Demonstration: Answer - Colorado State University

[Pages:12]ANALYSIS HW 3

CLAY SHONKWILER

1

(a) Find all smooth functions f : R R with the property f (x + y) = f (x) + f (y) for all real x, y.

Demonstration: Let f be such a function. Since f is smooth, f exists. Then

f (x + h) - f (x)

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

f (x) = lim

= lim

=

= c.

h0

h

h0

h

h

Then

f (x) = f (x)dx = cdx = cx + d

However, f (0) = f (0 + 0) = f (0) + f (0) = 2f (0), so f (0) = 0, meaning d = 0, so f (x) = cx.

(b) What can you say if f is only assumed to be continuous?

Answer: If f is merely continuous, the same result holds. Assume f is

continuous and f (x + y) = f (x) + f (y). Then, for all n Z, f (nx) = nf (x). Using this result, we see that if p, q Z,

p

1

x

f ( x) = pf ( x) = pf ( ).

q

q

q

Therefore,

p

x

f (x) = f ( x) = pf ( ),

p

p

so

f

(

x p

)

=

f

(x) p

.

Specifically,

p

x f (x) p

f ( x) = pf ( = p = f (x).

q

q

qq

Then for r R, we can find a Cauchy sequence rj of rational numbers that converge to r. Then, by the result proved above for rational numbers,

limj f (rjx) = limj rjf (x) which is to say f (rx) = rf (x)

Now,

f (x) = f (x ? 1) = xf (1),

so f is of the form f (x) = mx for some m R.

1

2

CLAY SHONKWILER

(c) Repeat this for g(x) satisfying g(x + y) = g(x)g(y).

Answer: Let g be a function such that g(x + y) = g(x)g(y). For n Z, g(nx) = g(x)n, so

x g(n

=

g( x )n,

nn

which

implies

g(

x n

)

=

[g(x)]1/n.

In

turn,

if

p, q

Z,

p

xp

p

g( x) = g( ) = [g(x)] q .

q

q

Now, if r R, let rj be a Cauchy sequence converging to r. Then

g(rx) = lim g(rjx) = lim g(x)rj = g(x)r.

j

j

Hence, g(x) = g(1 ? x) = g(1)x. Since g(1) is just a constant, g has the form g(x) = cx.

2

Let f (x) be a continuous function for 0 x 1.

Evaluate limn n

1 0

f

(x)xndx.

Evaluation: Let > 0. Then f can be approximated by smooth function

h such that ||f - h||unif < /3. Now, h exists and is bounded on [0, 1], so let M := maxx[0,1] |f (x)|. Let N > 2M . Now,

|n

1 0

f

(x)xndx

-

f

(1)|

= |n 01(f (x) - h(x) + h(x))xndx - f (1)|

= =

|||nn/300n11(f/01(3xxx)nn-ddxxh++(xnn))x0011n

dx + n h(x)xn h(x)xn

1 0

dx

dx

h(x)xndx - f (1)| - f (1)|

-

f

(1)|

=

|

/3

n n+1

+n

h(1) n+1

-

1 n+1

1 0

h

(x)xn+1dx

-

f

(1)|

| /3 + h(1) -

1 0

h

(x)xndx

-

f

(1)|

| /3 + h(1) - M

1 0

xndx

-

f

(1)|

=

|

/3

+

h(1)

-

M n+1

-

f (1)|

< | /3 + h(1) - /3 - f (1)|

epsilon/3 + |h(1) - f (1)| + /3

< /3 + /3 + /3

=

where we integrated by parts. Hence,

1

lim n f (x)xndx = f (1).

n 0

ANALYSIS HW 3

3

3

Let f : R R be uniformly continuous. Are there constants a, b such that |f (x)| a|x| + b for all x? Proof or counterexample.

Proof. Let > 0. Then, since f is uniformly continuous, there exists 1 > 0 such that |x - y| < implies

|f (x) - f (y)| < .

Then

|f (x)|

= |f (x) - f (0) + f (0)| = |(f (x) - f (x - )) + (f (x - ) - f (x - 2)) + . . . + (f (x - k) - f (0)) + f (0)| |f (x) - f (x - )| + . . . + |f (x - k) - f (0)| + |f (0)| < + . . . + + |f (0)| = k + |f (0)|

where k is the smallest integer such that x - k < . Hence, k is the smallest

integer

such

that

k

>

x

- 1.

We

see,

then,

that

|f (x)| < k + |f (0)| x + |f (0)|.

4

Let

Pj ,

j

= 1, 2, . . .

be

a

sequence

of

points

in

R3.

If

||Pj+1 - Pj||

1 j4

,

show that these points converge.

Proof. Let > 0. Then, since the series

1 j4

converges,

there

exists

N

N

such that, if n > m > N ,

n 1 m1 1

1

>

j4 -

j4

=

n4

+...+

. m+1

1

1

Then, for any n > m > N ,

||Pn+1 - Pm+1||

= ||(Pn+1 - Pn) + (Pn - Pn-1) + . . . + (Pm+2 - Pm+1)||

||Pn+1 - Pn|| + . . . ||Pm+2 - Pm+1||

1 n4

+

...+

1 (m+1)4

1, then we approximate f by a C1 function h. Then

limn nc

1 0

0h(x)e-nxdx

= limn nc

e-10n -n

h(10)

-

h(0) -n

+

1 n

1 0

0h

(x)e-nxdx

= =

limn limn

nc-1 h(0) - e-10nh(10) + nc-1h(0) - nc-1e-10nh(10)

nc-1

1 0

+ nc-1

h

1

0

(x)e-nxdx h (x)e-nxdx,

where we integrated by parts. Now, nc-1h(0) will get arbitrarily large, since c > 1, so this limit does not exist. Hence, we conclude that Qc(f ) exists only when c 1 and that, when it does exist, Qc(f ) = 0.

(b) What if you only assume that f L1(0, 1)?

Answer: If we only assume f L1(0, 1), we can approximate f arbitrarily close by a C1 funcion h, and the above analysis demonstrates that Qc(h)

exists only when c 1 and that, when it exists, Qc(h) = 0.

6

Let B2 be the open unit disk in R2. Give an example of function f

L1(B2) that is not in L2(B2). Justify your assertions.

Example:

Define

f (r, )

=

1 r

.

Then,

define

fn(r, )

=

1 q

.

r+

1 n

Let

>0

and

let

N1

>

2 r

.

Then,

if

n

>

N1,

|f

-

fn|

=

|

1 r

-

1

q

r+

1 n

|

=

|

q r+

1 n

-r

r2+

r n

|

|

q

r+

1 n

-r

r2+

r n

=

1 r2n2+rn

1 rn

< /2.

Hence, fk is Cauchy, meaning there exists N N such that, if m, n > N ,

|fn - fm| < /2.

That means that, for the same m, n,

|fn - fm|drd < /2drd = ,

R

R

ANALYSIS HW 3

5

so fk is Cauchy in L1. Now, we want to demonstrate that fk converges in L1 to f .

limn

2 0

1 0

1

q

r+

1 n

drd

= limn

2 0

2

1

r

+

1 n

d

0

=

2 0

2

1

+

1 n

-

2 n

d

= 4

1

+

1 n

-

4 n

= 4

= ||f ||L1.

Hence,

we

see

that,

indeed,

f (r, )

=

1 r

L1(B2).

However,

||f ||L2 = =

2 0

1 0

1 r

drd

1/2

2 0

[log

r]10

d

1/2

.

However, since log(0) is undefined, it is going to be impossible to construct a Cauchy sequence of continuous functions that converge to f in the L2 norm, meaning f / L2(B2).

7

For each of the following, give an example of a sequence of continuous

functions.

(a) See attached sheet.

(b) See attached sheet.

(c) See attached sheet.

(d) See attached sheet.

(e) Example:

Let fn(x) =

1 nx

,

let

> 0 and let N N such that

N > 2 . If n > N , then

||fn||L1 =

1

1

0 |fn(x)|dx = n

11

1

dx =

0x

n

2x

1 0

=

2 n

<

.

Hence, fn converges to zero in the L1 norm. On the other hand,

||fn||L2 =

1

|fn(x)|2dx

0

1/2

=

1 n

1 1 1/2 1

dx =

0x

n

[log x]10

1/2

.

However, since log(0) is undefined, fn clearly does not converge to zero in the L2 norm.

(f). Example: Let

1

4n x

fn(x) =

1-x n

0

x (0, 1]

x [0, 1] x>2

6

CLAY SHONKWILER

Then let > 0 and let N N such that N > 1 . If n > N , then

||fn||L1

=

0

fn(x)dx

=

1 0

fn(x)dx

+

2 1

fn(x)dx

+

2

fn(x)dx

=

1 0

1 4n x

dx

+

2 1

1-x n

+

0

=

x 2n

|10

+

1 n

x

-

x2 2

2 1

=

1 2n

+

1 2n

< /2 + /2

=,

so fn converges to zero in the L1 norm. On the other hand

||fk||L2 = = =

0

|fn(x)|2

dx

1/2

1 0

1 16n2x

+

2 (2-x)2 1 n2

1/2

1 16n2

log x

1 0

+

2 (2-x)2 1 n2

1/2

.

However, since logx is not defined at zero, this sequence cannot converge to zero.

(g). Example: Let

x

log n

1

gn(x) =

(log n)x (n+1)-x

n log n

0

x (0, 1] x [1, n]

x [n, n + 1] xn+1

31

Let > 0. Then, if N N such that N > max{1, e , e } and n > N , then

||gn||L2

=

0

|gn(x)|2

dx

1/2

=

1 0

x2 (log n)2

dx

+

n 1

(log

1 n)2

x2

dx

+

n+1 n

(n+1)2-2(n+1)x+x2 n2(log n)2

dx

1/2

=

1 3(log n)2

+

(

1 (log n)2

-

1 n(log

n)2

)

+

4+

1 n

-

1 n2

6(log n)2

1/2

<

2

3

+

1 (log n)2

+

1 (log n)2

1/2

<

2

3

+

2

3

+

2

3

1/2

=2

=.

ANALYSIS HW 3

7

Therefore, gk converges to zero in the L2 norm. However, for any n > e2,

||gn||L1

=

0

gn(x)dx

=

1 0

x log n

+

n 1

1 (log n)x

dx

+

n+1 n

(n+1)-x n log n

dx

=

1 log n

x2 2

1 0

+

1 log n

[log

x]n1

+

1 n log n

(n

+

1)

-

x2 2

n+1 n

=

1 2 log n

+

log n log n

+

-2-

1 n

2 log n

>

1

+

1 2 log n

-

3 2 log n

=

1

-

1 log n

>

1 2

,

so gk clearly does not converge to zero in the L1 norm.

8

Let : R R3 define a smooth curve.

1 0

||

(t)||dt.

Show

that

||

1 0

(t)dt||

Proof. Let v =

1 0

(t)dt.

Then

||v||2 = v, v

=

v,

1 0

(t)dt

= =

||00v11 |||v|v,0|1||(|||t)((tt))||||ddtt

where we arrive at the third line by the Cauchy-Schwarz Inequality, which tells us that v, (t) ||v||||(t)||. Hence, we can conclude that

1

1

|| (t)dt|| = ||v|| ||t||dt.

0

0

9

Let f (x), where a x b, be a smooth function. (a) If f (c) = 0 for some a c b, show that

b

|f (x)| |f (t)|dt b - a

b

1/2

|f (t)|2dt

a

a

and hence, using the uniform norm ||f ||unif := maxaxb |f (x)|,

b

b

1/2

||f ||unif |f (t)|dt b - a |f (t)|2dt .

a

a

8

CLAY SHONKWILER

Proof. By the Fundamental Theorem of Calculus,

x

x

b

|f (x)| = | f (t)dt| |f (t)|dt |f (t)|dt.

c

c

a

Also, by Holder,

b

|f (t)|dt

a

b

1/2

12dt

a

b

1/2

|f (t)| = b - a

a

b

1/2

|f (t)|2dt .

a

(b) If

b a

f (t)dt

=

0,

show

that

the

above

inequality

still

holds.

Proof. Let F (x) =

x a

f

(t)dt.

Then

F (a)

=

0

and

b

F (b) = f (t)dt = 0,

a

so, by the Mean Value Theorem, there exists c [a, b] such that 0 = F (c) = f (c). Now, by (a),

b

b

1/2

|f (x)| |f (t)|dt b - a |f (t)|2dt .

a

a

(c) Use the result in part (b) to show that for any smooth f

b

b

1/2

||f ||unif [|f (t)| + |f (t)|]dt [|f (t)|2 + |f (t)|2]dt ,

a

a

where the constants and depend only on the region of integration.

Proof.

Let

g

=

f

- f,

where

f

=

1 b-a

b a

f

(t)dt

is

the

average

of

f

in

the

region. Then,

b a

g(t)

=

b a

f (t)

-

1 b-a

b a

f

(t)dt

dt

=

b a

f (t)dt

-

1 b-a

b a

b a

f

(t)dtdt

=

b a

f (t)dt

-

1 b-a

(b

-

a)

b a

f

(t)dt

= 0.

Hence we can use the result in part (b) to conclude that

b

b

1/2

||g||unif |g (t)|dt b - a |g (t)|2dt .

a

a

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