Analysis II Home Assignment 4 Subhadip...

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Analysis II Home Assignment 4 Subhadip Chowdhury Problem 4.1 f L p (R N ) ⇐⇒ Z R N (1 + |x| α ) -1 (1 + | log |x|| β ) -1 p < Problem 4.2 Z |f | p Z (|f | p ) q p p q Z (1) 1/(1- p q ) (1- p q ) = kf k p q .|Ω| 1-p/q ⇒kf k p ≤|Ω| 1 p - 1 q kf k q Thus f L q f L p . Also kf - gk p |Ω| 1 p - 1 q ⇒kf - gk q . So the L p (Ω) L q (Ω) with continuous injection. Problem 4.3 4.3.1 Note that h = max{f,g} = |f - g| + f + g 2 Now L p is a vector space. So f + g L p . Then by definition, |f + g|∈ L p . So h L p . 4.3.2 Note that h n = |f n - g n | + f n + g n 2 Now f n f in L p (Ω) and g n g in L p (Ω) imply that |f n - g n |→|f - g| in L p (Ω). Thus h n h in L p (Ω). 1

Transcript of Analysis II Home Assignment 4 Subhadip...

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Analysis IIHome Assignment 4

Subhadip Chowdhury

Problem 4.1

f ∈ Lp(RN) ⇐⇒∫RN

∣∣(1 + |x|α)−1(1 + | log |x||β)−1∣∣p <∞

Problem 4.2

∫|f |p ≤

(∫(|f |p)

qp

) pq(∫

(1)1/(1− pq

)dµ

)(1− pq

)

= ‖f‖pq .|Ω|1−p/q

⇒‖f‖p ≤ |Ω|1p− 1q ‖f‖q

Thus f ∈ Lq ⇒ f ∈ Lp. Also ‖f − g‖p ≤ ε

|Ω|1p−

1q⇒ ‖f − g‖q ≤ ε. So the Lp(Ω) ⊆ Lq(Ω) with continuous

injection.

Problem 4.3

4.3.1 Note that

h = maxf, g =|f − g|+ f + g

2

Now Lp is a vector space. So f + g ∈ Lp. Then by definition, |f + g| ∈ Lp. So h ∈ Lp.

4.3.2 Note that

hn =|fn − gn|+ fn + gn

2

Now fn → f in Lp(Ω) and gn → g in Lp(Ω) imply that |fn − gn| → |f − g| in Lp(Ω). Thus hn → h inLp(Ω).

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Analysis Subhadip Chowdhury Assignment 4

4.3.3 Note that

|fngn − fg|p ≤ c (|fn − f |p|gn|p + |gn − g|p|f |p) (*)

for some constant c depending only on p. Now gpn → gp a.e. and (gn) is bounded in L∞, suppose gn ≤ kfor all n. Hence by DCT, we get

∫|gn − g|p → 0. Thus taking n→∞ in (∗), we get∫

|fngn − fg|p ≤ ckp∫|fn − f |p + |f |p

∫|gn − g|p → 0

So fngn → fg in Lp.

Problem 4.4

4.4.1 We prove the result by inducting on k. For the base case, k = 2. Then for 1p

= 1p1

+ 1p2

, we

have

‖f1f2‖p =

(∫|f1f2|p

) 1p

((∫|f1|p1

) pp1

.

(∫|f2|p.

p1p1−p

)1− pp1

) 1p

= ‖f1‖p1 .‖f2‖ pp1p1−p

Note that 1p

= 1p1

+ 1p2⇒ 1

p2= p1−p

pp1⇒ pp1

p1−p = p2.

Thus ‖f2‖ pp1p1−p

= ‖f2‖p2 ⇒‖f1f2‖p ≤ ‖f1‖p1‖f2‖p2

Suppose for k = m ∈ N, we have ‖g‖p ≤∏m

i=1 ‖gi‖pi whenever gi ∈ Lpi and 1p

=∑m

i=11pi

.

Then let 1q

=∑m+1

i=11qi≤ 1 and fi ∈ Lqi . We have, 1

q= 1

qm+1+∑m

i=11qi

. Then by induction hypothesis,

(f1f2 . . . fm) ∈ L1∑m

i=11qi

and

‖f‖q ≤ ‖fm+1‖qm+1 .‖f1f2 . . . fm‖ 1∑mi=1

1qi

≤ ‖fm+1‖qm+1 .m∏i=1

‖fi‖qi =m+1∏i=1

‖fi‖pi

Thus by induction principle the result is true for all k ∈ N.

4.4.2 Suppose α 6= 0, 1. Put p1 = pα, p2 = q

1−α . Then note that f ∈ Lp ⇒ fα ∈ L pα . Similarly,

f 1−α ∈ Lq

1−α . So applying 4.4.1 with p1, p2 defined as above and noting that 1r

= 1p1

+ 1p2≤ 1, We have

‖f‖r ≤ ‖fα‖p1 .‖f 1−α‖p2 =

(∫|fα|p/α

)αp

.

(∫|f 1−α|

q1−α

) 1−αq

= ‖f‖αp‖f‖1−αq

For the cases α = 0, 1, the inequality is trivially true.

Problem 4.5

4.5.2 We call the set C. Let fn be a sequence in C which converges to f in Lp(Ω). We want to

prove that f ∈ C.Now fn → f in Lp(Ω)⇒ ∃ a subsequence (fnk) and a function h ∈ Lp such that

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Analysis Subhadip Chowdhury Assignment 4

(a) fnk(x)→ f(x) a.e. on Ω

(b) |fnk(x)| ≤ h(x) ∀k, a.e. on Ω

Then applying Fatou’s lemma to (|fnk |q), we get,∫|f |q ≤ 1 ⇒ f ∈ C. Thus C contains all its limit

points. hence C is closed.

4.5.3 By 4.4.2, we have

‖fn − f‖r ≤ ‖fn − f‖αp‖fn − f‖1−αq

where α is defined as in 4.4.2. Note that by part 2, we have ‖f‖q ≤ C. Thus ‖fn − f‖q ≤ 2C. Hence

‖fn − f‖r ≤ ‖fn − f‖αp (2C)1−α → 0

SO fn → f in Lr(Ω).

Problem 4.6

4.6.1 Taking p→∞ in problem 4.2, we get that

lim supp→∞

‖f‖p ≤ ‖f‖∞ (1)

We know that there exist a sequence Cn → ‖f‖∞ such that |f(x)| ≤ Cn a.e on Ω for all n. Then letΩn = x ∈ Ω : |f(x)| > Cn. We have

‖f‖pp ≥∫

Ωn

|f |p ≥ µ(Ωn)(Cn)p ⇒ ‖f‖p ≥ Cn(µ(Ω))1/p

Taking p→∞, we then have lim infp→∞

‖f‖p ≥ Cn for all n. Taking n→∞ we then have

lim infp→∞

‖f‖p ≥ ‖f‖∞ (2)

Combining above inequalities (1) and (2), we get our result.

4.6.2 Fix a k > C. Consider the set S = x ∈ Ω : |f(x)| > k. Suppose µ(S) = ε > 0. Then

C ≥ ‖f‖p ≥ kε1/p

Taing p→∞ we then have C ≥ k,contradiction!! So µ(S) = 0 proving that ‖f‖∞ ≤ C i.e. f ∈ L∞(Ω).

4.6.3 Consider f(x) = ln x for x ∈ (0, 1). Then we know that for 1 ≤ p <∞

limt0

(ln t)12p = lim

t→0+

1t

− 12pt−

12p−1

= 0

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Analysis Subhadip Chowdhury Assignment 4

⇒ ∃ε > 0 such that for 0 < x < ε, | lnx|p < 1

x12

. Then∫ 1

0

| lnx|pdx <∫ ε

0

1√xdx+

∫ 1

ε

| lnx|pdx

=ε12

−12

+ a finite number

<∞

Thus f ∈ Lp(Ω) for 1 ≤ p <∞. But clearly | lnx| is unbounded on (0, 1). So f 6∈ L∞(Ω).

Problem 4.7

Let (un) be sequence in Lp(Ω) converging to u such that aun → f in Lq(Ω). By theorem 4.9, by passingto a subsequence if necessary, we may assume un(x) → u(x) a.e. Then aun(x) → au(x) a.e. Hencef = au a.e. Thus by closed graph theorem, the operator T : Lp → Lq given by u 7→ au is continuous. So∃C > 0 such that

‖au‖q ≤ C‖u‖p (*)

Case 1: If p <∞, then ∫Ω

|au|q ≤ Cq

(∫Ω

|u|p) q

p

Putting u = v1q for some v ∈ L

pq , we get that∫Ω

|a|q|v| ≤ Cq

(∫Ω

|v|pq

) qp

= Cq‖v‖ pq

for all v ∈ Lpq . Thus the map S : L

pq → R given by v 7→

∫Ω|a|q|v| is a continuous linear map.

Hence S ∈(Lpq

)∗⇒ aq ∈ L

11− qp i.e. a ∈ L

qpp−q .

Case 2: If p =∞, from equation (∗) we have ‖a‖q ≤ C‖1‖∞ ⇒ a ∈ Lq. [Note that 1 ∈ L∞].

Problem 4.8

4.8.1 For every integer n ≥ 1, consider the sets

Xn = f ∈ X ∩ L1+ 1n (Ω) : ‖f‖1+ 1

n≤ n

Note that if f ∈ X then f ∈ Lp for some p > 1. But also f ∈ L1. Hence f ∈ Lr for all 1 ≤ r ≤ p. Notethat from 4.4.2, ‖f‖r ≤ K for some constant K for all 1 ≤ r ≤ p. Hence f ∈ Xn for some n ∈ N.Thus ⋃

n∈N

Xn = X

Then since each Xn is closed, by Baire category theorem, there exists N ∈ N such that Int(XN) 6= ∅.Let x0 ∈ Int(XN). Then there exists r > 0 such that x0 +B(0, r) ⊆ XN ⊆ L1+ 1

n . But x0 ∈ L1+ 1n .

Hence B(0, r) ⊆ L1+ 1n ⇒ λB(0, r) ⊆ L1+ 1

n for all λ ∈ R⇒ X ⊆⋃λ

λB(0, r) ⊆ Lq for q = 1 + 1n.

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Analysis Subhadip Chowdhury Assignment 4

4.8.2 Consider the inclusion map T : X → L1 given by Tu = u. Suppose we have a sequence (un)

converging to u in X (note that X is closed) such that Tun → v in L1. Then by theorem 4.9, there is asubsequence (unk) such that unk(x)→ u(x) a.e. Then Tunk(x)→ u(x) a.e. implying u = v a.e. So byclosed graph theorem, T is continuous. Hence there exist some C such that

‖f‖p ≤ C‖f‖1; ∀f ∈ X

Problem 4.9

We may assume WLOG |Ω| = 1, by dividing the measure by |Ω| if necessary. Then∫Ω

j(f)dµ =

∫Ω

supa,b∈R

at+b≤j(t)∀t∈D(j)

(af + b)dµ since j is convex l.s.c.

≥ supa,b∈R

at+b≤j(t)∀t∈D(j)

(b+ a

∫Ω

fdµ

)

= j

(∫Ω

fdµ

)

Problem 4.10

4.10.1 Since j is convex and integration is linear, we have J is convex.

Problem 4.11

4.11.2 We want to show that∫|u+ v|α ≤

∫(|u|+ |v|)α ≤

∫|u|α +

∫|v|α

We claim that (|u|+ |v|)α ≤ |u|α + |v|α. Then it is enough to show that for x ≥ 0,

(x+ 1)α ≤ xα + 1 (*)

Let f(x) = (x+1)α−xα. Then f ′(x) = α((x+1)α−1−xα−1) < 0 implying that f is a decreasing functionof x ∈ [0,∞). Hence f(x) ≤ f(0) = 1⇒ (∗).Putting x = |u|

|v| , we get our required result.

4.11.2 Let

f =

∫|u|α g =

∫|v|α

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Analysis Subhadip Chowdhury Assignment 4

Then We have

[u]α + [v]α = f1α + g

1α = f

1α−1

∫|u|α + g

1α−1

∫|v|α

=

∫f

1α−1|u|α +

∫g

1α−1|v|α

=

∫f

(1−α)|u|α +

∫g

(1−α)|v|α

≤∫ (

f1α

)1−α(|u+ v|α) +

∫ (g

)1−α(|u+ v|α) since u, v ≥ 0

=

∫ (f

1α + g

)1−α(|u+ v|α)

⇒ ([u]α + [v]α)α =(f

1α + g

)α≤∫|u+ v|α = ([u+ v]α)α

⇒ [u]α + [v]α ≤ [u+ v]α

Thus Lα is a vector space but not a norm.

Problem 4.12

4.12.1 Since the inequality is homogeneous, we may divide by |b|p if necessary, to assume wlog

that b = 1. Let

f(a) =(|a|p + 1)1− p

2

(|a|p + 1− 2 |a+1|p

2p

) p2

|a− 1|p

be a function from a : |a+ 1| > 0 → R.

Problem 4.13

4.13.1

Case 1: |a+ b| ≥ |a| Then,

||a+ b| − |a| − |b|| ≤ ||a+ b| − |a||+ | − |b|| = |a+ b| − |a|+ |b| ≤ |b|+ |b| = 2|b|

Case 2: |a+ b| < |a| Then |a| − |a+ b| ≤ | − b| = |b| ⇒ ||a| − |a+ b|| ≤ |b|. Thus,

||a+ b| − |a| − |b|| = ||a| − |a+ b|+ |b|| ≤ |b|+ |b| = 2|b|

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Analysis Subhadip Chowdhury Assignment 4

4.13.2 Note that supn∫|fn| ≤M . Thus by Fatou’s lemma,

∫|f | ≤ lim

n→∞

∫|fn| ≤ ∞. So f ∈ L1(Ω).

Let a = fn − f, b = f . Consider the sequence ϕn = ||a+ b| − |a| − |b||. Then ϕn ≤ 2|b| = 2|f |.Also note that (|a+ b| − |a| − |b|)(x) = |fn(x)| − |f(x)|+ |fn(x)− f(x)| → 0.Then by dominated convergence theorem, ‖ϕn‖1 → 0. Now∫

(|fn| − |fn − f |)−∫|f | ≤

∫(||fn| − |fn − f | − |f ||) =

∫|ϕn| → 0

⇒∫

(|fn| − |fn − f |)→∫|f | (*)

4.13.3 Note that ‖fn‖ → ‖f | ⇒ ‖fn‖ < M for some constant M for all n. Then by part 2,

limn→∞

‖fn − f‖ = limn→∞

∫|fn − f |

by (∗)= lim

n→∞

∫|fn| −

∫|f | = lim

n→∞‖fn‖ − ‖f‖ = 0

Problem 4.14

4.14.1 Let us denote the measure by µ. See part 2.

4.14.2 Note that for a.e. x, fn(x)→ f(x)⇒ |fn(x)− f(x)| < α for all n bigger than a sufficiently

large N . Thus

x 6∈⋃k≥N

x : |fk(x)− f(x)| > α

for some N for a.e. x. Thus χSn(α)(x)→ 0 a.e.Also clearly |χSn(α)| < 1 ∈ L1(Ω) since µ(Ω) <∞. Thus by dominated convergence theorem,∫

|χSn(α)|n→∞−−−→ 0⇒ µ(Sn(α)) −−−→

n→∞0

Also µ[|fn − f | > α] ≤ µ(Sn(α))⇒ µ[|fn − f | > α] −−−→

n→∞0⇒ 4.14.1

4.14.3 Let α = 1m

for some fixed m ∈ N. Then by part 2,

µ(Sn(1

m))

n→∞−−−→ 0

Thus given δ > 0, there exists Nm ∈ N such that µ(Sn( 1m

)) < δ2m

for all n ≥ Nm.Now

x 6∈ SNm(1

m)⇒ |fk(x)− f(x)| ≤ 1

mfor all k ≥ Nm

Let Σm = SNm( 1m

). Let Σ =⋃m∈N

Σm. Then µ(Σ) < δ. Note that x 6∈ Σ⇒ x 6∈ Σm for all m.

Fix any ε > 0. Then there exists M ∈ N such that 1M< ε Then x ∈ Ω\Σ⇒ x 6∈ ΣM ⇒

|fk(x)− f(x)| ≤ 1

M< ε for all k ≥ NM

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Analysis Subhadip Chowdhury Assignment 4

i.e. fk → f uniformly on Ω\Σ. Taking A = Σ, we prove the theorem.

4.14.4 Fix an ε > 0. Then ∃δ > 0 such that

∫A

|fn|p < ε for all n and ∀A ⊆ Ω measurable with

µ(A) < δ.By part 3, there exists a certainAδ and an integerNδ > 0 such that µ(Aδ) < δ and |fn(x)−f(x)|p < ε

µ(Ω\Aδ)for all n ≥ Nδ for all x ∈ Ω\Aδ. So in particular,∫

|fn|p < ε and

∫Ω\Aδ|fn(x)− f(x)|p < ε

for all n ≥ Nδ. Then ∫|fn − f |p =

(∫Aδ

|fn − f |p +

∫Ω\Aδ|fn − f |p

)≤∫Aδ

|f |p +

∫Aδ

|fn|p + ε

= 2ε+

∫Aδ

|f |p

But |fn|p ∈ L1(Ω) and supn∫Aδ|fn|p < ∞ . Also |fn(x)|p → |f(x)|p for a.e. x ∈ Aδ. So by Fatou’s

lemma,∫Aδ|f |p ≤ lim inf

n→∞

∫Aδ|fn|p ≤ ε. Thus∫

|fn − f |p ≤ 3ε for all n ≥ Nδ

Hence ‖fn − f‖pn→∞−−−→ 0 i.e. fn → f in Lp(Ω).

Note that (fn − f) and fn both are in Lp(Ω), imply that f ∈ Lp(Ω).

Problem 4.15

4.15.1

4.15.1.(i) For x 6= 0,

limn→∞

ne−nx = limn→∞

n

enx= lim

n→∞

11n

+ x+ nx2/2 + n2x3/6 + . . .= 0

Thus fn → 0 a.e.

4.15.1.(ii)

‖fn‖1 =

∫ 1

0

|fn| =∫ 1

0

ne−nxdx =

∫ n

0

e−zdz = 1− e−n ≤ 1;∀n ∈ N

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Analysis Subhadip Chowdhury Assignment 4

4.15.1.(iii)‖fn − 0‖1 = 1− e−n 6→ 0

4.15.2

4.15.2.(i) For x 6= 0,

limn→∞

n1p e−nx = lim

n→∞

n1p

enx= lim

n→∞

1

n−1p + n1− 1

px+ n2− 1px2/2 + n3− 1

px3/6 + . . .= 0

Thus gn → 0 a.e.

4.15.2.(ii)

‖gn‖p =

∫ 1

0

|gn|p =

∫ 1

0

ne−pnxdx =1

p

∫ np

0

e−zdz =1

p(1− e−np) ≤ 1

p;∀n ∈ N

4.15.2.(iii)

‖gn − 0‖p =1

p(1− e−np) 6→ 0

4.15.2.(iv) Observe that∫ 1

0

gn(x)xmdx = n1p

∫ 1

0

e−nxxm =n

1p

nm+1

∫ n

0

e−yymdy ≤ n1p

nm+1Γ(m+ 1)

n→∞−−−→ 0

Thus for any polynomial ϕ(x) ∈ R[x], we have∫ 1

0

gnϕ→ 0

Since the polynomials are dense in Lp′(Ω), we have∫ 1

0

gnh→ 0

for all h ∈ Lp′(Ω) i.e. gn 0.

Problem 4.16

4.16.1 Let

Kn = conv (∪∞i=nfi)We claim that ∩∞n=1Kn = f. Indeed since fn’s are bounded, by Cantor’s intersection theorem,∩∞n=1Kn 6= ∅. if g ∈ ∩∞n=1Kn, then g(x) ∈ conv (∪∞i=nfi(x)) for all n i.e. fn(x) → g(x) a.e. Thusf(x) = g(x) a.e. So g ≡ f in Lp(Ω) and ∩∞n=1Kn = f. Then by exercise 3.13.2, fn f weakly inσ(Lp, Lp

′).

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Analysis Subhadip Chowdhury Assignment 4

4.16.2 By theorem 4.9 applied to the sequence (fn − f) in Lp, we get that there is a subsequence

(fnk − f) such that fnk − f → 0 a.e. Then by part 1, fnk f weakly. Indeed, we can say that everysubsequence of (fn− f) has a further subsubsequence (fnki − f)such that fnki f weakly. Thus fn fweakly.

4.16.3 fix an ε > 0. By Egorov’s theorem, ∃A ⊆ Ω such that |A| < ε and fn → f uniformly on

Ω\A. Now can write ∫|fn − f |q =

∫A

|fn − f |q︸ ︷︷ ︸I

+

∫Ω\A|fn − f |q︸ ︷︷ ︸J

Note that by uniform convexity for sufficiently large n,

J ≤ |Ω\A|εq

And by problem 4.2, we have

I1q ≤ |A|

1q− 1p‖fn − f‖p <⇒ I ≤ ε1−

qp‖fn − f‖qp

But ‖fn − f‖p is bounded by some constant for all n since fn, f ∈ Lp(Ω)⇒ fn − f ∈ Lp(Ω) and fn is

bounded in Lp. Thus I + J < C1εq + C2ε

1− qp for constants Ci not depending on p. So,∫

|fn − f |q → 0⇒ ‖fn − f‖q → 0

Problem 4.17

4.17.1 Note that the inequality is homogeneous of degree p. WLOG, let |a| ≥ |b|. Then we can

divide both sides by |a|p. Let t = b/a. Then it is enough to show that

||1 + t|p − 1− |t|p| ≤ C(|t|+ |t|p−1

)for some constant C depending on p for all |t| ≤ 1. Note that

limt→0+

||1 + t|p − 1− |t|p|(|t|+ |t|p−1)

= limt→0+

|p(1 + t)p−1 − p(t)p−1|(1 + (p− 1)tp−2)

= p

and

limt→0−

||1 + t|p − 1− |t|p|(|t|+ |t|p−1)

= limt→0−

|p(1 + t)p−1 − p(−t)p−1|(1 + (p− 1)(−t)p−2)

= p

Let

g(t) =

||1+t|p−1−|t|p|

(|t|+|t|p−1)t 6= 0

p t = 0

Then clearly g is a continuous function on [−1, 1]. So it is bounded above since [−1, 1] is compact. Thus

sup|t|≤1

||1 + t|p − 1− |t|p|(|t|+ |t|p−1)

< C

for some C depending only on p which proves the given inequality.

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Analysis Subhadip Chowdhury Assignment 4

4.17.2 Note that by problem 4.16.1, we have fn − f 0 in Lp. Also f ∈ Lp ⇒ fp ∈ L1 ⇒ fp−1pp ∈

Lpp−1 ⇒ fp−1 ∈ Lp′ . So ∫

|fn − f ||f |p−1 → 0

Also note that ∫ (|fn − f |p−1

) 1

1− 1p =

∫|fn − f |p <∞

So by 4.16.1, (fn − f)p−1 0 in Lp′. Thus∫

|fn − f |p−1|f | → 0

Put a = fn − f and b = f in question 1. Then∫||fn|p − |fn − f |p − |f |p| <

∫C(|fn − f |p−1|f |+ |fn − f ||f |p−1

)→ 0

The assertion in part 2 follows.

4.17.3 By part 2,

limn→∞

∫|fn − f |p = 0

So ‖fn − f‖p → 0.

Problem 4.18

4.18.1 Take any step function g ∈ L∞(0, 1) such that g =∑l

i=1 aiχAi , where ai ∈ R and Ai is an

interval (bi, bi+1) with⊔Ai = (0, 1). We have∫ 1

0

ung =

∫ 1

0

f(nx)g(x)dx

=1

n

∫ n

0

f(y)g(y/n)dy

=1

n

(l∑

i=1

ainbi+1 − nbi

T

∫ T

0

f(y)dy

)+

1

n

l∑i=1

ai

∫ nbi+1

nb bi+1−biT

cT+nbi

f(y)dy

(f

l∑i=1

ai

∫ bi+1

bi

dy

)+

1

n

l∑i=1

ai

∫ nbi+1

nb bi+1−biT

cT+nbi

|f(y)|dy

≤∫ 1

0

fg +1

n

l∑i=1

ai‖f‖1,[0,T ] [since f is periodic]

→∫ 1

0

fg

since f ∈ Lploc means the second term is of the form bounded termn

; and hence it goes to 0. Since functions

of above form are dense in Lp′(0, 1), we get that un f weakly.

11

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Analysis Subhadip Chowdhury Assignment 4

4.18.2 Let kT ≤ n ≤ (k + 1)T for k ∈ Z.∫ 1

0

|un − f |p =

∫ 1

0

∣∣f(nx)− f∣∣p dx

=1

n

∫ n

0

∣∣f(y)− f∣∣p dy

=1

n

(k

∫ T

0

∣∣f(y)− f∣∣p dy +

∫ n

kT

∣∣f(y)− f∣∣p dy)

≤ 1

T

∫ T

0

∣∣f(y)− f∣∣p dy +

1

n‖f − f‖pp,[0,T ]

⇒∫ 1

0

|un − f |p −1

T

∫ T

0

∣∣f(y)− f∣∣p dy → 0

So

limn→∞

‖un − f‖p =

(1

T

∫ T

0

∣∣f(y)− f∣∣p dy) 1

p

4.18.3

4.18.3.(i) un(x) = sin(nx)⇒ un ∈ Lp(0, 1) for all 1 ≤ p ≤ ∞.

By part 1, un 1

∫ 2π

0sin ydy = 0 weakly in σ(Lp, Lp

′).

Also ‖un − sin‖p →(

12π

∫ 2π

0|sin y|p dy

) 1p

4.18.3.(ii)∫ 1

0|un|pdx = n

2(αp+βp) <∞ for all 1 ≤ p <∞. Hence un ∈ Lp(0, 1) for all 1 ≤ p <∞.

Also sup |un| = maxα, β. Hence un ∈ Lp(0, 1) for all 1 ≤ p ≤ ∞.un

α+β2

weakly in σ(Lp, Lp′).

Also ‖un − f‖p →(∫ 1

0|f |p dy

) 1p

=(|α|p+|β|p

2

) 1p

Problem 4.19

4.19.1 Since Lp is uniformly convex for 1 < p <∞, we have by prop. 3.32, fn → f strongly in Lp.

4.19.2 Consider fn(x) = sin(2πnx) + 1. Then given g ∈ L∞(0, 1), we have∫ 1

0

sin(2πnx)g(x)dx+

∫ 1

0

g =1

n

∫ n

0

sin(2πy)g(y)dy +

∫ 1

0

g ≤ 1

n‖g‖∞ ×+

∫ 1

0

g →∫ 1

0

g

So, fn 1.

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Analysis Subhadip Chowdhury Assignment 4

Problem 4.20

4.20.1 Clearly by the given inequality, f ∈ Lp ⇒ Af ∈ Lq. Now take any sequence un → u

strongly in Lp. Then by theorem 4.9, we can get a subsequence unk and a function h ∈ Lp such thatunk(x)→ u(x) a.e. and |unk(x)| ≤ h(x) for all k, a.e. Then for some h′ ∈ Lq, we have Aunk(x)→ Au(x)a.e. and |Aunk(x)| ≤ h′(x) for all k, a.e. Thus by dominated convergence theorem applied to (Aunk)

q weget that ‖Aunk −Au‖q → 0 By uniqueness of limit we then get that Aun → Au strongly in Lq. Hence Ais continuous from Lp strong to Lq strong.

4.20.2 From 4.18.3(ii), we know that un(x) = f(nx) α+β2

where f is defined as in the problem

4.18.3(ii). Now by given hypothesis Aun → Au; i.e. a(un) a(α+β2

) weakly in σ(Lq, Lq′). Now

a(un(x)) = a(f(nx)) =

a(α) k

n< x <

k+ 12

n

a(β)k+ 1

2

n< x < k+1

n

Since 1 ∈ Lq′(0, 1), integrating Aun and Au against 1, we get

n1

2n(a(α) + a(β)) = a(

α + β

2)

i.e1

2(a(α) + a(β)) = a(

α + β

2)

for all α, β ∈ R. Clearly the only solution to the functional equation is a(x) = (a(1) − a(0))x + a(0)which is an affine function.

Problem 4.21

4.21.1 Take g = χ(a,b). Then∫Rung =

∫ b

a

u0(x+ n)dx =

∫ b+n

a+n

u0(x)dx =

∫ b+n

0

u0(x)dx−∫ a+n

0

u0(x)dx→ 0

as n → ∞. Thus for any step functions(with compact support) g, we have∫ung → 0. Since step

functions(with compact support) is dense in Lp(R). we get that un 0 weakly.

4.21.2 Take g = χI . Let I = (a, b). Fix a δ > 0. Let E = [|u0| > δ]. Then∫Rung =

∫I

u0(x+ n)dx =

∫I+n

u0(y)dy =

∫(I+n)∩E

u0(y)dy +

∫(I+n)∩Ec

u0(y)dy

Now since |E| <∞, there exists N such that for n > N , we have |(I + n) ∩ E| < δ. Then∣∣∣∣∫Rung

∣∣∣∣ ≤ ‖u0‖∞δ + δ|(I + n) ∩ Ec| ≤ ‖u0‖∞δ + δ|I|

Taking δ → 0, we get that∣∣∫

R ung∣∣→ 0. Thus for step function g with compact support,

∣∣∫R ung

∣∣→ 0.

Since step functions with compact support are dense in L1(R), we get that un∗ 0.

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Analysis Subhadip Chowdhury Assignment 4

4.21.3 Suppose, by contradiction, there is a subsequence unk which weakly converge. Then for any

g ∈ L∞,∫unkg → 0. Take g ∈ L∞ defined by

g(x) =

(−1)i ni < x < ni+1

0 o.w.

Then∫unkg = (−1)k which do not converge as k →∞. Thus we have a contradiction! Hence such a

subsequence does not exist.

Problem 4.22

4.22.1 A⇒ B Since χE ∈ Lp′

for any measurable E with |E| <∞, this implication is clear by

definition of weak convergence. [or weak* convergence in case p =∞.]

B ⇒ A Note that, since simple functions are dense in Lp′(Ω), the vector space spanned by characteristic

functions χA of subsets A of Ω with |A| <∞ are dense in Lp′(Ω) for 1 ≤ p′ <∞. Now for any element

g of the vector space, we have∫

Ωfng →

∫Ωfg. Thus

∫Ωfnh→

∫Ωfh for all h ∈ Lp′(Ω) i.e. fn f .

4.22.2 Again the implication A⇒ B is trivial. For the other direction observe that every simple

function belongs to L∞(Ω) if |Ω| <∞. Hence the vector space spanned by characteristic functions χA ofsubsets A of Ω with |A| <∞ are dense in L∞(Ω). Thus A follows by the same argument as in part 1.

4.22.3 By proposition 3.5, and definition of weak convergence, A⇒ B.

To see that the opposite implication is not true consider Ω = R. Consider the sequence fn = χ(n,n+1).then clearly ‖fn‖1 = 1 <∞ and

∫Efn → 0 =

∫E

0 for any E ⊆ R with |E| <∞. But clearly by problem4.21.3 fn does not weakly converge.

4.22.4 Given ε > 0, we can find a subset ω ⊆ Ω such that∫ωcf < ε and |ω| <∞.

Now ∫ωcfn +

∫ω

fn =

∫Ω

fn →∫

Ω

f =

∫ωcf +

∫ω

f

⇒∫ωcfn −

∫ωcf →

∫ω

f −∫ω

fn → 0 (*)

since |ω| <∞. Also we have, for any F ⊆ Ω with |F | ≤ ∞,∫F

fn =

∫F∩ω

fn +

∫F∩ωc

fn →∫F∩ω

f +

∫F∩ωc

fn

⇒∫F

(fn − f)→∫F∩ωc

(fn − f) ≤∫ωcfn +

∫ωcf → 2

∫ωcf < 2ε

since fn, f ≥ 0. Thus we get∫F

(fn − f)→ 0 for F measurable subset of Ω, |F | ≤ ∞. Since vector spacespanned by χF for such F ’s is dense in L∞(Ω), we get that fn f .

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Analysis Subhadip Chowdhury Assignment 4

Problem 4.23

4.23.1 Clearly by definition C is convex. Suppose (un) is a sequence in C which converge to u in

Lp. Then there is a subsequence (unk) such that unk(x)→ u(x) a.e.[since 1 ≤ p <∞] Thus u(x) ≥ f(x)a.e. Hence C is closed.Now C is convex and strongly closed ⇒ C is weakly closed.

4.23.2 Assume first that f ∈ L∞(Ω). Call the set mentioned in part 2 C ′. Then clearly, C ⊆ C ′.

Now suppose, u ∈ C ′. Thus∫fϕ ≥

∫uϕ ∀ϕ ∈ L1(Ω) with fϕ ∈ L1(Ω) and ϕ ≥ 0 a.e. Since Ω is

σ−finite we can find Ωn with ∪Ωn = Ω and |Ωn| <∞. Let Ω′n = Ωn ∩ |f | < n. Then⋃

Ω′n = Ω. LetA = f < u. Then choose ϕ = χA∩Ωn∫

A∩Ω′n

|f − u| ≤ 0⇒ |A ∩ Ω′n| = 0⇒ |A| = 0

So f ≥ ua.e. implying C ′ ⊆ C.

4.23.3 The second definition C clearly shows that C is closed in σ(L∞, L1) since it is the intersection

of all closed sets in σ(L∞, L1) of the formu ∈ L∞(ω)

∣∣∣∣∫ fϕ ≥∫uϕ and fϕ ∈ L1(Ω)

for a fixed 0 ≤ ϕ ∈ L1(Ω).

4.23.4 Note that a weak* closed and bounded set is compact in weak* topology, since we can scale

down the bounded set into a closed subset of the unit ball in weak* topology which is compact. Hencebeing a closed subset of a compact set, the scaled down set will be compact giving us that the originalset is compact. Now given set C is a weak*(σ((L1)∗, L1))closed and bounded set; hence it is compact.

Problem 4.24

4.24.1 Given ϕ ∈ L1(RN), we have by prop. 4.16,∫ϕ(ρn ? ζnu) =

∫ζnu(ρn ? ϕ)

Hence, ∫vnϕ−

∫vϕ =

∫ζnu(ρn ? ϕ)−

∫ζϕu

=

∫ζnu(ρn ? ϕ− ϕ) +

∫uϕ(ζn − ζ)

≤ ‖ζn‖∞‖u‖∞‖ρn ? ϕ− ϕ‖1 + ‖u‖∞‖(ζn − ζ)ϕ‖1

≤ ‖u‖∞‖ρn ? ϕ− ϕ‖1 + ‖u‖∞‖(ζn − ζ)ϕ‖1

→ 0

first term by thm 4.22. and second term by DCT.

15

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Analysis Subhadip Chowdhury Assignment 4

4.24.2 Let B = B(x0, R) and let χ := χB(x0,R+1). Then let wn = ρn ? (ζnuχ). Note that wn = vnon B since by proposition 4.18,

supp(wn − vn) ⊆ B(0,1

n) +B(x0, R + 1)c)

Thus ∫B

|vn − v| =∫B

|wn − χv|

≤∫RN|ρn ? (ζnuχ)− χζu|

≤∫RN|ρn ? (ζnuχ)− ρn ? (χζu)|+

∫RN|ρn ? (χζu)− χζu|

≤∫RN|(ζnuχ)− (χζu)|+

∫RN|ρn ? (χζu)− χζu|

→ 0

Problem 4.25

4.25.1 Let

u′(x) =

u(x) x ∈ Ω

0 x ∈ Ωc

Then u′ ∈ L∞(RN). LetΩn = x ∈ Ω|dist(x,Ωc) > 2/n; |x| < n

Defineζn := χΩn , ζ := χΩ

Then note that ζn → ζ a.e. on RN and ‖ζn‖∞ ≤ 1 for all n. Define vn = ρn ? (ζnu′) and v = ζu′ = u′.

Then by problem 4.24, vn∗ v in L∞(RN). Thus vn

∗ u = u′ on L∞(Ω). Also vn ∈ C∞c (Ω). and∫

B|vn − v| =

∫B|vn − u′| → 0 for every ball B. Thus we may find an subsequence (vnk) which converges

a.e. to u′(x) on B. By a diagonal process, we can find a further subsequence (vnki ) which converges a.e.

to u′ on RN . Renaming this subsequence (un), we see that un satisfies all given properties.

4.25.2 By our construction u ≥ 0 a.e. ⇒ ρn ? (ζnu′) ≥ 0 a.e.

4.25.3 By part 1, C∞c (Ω) is dense in L∞(Ω).

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Analysis Subhadip Chowdhury Assignment 4

Problem 4.26

4.26.1 If f ∈ L1(Ω) then A ≤ sup‖ϕ‖∞≤1

‖f‖1‖ϕ‖∞ ≤ ‖f‖1 <∞.

Conversely, if A <∞, Then for all ϕ ∈ Cc(Ω), we have∫f

ϕ

‖ϕ‖∞≤ A⇒

∫fϕ ≤ A‖ϕ‖∞

Fix any compact subset K of Ω. Let ψ ∈ Cc(Ω) be a function such that 0 ≤ ψ ≤ 1 and ψ ≡ 1 on K. Letu ∈ L∞(Ω) be any function. Then by problem 4.25 , we can find a sequence (un) in Cc(Ω) such that

‖un‖∞ ≤ ‖u‖∞ and un → u a.e. on Ω and un∗ u in σ(L∞, L1). Then taking ϕ = unψ we get∫

fψun ≤ A‖ψun‖∞ ≤ A‖u‖∞

By DCT, taking limit as n→∞, we then have∫fψu ≤ A‖u‖∞

Choose u = sign(f). Then ∫K

|f |.1 ≤∫f.u.ψ ≤ A

for all compact subset K of Ω. Thus ‖f‖1 ≤ A <∞ i.e. f ∈ L1(Ω).

In case f ∈ L1(Ω), by above proof we have A ≤ ‖f‖1 ≤ A⇒ ‖f‖1 = A.

4.26.2 If f+ ∈ L1(Ω) then B ≤ sup‖ϕ‖∞≤1ϕ≥0

∫(maxf, 0ϕ) ≤ sup

‖ϕ‖∞≤1ϕ≥0

‖f+‖1‖ϕ‖∞ ≤ ‖f+‖1 <∞.

Conversely if B <∞, we can proceed as in part 1, to get that∫fψu ≤ B‖u‖∞

for all u ∈ L∞(Ω) Put u = χf≥0∩K . Then∫Kf+ ≤ B for every compact subset K of Ω. Hence

‖f+‖1 ≤ B.

The proof clearly shows that ‖f+‖1 = B if f ∈ L1(Ω).

4.26.3 Note that the sequence constructed in problem 4.25 is actually in C∞c (Ω). Hence the proof

is same.

4.26.4 [∫fϕ = 0∀ϕ ∈ C∞c (Ω)

]⇒ A = ‖f‖1 = 0⇒ f = 0 a.e.[∫

fϕ ≥ 0∀ϕ ∈ C∞c (Ω), ϕ ≥ 0

]⇒ B = ‖f+‖1 ≥ 0⇒ f > 0 a.e.

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Analysis Subhadip Chowdhury Assignment 4

Problem 4.27

Fix a ϕ0 ∈ C∞c (Ω) such that∫uϕ0 = 0. Fix any ε > 0. Then∫

u(ϕ− (ϕ0

∫(uϕ)) + εϕ0) = 0 + ε > 0⇒

∫v(ϕ− (ϕ0

∫(uϕ)) + εϕ0) ≥ 0

Taking ε→ 0, we then have ∫vϕ ≥

∫uϕ

∫vϕ0 = λ

∫uϕ

Conversely, taking ε < 0 and replacing ϕ − (ϕ0

∫(uϕ)) + εϕ0 by ϕ − (ϕ0

∫(uϕ)) − εϕ0 we get that as

ε→ 0, ∫vϕ ≤

∫uϕ

∫vϕ0 = λ

∫uϕ

. Thus ∫(v − λu)ϕ = 0

for all ϕ ∈ C∞c (Ω). Hence v = λu a.e. Clearly by construction λ ≥ 0.

Problem 4.28

We can approximate ρ by a sequence of C∞c (Ω) functions (ψi) such that ‖ρ − ψi‖1 → 0. Defineψin(x) = nNψi(nx). Note that

∫ψin =

∫RN n

Nψi(nx)dx1 . . . dxN =∫RN ψ

i(y)dy1 . . . dyN →∫ρ = 1. Fix

K ⊆ RN a compact set and ε > 0. Then there exist n large enough such that suppψin ⊆ B(0, δ) where δis such that |f(x− y)− f(y)| < ε for all x ∈ K and for all y ∈ B(0, δ). Then for n sufficiently large,∫

|(ψin ? f)(x)− f(x)|pdx =

∫ ∣∣∣∣∫ |f(x− y)− f(x)|ψin(y)dy

∣∣∣∣p dx≤∫ ∫

|f(x− y)− f(x)|pψin(y)dydx

=

∫ψin(y)dy

(∫|f(x− y)− f(x)|pdx

)≤∫B(0,δ)

εpψi ≤ εp∫ψi → εp

Hence taking ε→ 0, we get that ψin ? fn→∞−−−→ f in Lp. Taking limit i→∞, we then have ρn ? f → f in

Lp.

Problem 4.29

Let χn := χK+B(0, 12n

) and take un = ρ2n ? χn where ρ(x) =

e

1|x|2−1 |x| < 1

0 |x| > 1and ρn(x) = 1∫

ρnNρ(nx).

Then all it remains to check is that un satisfies (a), (b), (c) and (d). Note that

‖un‖∞ ≤ ‖χn‖∞‖ρ2n‖1 = 1

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Analysis Subhadip Chowdhury Assignment 4

So (a) is true. Next on K,

ρ2n ? χn =

∫ρ2n = 1

So (b) holds. Also

supp(un) ⊆ K +B(0,1

2n) +B(0,

1

2n) ⊆ K +B(0,

1

n)

So (c) holds. Lastly,|Dαun(x)| = |Dαρ2n(x) ? χn| ≤ ‖Dαρ2n‖1

NowDα(CnNρ(2nx)) = CnN(2n)|α|(Dαρ)(2nx)

So

‖Dαρ2n‖1 = C(2n)|α|∫|(Dαρ)(2nx)|nNdx = C(2n)|α|

∫|(Dαρ)(2y)|dy = Cαn

|α|

So (d) follows.

Problem 4.30

4.30.2 Let 1p

+ 1p′

= 1 = 1q

+ 1q′

. Then 1r

= 1 − 1p′− 1

q′. Then set α = p

q′and β = q

p′. Note that

(1 − α)r = (1p− 1

q′)rp = 1

rrp = p, and similarly (1 − β)r = q. Then ϕ1(y) = f(x − y)α ∈ Lq′ , ϕ2(y) =

g(y)β ∈ Lp′ . Also for ϕ3(y) = f(x− y)1−αg(y)1−β, we have ϕr3(y) = f(x− y)pg(y)q ∈ L1 ⇒ ϕ3 ∈ Lr. Wehave,

|(f ? g)(x)| =∣∣∣∣∫ f(x− y)g(y)dy

∣∣∣∣ ≤ ∫ |f(x− y)g(y)| dy

=

∫|f(x− y)|α|g(y)|β

(|f(x− y)|1−α|g(y)|1−β

)By Holder’s inequality y 7→ f(x− y)g(y) is integrable. In fact,

|(f ? g)(x)| ≤ ‖ϕ1‖q′‖ϕ2‖p′‖ϕ3‖r = ‖f‖αp‖g‖βq ‖ϕ2‖r

⇒∫|f ? g|r ≤ ‖f‖αrp ‖g‖βrq

∫|f(x− y)(1−α)rg(y)(1−β)r|dy

= ‖f‖αrp ‖g‖βrq∫|f(x− y)pg(y)q|dy

≤ ‖f‖αrp ‖g‖βrq ‖f‖pp‖g‖qq= ‖f‖rp‖g‖rq

⇒ ‖f ? g‖r ≤ ‖f‖p‖g‖q

4.30.3 By previous part f ? g ∈ L∞(RN ) if r =∞. Also we know that convolution of a Lp function

and a Lp′

function is continuous. Hence f ? g ∈ C(RN) ∩ L∞(RN).

For 1 < p, q <∞, note that we can construct sequences (fn), (gn) in Cc(Ω) so that fn → f in Lp andgn → g in Lq. But clearly, supp(fn ? gn) ⊆ supp(fn) + supp(gn) ⇒ (fn ? gn)(x) = 0 as |x| → ∞. But‖fn ? gn − f ? g‖1 → 0. Hence (f ? g)(x)→ 0 as |x| → ∞.

19

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Analysis Subhadip Chowdhury Assignment 4

Problem 4.31

4.31.1 Let χr :=χB(0,r)

|B(0,r)| . Then

fr(x) =

∫χr(x− y)f(y)dy = (χr ? f)(x)

Now f ∈ Lp and χr ∈ L1. Hence f ? χr ∈ Lp. Also χr ∈ Lp′ ⇒ f ? χr ∈ C(RN) by problem 4.30. Hence

fr ∈ C(RN) ∩ Lp(RN). Also by 4.30, fr(x)→ 0 as |x| → ∞ for 1 < p <∞.

4.31.2 Note that∫χr = 1, and supp(χr) = B(0, r) Hebce χ 1

nis a sequence of mollifiers. Hence

f1/n → f in Lp as n→∞⇒ fr → f in Lp as r → 0.

Problem 4.32

4.32.1 By definition of convolution it is immediate that f ? g = g ? f . Next note that

(f ? g) ? h(u) =

∫(f ? g)(x)h(u− x)dx

=

∫ (∫f(y)g(x− y)dy

)h(u− x)dx

=

∫ ∫f(y)g(x− y)h(u− x)dydx

=

∫ ∫f(y)g(x− y)h(u− x)dxdy

=

∫f(y)

(∫g(x− y)h(u− x)dx

)dy

=

∫f(y)

(∫g(x+ y − y)h(u− x− y)dx

)dy

=

∫f(y)(g ? h)(u− y)dy

= (f ? (g ? h))(u)

Thus (f ? g) ? h = f ? (g ? h).

Problem 4.33

4.33.1 Note that compactly supported continuous functions are uniformly continuous. Hence

given ε > 0, there exist δ > 0 such that |h| < δ implies |ϕ(h + x) − ϕ(x)| < ε for any x ∈ R. Thusin fact |ϕ(x + n + h) − ϕ(x + n)| < ε for any x ∈ R and any n ∈ N. So for |h| < δ, we have‖τhϕn − ϕn‖p < ε.|supp(ϕ)|1/p for any n. Hence ∀ε > 0;∃δ > 0 such that ‖τhf − f‖p < ε ∀f ∈ F and∀h ∈ R with |h| < δ.

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Analysis Subhadip Chowdhury Assignment 4

4.33.2 Fix ε = 13. Consider the open balls B(f, ε) ⊆ Lp(R) for each f ∈ F . Then clearly they

form an open cover of the closure of F . If the closure is compact, this cover has a finite subcover.

Let it bek⋃i=1

B(ϕni , ε). Take N >> maxn1 . . . , nk so that supp(ϕn)⋂( k⋃

i=1

supp(ϕni)

)= ∅. Then

‖ϕN − ϕni‖p =(2∫ϕp) 1p = C, some constant. Hence if ε < C/2, ϕN 6∈

k⋃i=1

B(ϕni , ε). Contradiction!

Hence the closure is not compact.

Problem 4.34

Fix an ε > 0. Then there is a finite covering of F by open ε−balls. Let F ⊆n⋃i=1

B(fi, ε).

4.34.1 Given any f ∈ F , f ∈ B(fi, ε) for some i ∈ 1, . . . , n. Hence ‖f‖p ≤ ‖fi‖p+ε ≤ max‖fi‖p :

1 ≤ i ≤ n+ ε. Hence F is bounded.

4.34.2 By lemma 4.3, for each i ∈ 1, . . . , n;

limh→0‖τhfi − fi‖p = 0

Hence for the same ε as above, there exists δi > 0 such that

‖τhfi − fi‖p < ε ∀i and ∀h ∈ RN with |h| < δi

Put δ = minδi : 1 ≤ i ≤ n. Take any f ∈ F , suppose f ∈ B(fk, ε). Then

‖τhf − f‖p ≤ ‖τhf − τhfk‖p + ‖τhfk − fk‖p + ‖fk − f‖p = 2‖fk − f‖p + ‖τhfk − fk‖p ≤ 3ε

for all |h| < δ.

4.34.3 Note that fi ∈ Lp(RN)⇒ ∃Ωi ⊂ RN bounded,open, such that ‖fi‖Lp(RN )\Ωi < ε where ε is

as above. Let Ω =n⋃i=1

Ωi. Take any f ∈ F and let f ∈ B(fk, ε). Then

‖f‖Lp(RN\Ω) ≤ ‖fk‖Lp(RN\Ω) + ‖f − fk‖Lp(RN\Ω) ≤ ‖fk‖Lp(RN\Ωk) + ‖f − fk‖Lp(RN ) = 2ε

Comparing with corollary 4.27, we find that this problem is the converse of the corollary, and togetherthey characterise all compact sets in Lp(RN).

Problem 4.35

Clearly F|Ω is bounded in Lp(Ω). Now take f ∈ F and write f = G ? u for an u ∈ B. Then

‖τhf − f‖p = ‖(τhG−G) ? u‖p ≤ C‖τhG−G‖p → 0

by lemma 4.3. Hence by theorem 4.26, F|Ω has compact closure in Lp(Ω) for any measurable Ω withfinite measure.

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Analysis Subhadip Chowdhury Assignment 4

Problem 4.36

4.36.1 [(d) + (e)]⇒ [(a) + (b) + (c)]

Suppose (d) and (e) hold. Then given ε > 0, there exists N > 0 such that

supf∈F

∫[|f |>t]

|f | < ε ∀t ≥ N

⇒∫

[|f |>t]|f | < ε ∀f ∈ F , ∀t ≥ N

Now given E measurable with |E| < εN

, we have∫E

|f | =∫E∩[|f |>N ]

|f |+∫E∩[|f |≤N ]

|f | ≤∫

[|f |>N ]

|f |+N |E| ≤ 2ε

Thus given ε > 0, there exists δ > 0 such that (b) holds.

We also know that given ε > 0, there exists N ∈ N such that

supf∈F

∫Ω\Ωn

|f | < ε ∀n ≥ N

⇒∫

Ω\Ωn|f | < ε ∀f ∈ F ∀n ≥ N

So we may take ω = ΩN and thus (c) holds.

Clearly (d) implies (a) is true.

[(a) + (b) + (c)]⇒ [(d) + (e)]

Suppose (a), (b) and (c) hold. Then (a) implies there is an M ∈ N such that∫|f | < M . Fix an ε > 0.

Then (b) gives us an δ > 0. Let E = [|f | > Mδ

] Then |E| < δ since otherwise,∫E|f | > M >

∫|f | which

is not possible. Thus

(b)⇒ supf∈F

∫[|f |>M

δ]

|f | < ε

Thus given ε > 0, there exists a N > 0 such that supf∈F

∫[|f |>t] |f | < ε for all t > N implying

limt→∞

supf∈F

∫[|f |>t]

|f | = 0

By (c) we can produce an ω satisfying properties in (c). Then note that since |ω| <∞, and Ωn arenondecreasing with union the whole of Ω; we can find a N ∈ N such that |ω∆Ωn| < ε/M for all n > N .Then ∫

Ω\Ωn|f | ≤

∫Ω\ω|f |+

∫Ωn∆ω

|f | ≤∫

Ω\ω|f |+ ε ≤ 2ε ∀f ∈ F

Thus we have (e) i.e.

limn→∞

supf∈F

∫Ω\Ωn

|f | = 0

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Analysis Subhadip Chowdhury Assignment 4

Problem 4.37

4.37.1 We have∫I

un(x)ϕ(x)dx =

∫I

nf(nx)ϕ(x)dx

=

∫ n

−nf(t)ϕ(t/n)dt

=

∫ n

−nf(t) (ϕ(t/n)− ϕ(0)) dt+ ϕ(0)

∫ n

−nf(t)dt

But the first term converges to 0 by Lebesgue’s theorem and the second term goes to 0 by given condition:∫ +∞

−∞f = 0

4.37.2 ∫I

|un(x)|dx =

∫ n

−n|f(t)|dt ≤

∫ +∞

−∞|f | <∞

since f ∈ L1(R). Hence (un) is bounded.

Note that for all δ > 0, ∫ δ

0

|un| =∫ nδ

0

|f | →∫ ∞

0

|f | > 0

Hence there exist∫∞

0|f | > ε > 0 such that for all δ > 0, there exists E = (0, δ) such that

∫E|un| > ε for

sufficiently large n. Hence no subsequence of (un) is equi-integrable.

4.37.3 Suppose there exists such a function u ∈ L1(I). Then by part 1,∫I

uϕ = 0 ∀ϕ ∈ C([−1,+1])

Then by corollary 4.24, u ≡ 0 a.e. But for ϕ = χ(0,1), we have∫unϕ =

∫ n

0

f →∫ ∞

0

f > 0

Contradiction! Hence u does not exist.

4.37.4 By Dunford-Pettis thm, we have

[ No subsequence of (un) is equi-integrable]⇒ [no subseqence of (un) weakly converges]

which agrees with the fact that part 2 and part 3 hold simultaneously. The thm states that part 3 is adirect implication of part 2.

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Analysis Subhadip Chowdhury Assignment 4

4.37.5 ∫I

|un(x)|dx =

∫[−n−

12 ,n−

12 ]

|un(x)|dx+

∫[n−

12<|x|<1]

|un(x)|dx

=

∫ n12

−n12

|f(t)|dt+

∫[n

12<|t|<n]

|f(t)|dt

→ 0 + 0

since f ∈ L1. Thus un → 0 in L1. Hence there is a subsequence (unk) such that unk(x)→ 0 a.e.

Problem 4.38

4.38.1 By definition |supp(un)| =∑n−1

0 ( 1n2 ) = 1

n. Also∫

I

un =n−1∑j=0

∫ jn

+ 1n2

jn

n =n−1∑j=0

n

n2= 1⇒ ‖un‖1 = 1

4.38.2 Suppose ϕ ∈ C1([0, 1]). Then

limn→∞

∫I

unϕ−∫I

ϕ = limn→∞

n−1∑j=0

n

∫ jn

+ 1n2

jn

ϕ− limn→∞

n−1∑j=0

1

nϕ(xj) where xj ∈ (j/n, (j + 1)/n)

= limn→∞

n−1∑j=0

1

n

∫ nj+1

nj

(t

n2

)− ϕ

( yjn2

))dt where yj ∈ (nj, n(j + 1))

= limn→∞

n−1∑j=0

1

n3

∫ nj+1

nj

((t− yj)ϕ′(zj)) dt where zj ∈ (t, yj)

≤ limn→∞

n−1∑j=0

1

n3

∫ nj+1

nj

(|n(j + 1)− nj|ϕ′(zj)) dt

= limn→∞

O(n−1∑j=0

1

n3n)

= limn→∞

O(1

n)

= 0

But C1([0, 1]) is dense in C([0, 1]). Hence limn→∞

∫Iunϕ =

∫Iϕ for all ϕ ∈ C([0, 1]).

4.38.3 The sequence (un) is not equi-integrable because for En = supp(un); |En| → 0 but

1 =

∫I

un =

∫En

|un|

So (b) does not hold in definition of equi-integrability.

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Analysis Subhadip Chowdhury Assignment 4

4.38.4 If (unk) u, we have by part 2 and corollary 4.24, u ≡ 1 a.e. Choose a further subsequence

(un′k) such that∑

k |supp(un′k)| < 1. Let ϕ = χA where

A = I\

(⋃k

supp(un′k)

)

so that |A| > 0. We have∫Iun′kϕ = 0 for all k and hence 0 =

∫Iϕ = |A|. Contradiction!

4.38.5 Consider a subsequence (unk) such that∑k

|supp(unk)| <∞

Let Bk =⋃j≥k(supp(unj)) and B =

⋃k Bk. Clearly |Bk| =

∑∞j=k

1nj→ 0 as k →∞, and thus |B| = 0.

If x 6∈ B, there exists some k0 such that unk(x) = 0 for all k ≥ k0. Thus unk(x)→ 0 a.e. as k →∞.

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