2.2.13) complete the proof.

2.2.5 Embedding theorems and Trace theorems

Definition 2.2.30 Let (X, ‖ · ‖X), (Y, ‖ · ‖Y ) be Banach spaces. X is

(continuously) embedded into Y , denoted X → Y , if there exists an

injective linear map i : X → Y and a constant C such that

‖i(x)‖Y ≤ C‖x‖X , ∀x ∈ X.

X is compactly embedded into Y , denoted X →→ Y , if i maps

bounded subsets of X into precompact subsets of Y .

Theorem 2.2.31 (Exercise)

(i) For Ω ⊂ Rn with Lebesgue measure Ln(Ω) <∞, 1 ≤ p < q ≤ ∞,

we have Lq(Ω) → Lp(Ω). This ceases to be true if Ln(Ω) = ∞;

(ii) Suppose Ω is a precompact domain in Rn, let m = 0, 1, · · · , and

0 ≤ α < β ≤ 1. Then Cm,β(Ω) →→ Cm,α(Ω) compactly.

Theorem 2.2.32 (Sobolev embedding theorem) Let Ω ⊂ Rn be

a bounded domain with Lipschitz boundary, k ∈ N, 1 ≤ p ≤ ∞. Then

the following hold:

(i) If kp < n, then W k,p(Ω) → Lq(Ω) for 1 ≤ q ≤ p∗ = npn−kp ; the

embedding is compact, if q < npn−kp ;


(ii) If 0 ≤ m < k − np < m + 1, then W k,p(Ω) → Cm,α(Ω), for

0 ≤ α ≤ k−m− np ; the embedding is compact, if α < k−m− n

p .

Proof. 1. The embedding of the case where kp < n.

Assume that kp < n and u ∈ W k,p(Ω). Then Dαu ∈ Lp(Ω) for all

|α| = k. The Gagliardo-Nirenberg-Sobolev inequality implies

‖Dβu‖Lp1 (Ω) ≤ C‖u‖W k,p(Ω), for |β| = k − 1, p1 =np

n− p,

so u ∈ W k−1,p1(Ω). Similarly, we find u ∈ W k−2,p2(Ω), where p2 =np1

n− p1=

np

n− 2p. Continuing, we eventually discover after k steps

that W 0,pk(Ω) = Lpk(Ω), for pk =npn−1

n− pn−1=

np

n− kp, and

‖u‖Lpk (Ω) ≤ C‖u‖W k,p(Ω).

Since Ω is bounded, thus for any 1 ≤ q ≤ p∗ = npn−kp , W k,p(Ω) → Lq(Ω).

2. The compactness of embedding.

That is to show that if um∞m=1 is a bounded sequence in W k,p(Ω),

there exists a subsequence umj∞j=1 which converges in Lq(Ω). For

simplicity, we just prove the case k = 1. By induction argument, one

get the proof of general case.

By extension argument, without loss of generality assume that

Ω = Rn and the functions um∞m=1 all have compact support in some

bounded open set U ⊂ Rn, moreover,

supm

‖um‖W k,p(U) <∞. (2.2.16)

Let us first study the smooth functions:

uεm := ηε ∗ um (ε > 0,m = 1, 2, · · · ),

where ηε :=1

εnη(x

ε) denotes the usual mollifier and η satisfies

η ∈ C∞0 (Rn), η ≥ 0,

∫

Rn

η(x) dx = 1.

A concrete example of η is

η(x) :=

Cexp(1

|x|2 − 1), if |x| < 1,

0, if |x| ≥ 1,


and constant C > 0 is selected so that

∫

Rn

η(x) dx = 1. We may

suppose that the functions uεm∞m=1 all have support in U as well.

3. Claim

uεm → um in Lq(U) as ε→ 0, uniformly in m. (2.2.17)

To prove this claim, we first note that if um is smooth, then

uεm(x) − um(x) =

∫

B(0,1)η(y)(um(x− εy) − um(x)) dy

=

∫

B(0,1)η(y)

∫ 1

0

d

dtum(x− εty) dtdy

= −ε

∫

B(0,1)η(y)

∫ 1

0Dum(x− εty) · y dtdy.

Thus∫

U|uε

m(x) − um(x)| dx ≤ ε

∫

B(0,1)η(y)

∫ 1

0

∫

U|Dum(x− εty)| dxdtdy

≤ ε

∫

U|Dum(z)| dz.

By approximation this estimate holds if um ∈W 1,p(U). Hence

‖uεm − um‖L1(U) ≤ ε‖Dum‖L1(U) ≤ ε‖Dum‖Lp(U),

the latter inequality holds since U is bounded. Owing to (2.2.16) we

thereby discover

uεm → um in L1(U) as ε→ 0, uniformly in m. (2.2.18)

But then since 1 ≤ q < p∗, we using the Interpolation inequality for

Lp-norm that

‖uεm − um‖Lq(U) ≤ ‖uε

m − um‖θL1(U)‖u

εm − um‖1−θ

Lp∗(U),

where1

q= θ +

(1 − θ)

p∗, 0 < θ ≤ 1. Consequently, (2.2.16) and the

Gagliardo-Nirenberg-Sobolev inequality imply

‖uεm − um‖Lq(U) ≤ C‖uε

m − um‖θL1(U);


whence assertion (2.2.17) follows from (2.2.18).

4. Next we claim

for each fixed ε > 0, the sequence uεm∞m=1

is uniformly bounded and equi-continuous.(2.2.19)

Indeed, if x ∈ Rn, then

|uεm(x)| ≤

∫

B(x,ε)ηε(x− y)|um(y)|dy

≤ ‖ηε‖L∞(Rn)‖um‖L1(U) ≤C

εn<∞,

for m = 1, 2, · · · . Similarly

|Duεm(x)| ≤

∫

B(x,ε)|Dηε(x− y)| |um(y)|dy

≤ ‖Dηε‖L∞(Rn)‖um‖L1(U) ≤C

εn+1<∞,

for m = 1, 2, · · · . Thus the claim (2.2.19) follows these two estimates.

5. Now fix δ > 0. We will show that there exists a subsequence

umj∞j=1 of um∞m=1 such that

lim supj,k→∞

‖umj − umk‖Lq(U) ≤ δ. (2.2.20)

To see this, let us first employ assertion (2.2.17) to select ε > 0 so

small that

‖uεm − um‖Lq(U) ≤ δ/2, (2.2.21)

for m = 1, 2, · · · .

We now observe that since functions um∞m=1, and thus functions

uεm∞m=1, have support in some fixed bounded set U ⊂ Rn, we may

use (2.2.19) and the Arzela-Ascoli theorem, to obtain a subsequence

uεmj

∞j=1 ⊂ uεm∞m=1 which converges uniformly on U . In particular

therefore

lim supj,k→∞

‖uεmj

− uεmk

‖Lq(U) = 0. (2.2.22)

Then (2.2.21) and (2.2.22) imply (2.2.20).


6. We next employ assertion (2.2.20) with δ = 1,1

2,1

3, · · · and use

the standard diagonal argument to extract a subsequence uml∞l=1 of

um∞m=1 such that

lim supl,k→∞

‖uml− umk

‖Lq(U) = 0.

Conclusion (i) of theorem is proved by steps 1 - 6. Next to prove

conclusion (ii) of theorem.

7. Assume that 0 ≤ m < k − np < m+ 1. Then as above we see

u ∈W k−l,r(Ω), (2.2.23)

for r =np

n− lpprovided that lp < n. We choose the integer l0 = [

n

p].

Consequently, (2.2.23) holds for r =np

n− pl0> n. Hence (2.2.23) and

Morrey inequality imply that Dαu ∈ C0,1−nr (Ω) for all |α| ≤ k− l0 − 1.

Observe also that 1−n

r= 1−

n

p+ l0 = [

n

p]+1−

n

p. Thus u ∈ Cm, γ(Ω)

with m = k − l0 − 1 = k − [n

p] − 1, γ = 1 −

n

r= [

n

p] + 1 −

n

p, and the

stated estimate follows easily.

8. The Arzela-Ascoli theorem implies the compactness of embedding

W k,p(Ω) → Cm,α(Ω) provided that α < γ = k −m− np .

Denote (u)Ω = −∫

Ω u dx = average of u over Ω.

Theorem 2.2.33 (Poincare inequality for W 1,p(Ω)) Let Ω ⊂ Rn

be bounded and connected, with C1 boundary ∂Ω. Assume 1 ≤ p ≤ ∞.

Then there exists a constant C, depending only on p, n and Ω, such

that

‖u− (u)Ω‖Lp(Ω) ≤ C‖Du‖Lp(Ω), (2.2.24)

for all u ∈W 1,p(Rn).

Proof. Argue by contradiction. If the stated estimate is false, that is,

for each integer k = 1, 2, · · · , there would exist a function uk ∈W 1,p(Ω)

such that

‖uk − (uk)Ω‖Lp(Ω) > k‖Duk‖Lp(Ω). (2.2.25)


Normalize by defining

vk :=uk − (uk)Ω

‖uk − (uk)Ω‖Lp(Ω), k = 1, 2, · · · .

Then, there holds

(vk)Ω = 0, ‖vk‖Lp(Ω) = 1, (2.2.26)

and from (2.2.25), we have

‖Dvk‖Lp(Ω) <1

k, k = 1, 2, · · · . (2.2.27)

Particularly, vk∞k=1 is a bounded in W 1,p(Ω). From the compact em-

bedding, there exists a subsequence vkj∞j=1 ⊂ vk

∞k=1, and a function

v ∈ Lp(Ω) such that

vkj→ v in Lp(Ω).

From (2.2.26), it follows that

(v)Ω = 0, ‖v‖Lp(Ω) = 1.

On the other hand, (2.2.27) implies that for every ϕ ∈ C∞0 (Ω), there

holds∫

Ωvϕxi dx = lim

j→∞

∫

Ωvkjϕxi dx = − lim

j→∞

∫

Ω(vkj

)xiϕdx = 0,

consequently, v ∈ W 1,p(Ω) and Dv = 0 a.e. in Ω. Thus v is constant,

since Ω is connected, furthermore, (v)Ω = 0, hence v ≡ c = 0, which

contradicts the fact that ‖v‖Lp(Ω) = 1.

Theorem 2.2.34 (Poincare inequality for a ball) Suppose 1 ≤ p

< n. Then there exists a constant C, depending only on p and n, such

that

‖u− (u)x,r‖Lp(B(x,r)) ≤ Cr‖Du‖Lp(B(x,r)), (2.2.28)

holds for each ball B(x, r) ⊂ Rn and each function u ∈W 1,p0 (B(x, r)).

Proof. The case Ω = B(0, 1) follows from Theorem 2.2.33. In general,

if u ∈W 1,p0 (B(x, r)), set

v(y) := u(x+ ry), (y ∈ B(0, 1)).


Then v ∈W 1,p0 (B(0, 1)), and there holds

‖v − (v)0,1‖Lp(B(0,1)) ≤ C‖Dv‖Lp(B(0,1)),

Changing variables, we recover estimate (2.2.28).

Lemma 2.2.35 (Poincare-Sobolev inequality) For any ε > 0 there

exists a C = C(ε, n) such that for u ∈ H1(B1) with∣

∣x ∈ B1;u = 0∣

∣ ≥ ε∣

∣B1

∣

∣,

there holds∫

B1

u2 ≤ C

∫

B1

|Du|2.

Proof. Suppose not. Then there exists a sequence um ⊂ H1(B1)

such that∣

∣x ∈ B1;um = 0∣

∣ ≥ ε∣

∣B1

∣

∣,

and∫

B1

u2m = 1,

∫

B1

|Dum|2 → 0 as m→ ∞.

Hence we may assume um → u0 ∈ H1(B1) strongly in L2(B1) and

weakly in H1(B1). Then limm→∞

(

|Dum|2 −|D(um −u0)|2)

= |Du0|2 and

|Du0|2 ≤ lim

m→∞|Dum|2 = 0. Thus u0 is a nonzero constant. So

0 = limm→∞

∫

B1

|um − u0|2 ≥ lim

m→∞

∫

um=0|um − u0|

2

≥ |u0|2 inf

m

∣

∣um = 0∣

∣ > 0.

This contradiction completes the proof.

Definition 2.2.36 For a domain Ω with Ck-boundary ∂Ω = Γ, k ∈

N, 1 < p <∞, denote W k− 1p,p(Γ) as the set of equivalent classes

u+

W k,p0 (Ω); u ∈W k,p(Ω)

, endowed with norm

‖u|Γ‖W

k− 1p ,p

(Γ)= inf

‖v‖W k,p(Ω) : u− v ∈W k,p0 (Ω)

.

The elements of W k− 1p,p(Γ) is call the traces u|Γ of functions u ∈

W k,p(Ω).


Theorem 2.2.37 (Trace Theorem) Assume Ω is a domain with Ck-

boundary ∂Ω = Γ, k ∈ N, 1 < p < ∞. Then W k− 1p,p(Γ) is a Banach

space.

Theorem 2.2.38 (Extension Theorem) For any domain Ω with Ck-

boundary ∂Ω = Γ, k ∈ N, 1 < p < ∞, there exists a continuous linear

extension operator ext : W k− 1p,p(Γ) →W k,p(Ω) such that

(

ext(u))∣

∣

Γ=

u, for all u ∈W k− 1p,p(Γ).

Theorem 2.2.39 Suppose Ω is a domain with Ck-boundary ∂Ω =

Γ, k ∈ N, 1 < p < ∞. Then W k,p(Γ) → Wk− 1

p,p(Γ) → W k−1,p(Γ)

and both embeddings are compact.

EXERCISES

1. Prove The Young inequality with ε - Lemma 2.2.2.

2. Prove the generalized Holder inequality - Lemma 2.2.4.

3. Complete Example 2.2.15.




7. Verify ‖ · ‖Ck,γ(Ω), ‖ · ‖Lp(Ω) and ‖ · ‖W k,p(Ω) are norms.

8. Prove Theorem 2.2.31.

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