201A, Fall 10, ThomasesHomework 4 - Solutions

Mihaela Ifrim

1. Suppose fn C([0, 1]) is a monotone decreasing sequence that con-verges pointwise to f C([0, 1]). Prove that fn converges uniformlyto f . This result is called Dini s monotone convergence theorem.

Proof. Put gn(x) := fn(x) f(x). The reason I am doing this isbecause it will be easier for me to prove that gn converges uniformlyto zero rather than prove that fn converges uniformly to f (note thatthis is just a rewriting of the our problem). The new sequence we haveconstructed gn is in C([0, 1]) since fn C([0, 1]) and f C([0, 1]).Also, note that gn 0 pointwise and also gn gn+1.

This last inequality is true since we know that fn fn+1, which im-plies that fn f fn+1 f i.e., gn gn+1.

So, our goal is to prove that gn 0 uniformly on [0, 1].Let > 0. For each x [0, 1] there is an integer Nx such that

0 gNx(x)

2.

By the continuity and by the monotocity of the sequence {gn}n (wesaid above that a monotone decreasing sequence), there exists an openset I(x) that contains x, such that

0 gn(t) (1)

if t I(x) and if n Nx. Since [0, 1] is compact, we know that wecan find a finite open cover of it; in particular there exists a finite setof points x1, x2 , xm such that

[0, 1] I(x1) I(x2) I(x2) I(xm). (2)

Choosing N := max {Nx1 , Nx2 , Nx3 , , Nxm}

it follows from (1) and (2) that

0 gn(t) ,if t I(x) and if n N . This prove the uniformity of convergence.

Note The compactness of [0, 1] is necessarily ! For instance look at

fn(x) =1

nx+ 1,

where x (0, 1) and n N. Then fn(x) 0 monotonically in (0, 1),but the convergence is not uniform.

2. Consider the space ofcontinuous differentiable functions

C1([a, b]) ={f : [a, b] R | f, f are continous }

with the C1 - norm,

f C1= supaxb |f(x)|+ supaxb |f (x)|.Proove that C1([a, b]) is a Banach space with respect to the givennorm.

Proof. Set I = [a, b]. Since we are asked to prove that C1([a, b]) is aBanach space, let {fn}n be a Cauchy sequence in C1([a, b]) .Since fn fm fn fm C1 the sequence {fn}n is Cauchy inC (I) such that fn f uniformly.

Next set gn = f n.Then

gn gm = f n fm fn fm C1 ,

so {gn}n is Cauchy in C (I).Therefore, there exists a function g C(I) such that

gn g

uniformly.

It remains to prove that f C1(I) and that fn f in C1(I). Fixany x I and any h R such that x+ h I.

Then

1h

(f(x+ h) f(x)) = limn

1h

(fn(x+ h) fn(x))

= limn

1h

h0f

n(x+ t) dt.

(3)

Therefore we get that

1h

(f(x+ h) f(x)) = limn

1h

h0gn(x+ t) dt. (4)

Now, recall that uniform convergence on a finite interval impliesconvergence of integralswe have proved this in homework no 3. Sincegn g uniformly we find that

1h

(f(x+ h) f(x)) = limn

1h

h0gn(x+ t) dt =

1h

h0g(x+ t) dt.

Since g is a continuous function, the limit as h 0 exists and so

f(x) = lim

h01h

(f(x+ h) f(x))

= limh0

1h

h0g(x+ t) dt

= g(x).

(5)

This proves that f C1(I).

To prove that fn f in C1(I) we note that

f fn C1 = f fn + f f n

= f fn + g gn .(6)

By the construction of f and g it follows that f fn C1(I) 0.

Hence the proof is done, since our random Cauchy sequence {fn}n itturn out to be a convergent sequence C1([a, b]) is a Banach spacewith respect to the given norm.

3. Let F C([0, 1] [0, 1]) and consider the map

F : C([0, 1]) C([0, 1])

given by

(Ff) (x) = 10 F (x, y)f(y) dy.Prove that {Ff | f 1} is an equicontinuous family so that anygiven sequence {fn}n with fn 1 for all ns has a subsequence{fn(i)

}n(i)

with{Ffn(i)}n(i) uniformly convergent.

Proof. First lets observe that [0, 1] is a compact set, and by Ty-chonoffs Theorem, [0, 1] [0, 1] is also a compact set. So, F is uni-formly continuous, which by definition means that for all > 0 thereexists some > 0 such that if |x1 x2|+ |y1 y2| < then

|F (x1, y1) F (x2, y2)| < .

We have to prove that {Ff | f 1} is an equicontinuous familyof functions, so lets apply the definiton to see if indeed it is verified.Then for > 0, take > 0 as above, and if |x1 x2| <

|Ff(x1)Ff(x2)| = | 10

(F (x1, y) F (x2, y)) f(y) dy|

10|F (x1, y) F (x2, y)||f(y)| dy

10 dy

= (1 0)= .

(7)

So, indeed the family is equicontinuous.

So the family {Ff : f 1} is equicontinuous. Note also thatthe family is bounded. It is also closed. To see this, suppose {Ffn}in the family converges to Ff . Since we can pass the limit inside theintegral, and {f C([01, ]) : fn 1} is closed. So, by Arzela-AscoliTheorem, it is compact. In particular, it is sequentially compact, soany sequence {fn} with fn 1 for all n has a subsequence {fn(i)}such that {Ffn(i)} is uniformly convergent.

4. Suppose

K = {f C([0, 1]) : Lip(f) 1, and 10 f(x)dx = 0}Prove that K is compact in C([0, 1]).

Proof. In order to solve the problem I will use Arzela - Ascoli Theo-rem. This result will give us the desired result.

K is equicontinuousThe Lipschitz condition implies that K is equicontinuous. To provethis fix > 0. Set = . Then for any f K, and for any x and y in[0, 1] that satisfy |x y| < we have

|f(x) f(y)| Lip(f)|x y| |x y| .

K is boundedTo pove that K is bounded, note that if

f = 0, and f is continuous,

then there must exist an x0 [0, 1] such that f(x0) = 0. Then for anyx [0, 1] and any f K, we have:

|f(x)| = |f(x) f(x0)| Lip(f)|x x0| |x x0| 1. (8)Hence f 1.

K is closedLet {fn}n be a Cauchy sequence in K. Since C([0, 1]) is a completemetric space, there exists an f C([0, 1]) such that fn f uniformly.

We need to prove that indeed f belongs to K.Since fn f uniformly we know that

Lip(f) lim supn

Lip(fn) 1f = lim

n

fn = 0.

(9)

This proves that f K.

This finishes the proof of the fact that K is compact in C([0, 1]).

5. Prove that C([a, b]) is separable.

Proof. Without loss of generality, we may assume that a = 0, b = 1.

Let P (Q) denote the set of all polynomials over a Q, and P (R) denotethe set of all polynomials over R. It is obvious that P (Q) is a countableset.

Suppose q P (R), and

q(x) =n

k=0

kxk, k R. (10)

Since Q is dense in R, then for each > 0, we can find aj Q so that

|aj j | < 3(n+ 1) , 1 j n. (11)

Hence

|p(x) q(x)| n

k=0

|aj j ||x|j n

k=0

|aj j | < 2 . (12)

Therefore p q < /2. This implies that any q P (R), and any > 0, we can find p P (R), so that q p < .For any f C([a, b]), and any > 0, we can find a polynomial q P (R), so that f q < /2 by Weietrass theorem. By the aboveobservation, for this q, we can find p P (R) so that p q < /2.

Therefore f p < . Hence P (Q) is dense in C([0, 1]). ThereforeC([0, 1]) is separable since P (Q) is countable.Let T : C([a, b]) C([0, 1]), by

(Tf)(x) = f (a+ (b a)x) , (13)

then T is an invertible isometry. Since P (Q) is a countable densesubset of C([0, 1]), so is T1 (P (Q)). This implies that C([a, b]) isseparable.

6. (a) Suppose T : X X is a contraction mapping. Show that Tn isalso a contraction mapping for every n N.

(b) Give an example to show that Tn is a contraction mapping forn 2 does not necessarily imply that T is a contraction mapping.

Proof. (a) Since T : X X is a contraction, then we can find a nonnegativeconstant C with 0 C < 1 so that

d(Tx, Ty) Cd(x, y),x, y X. (14)

By induction, we can prove that, for any n N,

d(Tnx, Tny) Cnd(x, y), x, y X. (15)

Since 0 C < 1, then0 Cn < 1. (16)

(14) and (16) implies Tn : X X is a contraction mapping.

Proof of the induction.For n = 1, (14) holds since T is a contraction. Suppose (14) holds forn = k, then for n = k + 1, we find

d(T k+1x, T k+1y) = d(T k(Tx), T k(Tx))

Ckd(Tx, Ty) Ck Cd(x, y)= Ck+1d(x, y),

for any x, y X. By induction hypothesis (14) holds for n = k + 1.Then by mathematical induction, (14) holds for any n N.

(b) Let X = C([0, 1]) and define T : X X by

T (f)(x) = x0f(t)dt. (17)

Then for any f, g X, we find

|(Tf)(x) T (g)(x)| x0|f(t) g(t)|dt f g. (18)

HenceTf Tg f g. (19)

If we take f(x) = ex, then

T (f)(x) = x0etdt = ex = f(x). (20)

In this case,Tf 0 = f 0. (21)

Hence T is not a contraction mapping. Since

(T 2f)(x) = x0T (f)(x2)dx2

= x0

x20

f(x1)dx1dx2.

By Fubini theorem, we find

(T 2f)(x) = x0f(x1)(x x1)dx1. (22)

Using Fubini theorem, we find

(T 3f)(x) = x0

(x x1)(Tf)(x1)dx1

= x0

(x x1) x10

f(x2)dx2dx1

= x0f(x2)

xx2

(x x1)dx1dx2

= x0f(x2)

(x x2)22!

dx2.

Similarly, we can prove that

(Tnf)(x) =1

(n 1)! x0f(t)(x t)n1dt,n N (23)

Hence

|(Tnf)(x) (Tng)(x)| 1(n 1)!

( x0

(x t)n1dt)f g

=xn

n!f g

1n!f g.

Therefore

Tnf Tng 1n!f g,f, g X,n N. (24)

This implies that Tn is always a contraction mapping from X to Xitself when n 2, but T is not a contraction mapping.

Proof. due to Tim and Amanda Let X = Z Z under the discretemetric. Then the map defined by

A =(

0 10 0

)is not a contraction. Note that that A2 = 0, which is a contraction.

7. (a) Suppose D is a bounded, closed subset of Rn with the standardmetric. Suppose T : D D satisfies

Tx Ty < x y, (25)

for every x, y D with x 6= y. Prove that T has exactly one fixedpoint in D.(b) Give an example of a mapping T : R R satisfying (25) that doesnot have a fixed point.

Proof.(a) Since D is closed and bounded in Rn, Heine-Borel Theorem impliesD is compact. Define : D [0,) by

(x) = d(x, Tx) = x Tx, x X. (26)Then for any x, y D, we have, by triangle inequality,

|(x) (y)| = |d(x, Tx) d(y, Ty)| d(x, y) + d(Tx, Ty) 2d(x, y),

sinced(Tx, Ty) = Tx Ty < x y = d(x, y). (27)

Therefore is a Lipshchitz continuous function. Since D is compactand is continuous, then attains its minimum on D. Hence we canfind x D so that

(x) = minxD

(x). (28)

If (x) = d(x, Tx) 6= 0, then(Tx) = T 2x Tx < Tx x = (x). (29)

This implies that (x) is no longer the minimum of f on D whichwill lead a contradiction that (x) is the minimum of f on D. Thus(x) = d(x, Tx) = 0, which implies, by metric property of d, that

Tx = x. (30)

Hence x is a fixed point of T in D. If we have another fixed point yof T in D, and y 6= x, then

x y = Tx Ty < x y. (31)This is impossible. We conclude that T has a unique fixed point x inD.

(b)

Proof. One nice example of a such function is T (x) = 1 + log(1 + ex).To verify that indeed ( 25) is satisfied you just need to use some el-emetary calculus and see that

T(x) =

ex

1 + ex,

which is obviously strictly less than 1 and strictly greater than 0.

By the Mean Value Theorem we have that if x 6= y than

T (x) T (y) = T ()(x y),

for between x and y. So, indeed ( 25) is satisfied.

To check that T has no fixed point look at

T (x)x = 1+log(1+ex)x = 1+log(1+ex)log ex = 1+log(1+ex) > 1.Hence there is no fixed point for T .

Why doesnt this example contradict the contraction mapping theo-rem? Let me know if you do not know the answer to this question!