Analysis II Home Assignment 4 Subhadip subhadip/coursedocuments/Homework_shared...  Analysis II...

download Analysis II Home Assignment 4 Subhadip subhadip/coursedocuments/Homework_shared...  Analysis II Home

of 25

  • date post

    26-Jun-2018
  • Category

    Documents

  • view

    215
  • download

    1

Embed Size (px)

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

  • Analysis IIHome Assignment 4

    Subhadip Chowdhury

    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

    2

  • 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.

    4

  • 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|

    5

  • 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|

    6

  • 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

    7

  • 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

    ).

    9

  • 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