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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