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### Transcript of Analysis II Home Assignment 4 Subhadip subhadip/coursedocuments/Homework_shared...  Analysis II...

• Analysis IIHome Assignment 4

Problem 4.1

f Lp(RN) RN

(1 + |x|)1(1 + | log |x||)1p

• 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

f1f2p =(|f1f2|p

) 1p

((|f1|p1

) pp1

.

(|f2|p.

p1p1p

)1 pp1

) 1p

= f1p1 .f2 pp1p1p

Note that 1p

= 1p1

+ 1p2 1

p2= p1p

pp1 pp1

p1p = p2.

Thus f2 pp1p1p

= f2p2 f1f2p f1p1f2p2

Suppose for k = m N, we have gp m

i=1 gipi whenever gi Lpi and1p

=m

i=11pi

.

Then let 1q

=m+1

i=11qi 1 and fi Lqi . We have, 1q =

1qm+1

+m

i=11qi

. Then by induction hypothesis,

(f1f2 . . . fm) L1m

i=11qi

and

fq fm+1qm+1 .f1f2 . . . fm 1mi=1

1qi

fm+1qm+1 .mi=1

fiqi =m+1i=1

fipi

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

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

1 . Then note that f Lp f L p . Similarly,

f 1 Lq

1 . So applying 4.4.1 with p1, p2 defined as above and noting that1r

= 1p1

+ 1p2 1, We have

fr fp1 .f 1p2 =(|f|p/

)p

.

(|f 1|

q1

) 1q

= fpf1q

For the cases = 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(). We want toprove that f C.Now fn f in Lp() 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

(b) |fnk(x)| h(x) k, a.e. on

Then applying Fatous 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 havefn fr fn fpfn f1q

where is defined as in 4.4.2. Note that by part 2, we have fq C. Thus fn fq 2C. Hence

fn fr fn fp (2C)1 0

SO fn f in Lr().

Problem 4.6

4.6.1 Taking p in problem 4.2, we get that

lim supp

fp f (1)

We know that there exist a sequence Cn f such that |f(x)| Cn a.e on for all n. Then letn = {x : |f(x)| > Cn}. We have

fpp

n

|f |p (n)(Cn)p fp Cn(())1/p

Taking p, we then have lim infp

fp Cn for all n. Taking n we then have

lim infp

fp f (2)

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

4.6.2 Fix a k > C. Consider the set S = {x : |f(x)| > k}. Suppose (S) = > 0. Then

C fp k1/p

Taing p we then have C k,contradiction!! So (S) = 0 proving that f C i.e. f L().

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

• Analysis Subhadip Chowdhury Assignment 4

> 0 such that for 0 < x < , | lnx|p < 1x12

. Then 10

| lnx|pdx <

0

1xdx+

1

| lnx|pdx

=12

12

+ a finite number

0 such that x0 +B(0, r) XN L1+

1n . But x0 L1+

1n .

Hence B(0, r) L1+ 1n B(0, r) L1+ 1n for all R X

B(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

fp Cf1; f X

Problem 4.9

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

j(f)d =

supa,bR

at+bj(t)tD(j)

(af + b)d since j is convex l.s.c.

supa,bR

at+bj(t)tD(j)

(b+ a

fd

)

= j

(

fd

)

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|

(|u|+ |v|)

|u| +

|v|

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

(x+ 1) x + 1 (*)

Let f(x) = (x+1)x. Then f (x) = ((x+1)1x1) < 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| g =

|v|

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• Analysis Subhadip Chowdhury Assignment 4

Then We have

[u] + [v] = f1 + g

1 = f

11|u| + g

11|v|

=

f

11|u| +

g

11|v|

=

f

1

(1)|u| +g

1

(1)|v|

(

f1

)1(|u+ v|) +

(g

1

)1(|u+ v|) since u, v 0

=

(f

1 + g

1

)1(|u+ v|)

([u] + [v]) =(f

1 + g

1

)|u+ v| = ([u+ v])

[u] + [v] [u+ v]

Thus L 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 wlogthat b = 1. Let

f(a) =(|a|p + 1)1 p2

(|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 Fatous lemma,

|f | lim

n

|fn| . So f L1().

Let a = fn f, b = f . Consider the sequence n = ||a+ b| |a| |b||. Then n 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, n1 0. Now

(|fn| |fn f |)|f |

(||fn| |fn f | |f ||) =

|n| 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

nfn f = 0

Problem 4.14

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

4.14.2 Note that for a.e. x, fn(x) f(x) |fn(x) f(x)| < for all n bigger than a sufficientlylarge N . Thus

x 6kN

{x : |fk(x) f(x)| > }

for some N for a.e. x. Thus Sn()(x) 0 a.e.Also clearly |Sn()| < 1 L1() since () ] (Sn()) [|fn f | > ]

n0 4.14.1

4.14.3 Let = 1m

for some fixed m N. Then by part 2,

(Sn(1

m))

n 0

Thus given > 0, there exists Nm N such that (Sn( 1m)) 0. Then there exists M N such that 1M< Then x \ x 6 M

|fk(x) f(x)| 1

M< for all k NM

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• Analysis Subhadip Chowdhury Assignment 4

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

4.14.4 Fix an > 0. Then > 0 such thatA

|fn|p < for all n and A measurable with

(A) < .By part 3, there exists a certainA and an integerN > 0 such that (A) < and |fn(x)f(x)|p < (\A)for all n N for all x \A. So in particular,

A

|fn|p < and

\A|fn(x) f(x)|p <

for all n N. Then |fn f |p =

(A

|fn f |p +

\A|fn f |p

)A

|f |p +A

|fn|p +

= 2+

A

|f |p

But |fn|p L1() and supnA|fn|p < . Also |fn(x)|p |f(x)|p for a.e. x A. So by Fatous

lemma,A|f |p lim inf

n

A|fn|p . Thus

|fn f |p 3 for all n N

Hence fn fpn 0 i.e. fn f in Lp().

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

Problem 4.15

4.15.1

4.15.1.(i) For x 6= 0,

limn

nenx = limn

n

enx= lim

n

11n

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

Thus fn 0 a.e.

4.15.1.(ii)

fn1 = 1

0

|fn| = 1

0

nenxdx =

n0

ezdz = 1 en 1;n N

8

• Analysis Subhadip Chowdhury Assignment 4

4.15.1.(iii)fn 01 = 1 en 6 0

4.15.2

4.15.2.(i) For x 6= 0,

limn

n1p enx = lim

n

n1p

enx= lim

n

1

n1p + n1

1px+ n2

1px2/2 + n3

1px3/6 + . . .

= 0

Thus gn 0 a.e.

4.15.2.(ii)

gnp = 1

0

|gn|p = 1

0

nepnxdx =1

p

np0

ezdz =1

p(1 enp) 1

p;n N

4.15.2.(iii)

gn 0p =1

p(1 enp) 6 0

4.15.2.(iv) Observe that 10

gn(x)xmdx = n

1p

10

enxxm =n

1p

nm+1

n0

eyymdy n1p

nm+1(m+ 1)

n 0

Thus for any polynomial (x) R[x], we have 10

gn 0

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

0

gnh 0

for all h Lp() i.e. gn 0.

Problem 4.16

4.16.1 LetKn = conv (i=n{fi})

We claim that n=1Kn = {f}. Indeed since fns are bounded, by Cantors intersection theorem,n=1Kn 6= . if g n=1Kn, then g(x) conv (i=n{fi(x)}) for all n i.e. fn(x) g(x) a.e. Thusf(x) = g(x) a.e. So g f in Lp() and n=1Kn = {f}. Then by exercise 3.13.2, fn f weakly in(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. Th