Assignment-6 - University of California, Berkeley

Assignment-6

(Due 07/30)

1. Let sequences fn and gn converge uniformly on some set E R to f and g respectively (a) Construct an example such that fngn does not converge uniformly on E.

Solution: Take fn = gn = x + 1/n and E = R. Clearly fn, gn x uniformly on R. Now fngn = (x + 1/n)2, and we claim that this does not converge uniformly to x2. To see this, we

let hn(x) be the sequence of the squares and expand

hn(x) =

1 x+

n

2

=

x2

+

2x n

+

x2 n2

.

But then we see that

|hn(n)

- x2|

=

2+

1 n

2,

no matter how big of an n we choose. This contradicts uniform convergence.

(b) Prove that fngn does converge uniformly if f and g are bounded on E.

Solution: Let |f (t)|, g(t)| < M for all t E. Without loss of generality, we can assume that M > 1. Given > 0, choose N such that for all t E and all n > N ,

|f (t) - fn(t)|,

|g(t) - gn(t)| <

. 3M

Note also that if is small enough, then for all n > N and all t E,

|fn(t)| |fn(t) - f (t)| + |f (t)| 3M + M < M + 1.

Next, note that

fn(t)gn(t) - f (t)g(t) = fn(t)gn(t) - fn(t)g(t) + fn(t)g(t) - f (t)g(t), and so by triangle inequality, for all t E and all n > N ,

|fn(t)gn(t) - f (t)g(t)| |fn(t)gn(t) - fn(t)g(t)| + |fn(t)g(t) - f (t)g(t)|

= |fn(t)||gn(t) - g(t)| + |g(t)||fn(t) - f (t)|

(M + 1) + M < .

3M

3M

2. Show that the sequence of functions

nx fn(x) = n + 1

does not converge uniformly on all of R, but does converge uniformly on bounded intervals (a, b). What is the point-wise limit on R?

1

Solution: We show that fn(x) x point-wise on R.

nx

|x|

|fn(x) - x| =

-x n+1

=

.

n+1

At x = 0, fn(0) = 0 for all n, and so there is nothing to prove. If x = 0, given any > 0, we can pick N large enough so that |x|/(N + 1) < . THen for n > N we see that

|fn(x) - x| < .

Of course the N depends on x and hence we have only proved point-wise. But notice that if x (a, b) in a bounded interval, then there is an M such that |x| < M .Then we can simply choose N > M/, then for any n > N and x (a, b), it is easy to see that

|fn(x) - x| < ,

and hence the convergence is uniform on bounded intervals.

We now claim that the convergence is NOT uniform on all of R.

That is we need to show that there exists > 0, a subsequence nk and points xnk R such that

|fnk (xnk ) - xnk | > .

For this, let = 1, nk = k and xk = k + 2. Then by the calculation above

|fk(xk) - xk|

=

|xk | k+1

=

k k

+2 +1

>

1.

Note that we cannot use the method of Mn, since fn is not a bounded function on R.

3. If g : [0, 1] R is a continuous function such that g(1) = 0, show that the sequence of functions {g(x)xn} n=1 converges uniformly on [0, 1].

Solution: We claim that the sequence converges uniformly to zero. Let > 0. Since g(1) = 0 and g is continuous at 1, there exists a such that

|g(x)| <

for all x [1 - , 1]. So for x [1 - .1], |g(x)xn| < |g(x)| < .

Since g is also bounded on [0, 1], letting M = supt[0,1] |g(t)|, for x [0, 1 - ] we have |g(x)xn| M |x|n M (1 - )n.

Choose N big enough so that (1 - )n < /M for all n > N . Then for all n > N and all x [0, 1], |g(x)xn| < ,

showing that on [0, 1].

g(x)xn -u-.c 0

2

4. (a) Define a sequence of functions by

fn(x) =

1,

x

=

1,

1 2

,

?

?

?

,

1 n

0, otherwise.

Calculate the pointwise limit function f . Is each fn continuous at zero? Does fn f uniformly on R. Is f continuous at 0?

Solution: The limit function is

f (x) =

1,

x

=

1,

1 2

,

1 3

?

?

?

0, otherwise.

The function fn is continuous everywhere except at x = 1, 1/2, ? ? ? , 1/n, whereas the limit function is discontinuous at the reciprocals of all natural numbers. The theorem on uniform convergence and continuity does not automatically imply that the convergence is not uniform, but we nevertheless claim that fn does not converge uniformly. Suppose the convergence is uniform, then there exists an N such that for all x R,

1

|fN (x)

-

f (x)|

. 2

But

then

if

x

=

1 N +1

,

we

have

that

1

1

fN N + 1 - f N + 1 = 1,

a contradiction.

(b) Repeat the exercise with the functions

gn(x) =

x,

x

=

1,

1 2

,

?

?

?

,

1 n

0, otherwise,

and

x,

x

=

1,

1 2

,

?

?

?

,

1 n-1

hn(x) =

1,

x=

1 n

0, otherwise.

Solution:

? The sequence gn(x). The pointwise limit is clearly

g(x) =

x,

x

=

1,

1 2

,

?

?

?

0, otherwise

Again, we note that gn is continuous everywhere except at x = 1, 1/2, ? ? ? , 1/n while g(x) is continuous everywhere except the reciprocals of natural numbers. Claim. gn -u-.c g on R.

Proof. Let us consider Mn = supR |gn(x) - g(x)|. Now, |gn(x) - g(x)| = 0 everywhere, except when x = 1/m for m > n, in which case the difference is 1/m. So

Mn

=

1 n+1

-n---

0,

3

and hence gn -u-.c g on R. ? The sequence hn(x). The pointwise limit in this case is

h(x) =

x,

x = 1,

1 2

,

?

?

?

0, otherwise,

exactly as above. Again hn is continuous everywhere except at x = 1, 1/2, ? ? ? , 1/n while h(x) is continuous everywhere except the reciprocals of natural numbers. But in this case we claim.

Claim. hn does not converge uniformly to h on R.

Proof. To see this, we again consider Mn = supR |hn(x) - h(x)|. Now, |gn(x) - g(x)| = 0 everywhere, except when x = 1/m for m n. Moreover, we have that

1 hn m

1 -h |=

m

1 m

,

m>n

1-

1 n

,

m = n,

and so Mn = 1 - 1/n for n > 2 and hence

lim

n

Mn

=

1

=

0.

.

5. Let {fn} be a sequence of real-valued continuous functions fn : E R for some subset E R. Suppose fn -u-.c f on E. Show that fn(xn) f (x)

for every sequence of points xn x in E. Is the conclusion true if fn f only pointwise? Either provide a proof, or a counter example. Is the converse of the above statement true?

Solution: Given > 0, there exists N1 such that for any n > N1 and any y E,

|f (y)

-

fn(y)|

<

. 2

Since f is continuous, there also exists N2 such that for all n > N2,

|f (xn)

-

f (x)|

<

. 2

Now let N = max(N1, N2). If n > N we then have

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

The conclusion does not hold assuming only pointwise convergence. Consider for instant fn(x) = xn on [0, 1]. Then fn f , pointwise, where

1, x = ` f (x) =

, 0, otherwise.

Now let xn = 1 - 1/n 1. Then

fn(xn) e-1 = f (1) = 1.

4

The converse is also not true. Consider the same function as above, fn(x) = xn, but on (0, 1). Then fn f = 0 pointwise but not uniformly. For any x (0, 1) and any sequence xn x, we claim that fn(xn) 0 = f (x). To see this, let x < < 1. Then since xn x, for N large enough, xn < . but then fn(xn) < n -n--- 0. This proves the claim. But since fn does not converge uniformly to f , we see that the converse does not hold.

6. For n = 1, 2, ? ? ? and x R, define

x fn(x) = 1 + nx2 .

Show that {fn} converges uniformly to a differentiable function f on R, and that the equation

f

(x)

=

lim

n

fn(x)

is correct for all x = 0 but false at x = 0. Why does this not contradict the theorem on uniform convergence and differentiation?

Solution: 1 + nx2 >

Itis clear that 2x n, and so

the

pointwise

limit

is

0.

Now

by

completing

squares

it

is

easy

to

see

that

x

1

|fn(x)|

=

1 + nx2

<

. 2n

Given > 0 choosing N > 22, we see that the sequence converges to zero uniformly on R. Using

quotient rule, (1 + nx2) - 2nx2 1 - nx2

fn(x) = (1 + nx2)2 = (1 + nx2)2 .

It is easy to see (by pulling out an n2 from the denominator and n from the numerator, or by using

L'Hospital's rule) that if x = 0 then fn(x) 0 = f (x). If x = 0, then f (0) = 0 but fn(0) = 1 for all n, and so

lim

n

fn(0)

=

f

(0).

7. Let fn : [0, 1] R be a sequence of bounded functions. That is for each n, there exists an Mn such that |fn(t)| < Mn

for all t [0, 1]. Suppose in addition that fn -u-.c f . (a) Show that f is bounded on [0, 1]. Hint. Apply the definition of uniform convergence with = 1.

Solution: There exists an N such that |f (t) - fN (t)| < 1

for all t [0, 1]. But then by triangle inequality, |f (t)| |f (t) - fN (t)| + |fN (t)| < 1 + MN ,

and so f is bounded on [0, 1].

(b) Show that the sequence of functions is uniformly bounded. That is, show that there is an M such that |fn(t)| < M

5

for all t [0, 1] and all n. Hint. Show that for large n, fn can be bounded essentially by the bound for f .

Solution: Let M0 = supt[0,1] |f (t)|. By the first part, we know that M0 < . There exists an N such that for all n N and all t [0, 1],

|fn(t) - f (t)| < 1. BUt then for n N and t [0, 1],

|fn(t)| |fn(t) - f (t)| + |f (t)| < 1 + M0. Now let M = maxM0 + 1, M1., ? ? ? , MN-1. Then clearly for any fn and any t [0, 1],

|fn(t)| < M.

(c) Suppose additionally fn R[0, 1] for all n. Prove or disprove -

1-1/n

1

lim

n 0

fn(t) dt = f (t) dt.

0

Hint.

1-1/n

1

1/n

fn(t) dt = fn(t) dt -

fn(t) dt

0

0

0

Solution: As in the hint,

Since fn -u-.c f ,

1-1/n

1

1/n

fn(t) dt = fn(t) dt -

fn(t) dt.

0

0

0

1-1/n

1

1/n

lim

n 0

fn(t) dt = f (t) dt - lim

fn(t) dt.

0

n 0

To compute the second term we estimate

1/n

1/n

M

0

fn(t) dt

0

|fn(t)| dt n 0

as n . This completes the proof.

8. Find the radius and interval of convergence of the following series.

1.

(2x+1)n n=1 n

Solution: We can rewrite this as

2n

1n

x+ .

n

2

n=1

Written this way, the center is clearly a = -1/2. To find the radius of convergence we compute

2n 1/n

lim sup

= 2,

n n

6

and so R = 1/2. So the series definitely converges on (-1, 0). Next, we check boundary points.

? x = -1. The series is then

(-1)n

n

n=1

which converges by the alternating series test.

? x = 0. The series is then the harmonic series which diverges.

So I = [-1, 0).

2.

n=1

2n n2

xn

.

Solution: Apply root test to conclude that radius of convergence is 1/2 and interval of convergence is [-1/2, 1/2].

3.

n=0

2n n!

xn

.

Solution: Apply ratio test to conclude that radius of convergence is infinity and hence I = R.

9. Decide whether each proposition is true or false, providing a complete proof, or a counter example, as appropriate. (a) If fn converges uniformly, then fn converges uniformly to zero.

Solution: True. Apply uniform Cauchy criteria for series with n = m + 1.

(b) If 0 fn(x) gn(x) and gn converges uniformly, then fn also converges uniformly. Solution: True. Again uniform Cauchy criteria for series.

(c) If fn converges uniformly on E, then there exists constants Mn such that |fn(x)| < Mn for all x E and Mn converges.

Solution: Not necessarily true. Let fn : R R be defined by

fn(x) =

1 n

,

x [n, n + 1)

0, otherwise

Then for any n > m and for any x R,

n

1

fk (x)

<

, m

k=m

and so the series satisfies uniform Cauchy criteria, and hence is uniformly convergent. But supR |fn| = n-1 and n-1 diverges.

(d) If each fn is uniformly continuous on E and fn -u-.c f on E, then f is also uniformly continuous on E.

7

Solution: True. Given any > 0, let N N such that

sup |fN (x) -

E

f (x)|

<

. 3

Since fN is uniformly continuous, there exists > 0 such that for any x, y E,

|x - y| <

=

|fN (x)

-

fN (y)|

<

. 3

So if x, y E with |x - y| < , then

|f (x) - f (y)| |f (x) - fN (x)| + |fN (x) - fN (y)| + |fN (y) - f (y)|

< + + = . 333

So f is also uniformly continuous.

(e) If fn has a finite number of discontinuities on E and fn -u-.c f , then f has a finite number of discontinuities on E.

Solution: False. See example in 4(a).

(f) If fn has at most M number of discontinuities on E (where M is fixed and independent of n) and fn -u-.c f , then f has at most M number of discontinuities on E.

Solution: This is true. To prove, this suppose f has at least M +1 discontinuities p1, ? ? ? , pM+1. We first have the following observation, which follows from a simple counting argument.

Claim. There exists a k {1, 2, ? ? ? , M + 1} such that for all N N, there exists n N such that fn is continuous at pk. Proof If not, then for all k {1, 2, ? ? ? , M +1}, there exists an Nk N such that for all n Nk, fn is discontinuous at pk. Let N = max(N1, ? ? ? , NM+1). Then fN is discontinuous at each p1, p2, ? ? ? , pM+1 contradicting the hypothesis that fN can have at most M discontinuities.

Without loss of generality, let p1 be such that for all N N, there exists n N such that fn is continuous at p1. Then there exists a subsequence {fnj } such that each fnj is continuous at p1 To see this, simply apply the claim to pick an n1 1 such that fn1 is continuous at p1, and then having chosen n1 < n2 < ? ? ? < nj-1, apply claim with N = nj-1, to obtain nj > nj-1 such that fnj is continuous at p1. But then since fn -u-.c f , we also must have fnj -u-.c f . This contradicts the theorem on uniform convergence and continuity, since each fnj is continuous at p1 but f is discontinuous at p1.

10. Define

x2n g(x) = x2n + 1 .

n=0

Find the values of x where the series converges, and show that we get a continuous function on this set.

Solution:

Let fn(x) =

. x2n

1+x2n

If |x| 1, clearly limn fn(x) = 0.

On the other hand, for any

x [-r, r] with r < 1, since 1 + x2n > 1, we have

|fn(x)| r2n.

8

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