Math 318 Exam #1 Solutions

[Pages:7]Math 318 Exam #1 Solutions

1. (a) Suppose (fn) and (gn) are two sequences of functions that converge uniformly on a subset A R. Is it true that the sequence (fngn) converges uniformly on A? Prove or give a counterexample.

Answer. No. For each n, let

1

1

fn(x) = x and gn(x) = n

on the interval (0, 1).

Then

certainly

fn

1 x

uniformly

(since

it's a constant sequence) and gn 0 uniformly (since the value

of gn is independent of x). However, I claim that the sequence

1 ((fngn)(x)) = nx

does not converge uniformly. Notice that, for any given > 0 and

any

fixed x (0, 1), we can pick N

>

1 x

so that n N

implies

1

11

-0 = < ,

nx

nx N x

so the sequence (1/(nx)) converges to the zero function pointwise.

However, for any fixed N N, we can choose x such that 0 <

x < 1/N so that

1

1

- 0 = > 1,

Nx

Nx

so the convergence cannot be uniform.

(b) If your answer to (a) was "yes", does the result still hold if one of the sequences does not converge uniformly? If your answer to (a) was "no", what additional assumptions on (fn) and (gn) will yield uniform convergence of (fngn)?

Answer. Notice that the limiting function 1/x above was not bounded (though the limit of the gn obviously is). Here's the modified claim: Suppose fn f uniformly and gn g uniformly on A R and that both f and g are bounded. Then (fngn) converges uniformly on A.

Proof. Let > 0. Suppose M1 > 0 is an upper bound for f (meaning |f (x)| M1 for all x A) and that M2 is an upper

bound for g. Since fn f uniformly, there exists N1 N such that n N1 implies

|fn(x) - f (x)| < 1

for all x A. Equivalently, |fn(x)| < |f (x)| + 1 M1 + 1 for all x A.

In turn, since gn g uniformly, there exists N2 N such that n N2 implies

|gn(x)

-

g(x)|

<

2(M1

+

. 1)

Finally, since fn f uniformly, there exists N3 N such that

n N3 implies

|fn(x)

-

f (x)|

<

. 2M2

Now, if N = max{N1, N2, N3} and n N , we have

|fn(x)gn(x) - f (x)g(x)| = |fn(x)gn(x) - fn(x)g(x) + fn(x)g(x) - f (x)g(x)| |fn(x)gn(x) - fn(x)g(x)| + |fn(x)g(x) - f (x)g(x)| = |fn(x)||gn(x) - g(x)| + |g(x)||fn(x) - f (x)| (M1 + 1)|gn(x) - g(x)| + M2|fn(x) - f (x)|

<

(M1

+

1) 2(M

+

1)

+

M2 2M2

=

for all x A, so we see that, indeed, (fngn) (f g) uniformly on A.

2. (a) Let (fn) be a sequence of continuous functions. Suppose fn f uniformly on A R. Prove that

lim

n

fn(xn)

=

f

(x)

for all x A and all sequences (xn) in A converging to x.

Proof. Fix x A and let (xn) be a sequence converging to x. Let > 0. The fact that fn f uniformly on A implies that there

exists N1 N such that n N1 implies

|fn(y) - f (y)| < /2

for all y A. In turn, since fn f uniformly and each fn is continuous, the limit function f is also continuous. Hence, by Theorem 4.3.2(iv), limn f (xn) = f (x), so there exists N2 N such that n N2 implies

|f (xn) - f (x)| < /2.

Therefore, if N = max{N1, N2} and n N , then

|fn(xn) - f (x)| = |fn(xn) - f (xn) + f (xn) - f (x)| |fn(xn) - f (xn)| + |f (xn) - f (x)| < /2 + /2 =.

Since the choice of > 0 was arbitrary, we see that limn fn(xn) = f (x). In turn, since the choice of x A was arbitrary, we see that this is true for all x A, as desired.

(b) Is the converse true?

Answer.

No.

Consider

the

functions

fn(x)

=

x n

on

all

of

R.

Then

(fn(x)) 0, but not uniformly. However, the conclusion of part

(a) will hold. To see this, let x R and (xn) x. Let > 0.

Since (xn) x, there exists N1 N such that n N1 implies

|xn - x| < 1, meaning that |xn| < |x| + 1. Also, we can certainly pick N2 such that N2 > |x|+1 .

Therefore, if N = max{N1, N2} and n N , we see that

|fn(xn)-f (x)|

=

|fn(xn)-0|

=

|xn| n

|x| + n

1

|x| + N

1

|x| + N2

1

<

.

Since the choice of > 0 was arbitrary, we see that, indeed, limn fn(xn) = f (x). Since the choice of sequence (xn) x was arbitrary, the same holds for any sequence converging to x. Finally, since the choice of x was arbitrary, this is true for any x R.

3. Let fn be a monotone increasing, continuous function on [0, 1] for each

n N. Suppose f (x) =

n=0

fn(x)

converges

for

every

x

[0, 1].

Show that the function f is continuous on [0, 1].

Proof. Let > 0. The goal is to show that the series satisfies the hypotheses of the Cauchy Criterion. To that end, note that, since

fn(0) converges, there exists N0 N such that n > m N0 implies

|fm+1(0) + . . . + fn(0)| < /2.

Likewise, since fn(1) converges, there exists N1 N such that n > m N1 implies

|fm+1(1) + . . . + fn(1)| < /2.

Let N = max{N0, N1} and let n > m N . Since each fi is increasing, the function fm+1 + . . . + fn is also increasing, so Lemma 3.1, stated below, implies that

|fm+1(x) + ? ? ? + fn(x)| |fm+1(0) + . . . + fn(0)| + |fm+1(1) + . . . + fn(1)| < /2 + /2 =

for any x [0, 1]. Therefore, the series fn satisfies the hypotheses of the Cauchy Criterion for Uniform Convergence, so we see that fn converges uniformly. Since each fn is continuous, Theorem 6.4.2 implies that f is also continuous.

Lemma 3.1. Suppose g : [a, b] R is monotone increasing. Then

|g(x)| |g(a)| + |g(b)|

for any x [a, b].

Proof. The fact that g is increasing implies that

g(a) g(x) g(b)

(1)

for any x [a, b]. Therefore,

|g(x)| max{|g(a)|, |g(b)|} |g(a)| + |g(b)|,

as desired.

4. Consider a series and define

n=1

an.

Let

Sn

=

n k=1

ak

be

the

nth

partial

sum

n =

n k=1

Sk

.

n

We say that the series

n=1

an

is

Cesaro

summable

to

L

if

lim

n

=

L.

A consequence of HW 4 Problem #5 from last semester is that if

an = L, then the series is Cesaro summable to L.

(a) Show that the series

n=1(-1)n+1

is

Cesaro

summable

to

1 2

.

(You showed on HW 3 Problem #5 that this is also the Abel

sum of the series.)

Proof. Clearly,

S1 = 1 S2 = 1 - 1 = 0 S3 = 1 - 1 + 1 = 1 S4 = 1 - 1 + 1 - 1 = 0

...

1 (-1)k+1 Sk = 2 + 2 so

n =

n k=1

Sk

=

n

n k=1

1 2

+

(-1)k+1 2

n

=

n 2

+

Sn 2

=

1 + Sn .

n

2 2n

Since

Sn

is

always

either

0

or

1,

the

term

Sn 2n

certainly

goes

to

zero as n , so

lim

n

n

=

lim

n

1 + Sn 2 2n

1 =.

2

Hence,

the

series

is

indeed

Cesaro

summable

to

1 2

.

(b) Does there exist a series

n=1

an

that

is

Abel

summable

but

not

Cesaro summable?

Answer. Yes. Consider an = (-1)nn. Then on [0, 1) we have

that

(-1)nnxn = x (-1)nnxn-1 = x d

(-1)nxn

d =x

1

dx

dx 1 + x

n=1

n=1

n=0

-x = (1 + x)2 .

Therefore, since the function on the right is continuous at 1, we see that

lim

x1-

(-1)nnxn

=

lim

x1-

(1

-x + x)2

1 =-

4

n=1

is the Abel sum of the series. However, the partial sums look like

S1 = -1 S2 = -1 + 2 = 1 S3 = -1 + 2 - 3 = -2 S4 = -1 + 2 - 3 + 4 = 2

... S2k-1 = -k

S2k = k

Therefore,

0

if n is even

n =

-

n+1 2n

if n is odd.

The even terms go to zero, but the odd terms go to -1/2, so lim n does not exist, so the series is not Cesaro summable.

5. Let

n=0

anxn

be

a

power

series

with

each

an

0.

Suppose

that

the

radius of convergence is 1, so that the power series defines a function

f (x) =

n=0

anxn

at

least

on

(-1, 1).

Prove that the power series converges at x = 1 (meaning f (1) makes

sense) if and only if f is bounded on [0, 1).

Proof. () If the power series converges at x = 1, then Abel's Theorem implies that f is continuous on the compact set [0, 1] and, therefore, bounded on that set. Hence, f is bounded on [0, 1).

() On the other hand, suppose f is bounded on [0, 1). My goal is to

show that the sequence of partial sums sk(1) =

k n=0

an

is

bounded

above; since each an 0, the sequence (sk(1)) is monotone increasing,

so boundedness plus the Monotone Convergence Theorem will imply

convergence.

To see that the sk(1) are bounded, let M > 0 be an upper bound for f (x) = anxn on [0, 1), meaning that |f (x)| M for all x [0, 1).

Now, I claim that M is an upper bound for sk(1) for all k N. Suppose not. Then there exists K N such that sK(1) > M . But then, since sK(x) f (x) M for all x [0, 1), we see that

|sK (1) - sK (x)| > sK (1) - M

for any x [0, 1).

But then this means that sK cannot possibly be continuous at x = 1, since there is no that will work for = sK(1) - M > 0. On the other hand, sK is clearly continuous since it is just a polynomial. From this contradiction, then, we can conclude that M is indeed an upper bound for sk(1) for every k and, therefore, that the sequence (sk(1)) converges, which is exactly what it means to say that the power series converges at x = 1.

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