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 0 exists.
Then
f (x) + f (h) − f (x)
f (h)
f (x + h) − f (x)
= lim
=
= c.
f 0 (x) = lim
h→0
h→0
h
h
h
Then
Z
Z
f (x) =
f 0 (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,
1
x
p
f ( x) = pf ( x) = pf ( ).
q
q
q
Therefore,
p
x
f (x) = f ( x) = pf ( ),
p
p
so f ( xp ) =
f (x)
p .
Specifically,
p
x
f (x)
p
f ( x) = pf ( = p
= f (x).
q
q
q
q
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 (rj x) = limj→∞ rj f (x)
f (rx) = rf (x)
which is to say
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
x
g(n = g( )n ,
n
n
which implies g( nx ) = [g(x)]1/n . In turn, if p, q ∈ Z,
p
x p
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(rj x) = 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
R1
Let f (x) be a continuous function for 0 ≤ x ≤ 1. Evaluate limn→∞ n 0 f (x)xn dx.
Evaluation: Let > 0. Then f can be approximated by smooth function
h such that ||f − h||unif < /3. Now, h0 exists and is bounded on [0, 1], so
let M := maxx∈[0,1] |f 0 (x)|. Let N > 2M
. Now,
R1
R1
|n 0 f (x)xn dx − f (1)| = |n 0 (f (x) − h(x) + h(x))xn dx − f (1)|
R1
R1
= |n 0 (f (x) − h(x))xn dx + n 0 h(x)xn dx − f (1)|
R
R1
1
≤ |n 0 /3xn dx + n 0 h(x)xn dx − f (1)|
R1
R1 n
n
= |/3n 0 x dx
+ n 0 h(x)x
R dx − f (1)|
1 0
h(1)
n
= |/3 n+1
+ n n+1
− 1
h (x)xn+1 dx − f (1)|
R 1 0 n+1n 0
≤ |/3 + h(1) − 0 h (x)x dx − f (1)|
R1
≤ |/3 + h(1) − M 0 xn dx − f (1)|
M
− f (1)|
= |/3 + h(1) − n+1
< |/3 + h(1) − /3 − f (1)|
≤ epsilon/3 + |h(1) − f (1)| + /3
< /3 + /3 + /3
=
where we integrated by parts. Hence,
Z 1
lim n
f (x)xn dx = 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 δ1 > 0
such that |x − y| < δ implies
|f (x) − f (y)| < .
Then
|f (x)| = |f (x) − f (0) + f (0)|
= |(f (x) − f (x − δ)) + (f (x − δ) − f (x − 2δ)) + . . . + (f (x − kδ) − f (0)) + f (0)|
≤ |f (x) − f (x − δ)| + . . . + |f (x − kδ) − f (0)| + |f (0)|
< + . . . + + |f (0)|
= k + |f (0)|
where k is the smallest integer such that x − kδ < δ. Hence, k is the smallest
integer such that k > xδ − 1. We see, then, that
|f (x)| < k + |f (0)| ≤ x + |f (0)|.
δ
4
Let Pj , j = 1, 2, . . . be a sequence of points in R3 . If ||Pj+1 − Pj || ≤ j14 ,
show that these points converge.
P 1
Proof. Let > 0. Then, since the series
converges, there exists N ∈ N
j4
such that, if n > m > N ,
>
n
m
X
X
1
1
1
1
−
= 4 + ... +
.
4
4
j
j
n
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
≤ n14 + . . . + (m+1)
4
< .
Hence, {Pj } is a Cauchy sequence and, since R3 is complete, converges. 5
(a) Let f (x) be a continuous function for 0 ≤ x ≤ 10. Find all real
R1
numbers c for which Qc (f ) := limn→∞ nc 0 0f (x)e−nx dx exists. If the limit
4
CLAY SHONKWILER
Qc (f ) exists, compute it. Answer: Let M = maxx∈[0,10] |f (x)|. If c ≤ 1,
then
R1
R1
nc 0 0f (x)e−nx dx ≤ nc M 0 0e−nx dx
−10n
1
= nc M ( e −n − −n
)
= nc−1 M (e−10n + 1).
Now, e−10n vanishes, as does nc−1 M (since c ≤ 1), so, for c ≤ 1, Qc (f ) exists
and is equal to 0.
Now, if c > 1, then we approximate f by a C 1 function h. Then
−10n
R
R1
h(0)
1 1
e
0 (x)e−nx dx
h(10)
−
+
0h
limn→∞ nc 0 0h(x)e−nx dx = limn→∞ nc
−n
−n
0
n c−1
R1 0
c−1
−10n
−nx dx
= limn→∞ n
h(0) − e
h(10) + n
0 Rh (x)e
1 0
−nx dx,
c−1
c−1
−10n
c−1
= limn→∞ n h(0) − n e
h(10) + n
0 h (x)e
where we integrated by parts. Now, nc−1 h(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 C 1 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, θ) = √1r . Then, define fn (r, θ) = q 1 1 . Let > 0
r+ n
and let N1 >
2π
r .
Then, if n > N1 ,
|f − fn | =
| √1r
−
q1
|
1
r+ n
q
1 √
r+ n
− r
=| √
≤|
|
r
r2 + n
√ q1 √
r+ n − r
√
r
r2 + n
1
r2 n2 +rn
=
1
≤ rn
< /2π.
√
Hence, fk is Cauchy, meaning there exists N ∈ N such that, if m, n > N ,
|fn − fm | < /2π.
That means that, for the same m, n,
Z
Z
|fn − fm |drdθ <
/2πdrdθ = ,
R
R
ANALYSIS HW 3
5
so fk is Cauchy in L1 . Now, we want to demonstrate that fk converges in
L1 to f .
i1
R 2π R 1
R 2π h q
limn→∞ 0 0 q 1 1 drdθ = limn→∞ 0 2 r + n1 dθ
0
r+ n
R 2π q
= 0 2 1 + n1 − √2n dθ
q
4π
= 4π 1 + n1 − √
n
= 4π
= ||f ||L1 .
Hence, we see that, indeed, f (r, θ) = √1r ∈ L1 (B2 ). However,
i1/2
hR R
2π 1
||f ||L2 = 0 0 1r drdθ
i1/2
hR
2π
.
= 0 [log r]10 dθ
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.
1
(e) Example: Let fn (x) = n√
, let > 0 and let N ∈ N such that
x
N > 2 . If n > N , then
Z 1
Z
1 √ 1
2
1 1 1
√ dx =
||fn ||L1 =
|fn (x)|dx =
2 x 0 = < .
n
n
n
x
0
0
Hence, fn converges to zero in the L1 norm. On the other hand,
Z 1
1/2
Z 1
1/2
i1/2
1
1h
1
2
dx
=
[log x]10
.
||fn ||L2 =
|fn (x)| dx
=
n 0 x
n
0
However, since log(0) is undefined, fn clearly does not converge to zero in
the L2 norm.
♣
(f). Example: Let
 1
 4n√x x ∈ (0, 1]
1−x
fn (x) =
x ∈ [0, 1]
 n
0
x>2
6
CLAY SHONKWILER
Then let > 0 and let N ∈ N such that N > 1 . If n > N , then
||fn ||L1
R∞
= 0 fn (x)dx
R1
R2
R∞
= 0 fn (x)dx + 1 fn (x)dx + 2 fn (x)dx
R1
R2
= 0 4n1√x dx + 1 1−x
+0
h
in2
√
2
= 2nx |10 + n1 x − x2
1
1
= 2n
+ 2n
< /2 + /2
= ,
1
so fn converges to zero in the L1 norm. On the other hand
||fk ||L2
1/2
R ∞
|fn (x)|2 dx
0
hR
i
R2
2 1/2
1
= 0 16n12 x + 1 (2−x)
n2
h
i
R
1
2 (2−x)2 1/2
1
= 16n
.
2 log x 0 + 1
2
n
=
However, since logx is not defined at zero, this sequence cannot converge to
zero.
♣
(g). Example: Let
gn (x) =









x
log n
1
(log n)x
(n+1)−x
n log n
0
x ∈ (0, 1]
x ∈ [1, n]
x ∈ [n, n + 1]
x≥n+1
3
1
Let > 0. Then, if N ∈ N such that N > max{1, e , e } and n > N , then
||gn ||L2
R ∞
1/2
|gn (x)|2 dx
0
hR
Rn
R n+1 (n+1)2 −2(n+1)x+x2 i1/2
2
1
1
= 0 (logx n)2 dx + 1 (log n)
dx
2 x2 dx + n
n2 (log n)2
1/2
1
4+ n
− 12
n
= 3(log1 n)2 + ( (log1n)2 − n(log1 n)2 ) + 6(log
n)2
h 2
i1/2
< 3 + (log1n)2 + (log1n)2
h 2
i
2
2 1/2
< 3 + 3 + 3
√
= 2
= .
=
ANALYSIS HW 3
7
Therefore, gk converges to zero in the L2 norm. However, for any n > e2 ,
R∞
||gn ||L1 = 0 gn (x)dx
R1
R n+1 (n+1)−x
Rn
= 0 logx n + 1 (log1n)x dx + n
n log n dx
h 2 i1
h
in+1
1
x2
(n
+
1)
−
= log1 n x2 + log1 n [log x]n1 + n log
n
2
0
n
−2− 1
log n
1
n
= 2 log
n + log n + 2 log n
1
3
> 1 + 2 log
n − 2 log n
= 1 − log1 n
> 12 ,
so gk clearly does not converge to zero in the L1 norm.
♣
8
Let γ : R →
R1
0 ||γ(t)||dt.
Proof. Let v =
R1
0
R3
Show that ||
define a smooth curve.
R1
0
γ(t)dt|| ≤
γ(t)dt. Then
R1
||v||2 = hv, vi = hv, 0 γ(t)dti
R1
= 0 hv, γ(t)i
R1
≤ 0 ||v||||γ(t)||dt
R1
= ||v|| 0 ||γ(t)||dt
where we arrive at the third line by the Cauchy-Schwarz Inequality, which
tells us that hv, γ(t)i ≤ ||v||||γ(t)||. Hence, we can conclude that
Z 1
Z 1
||
γ(t)dt|| = ||v|| ≤
||γt||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
Z b
1/2
Z b
√
0
0
2
|f (x)| ≤
|f (t)|dt ≤ b − a
|f (t)| dt
a
a
and hence, using the uniform norm ||f ||unif := maxa≤x≤b |f (x)|,
Z
||f ||unif ≤
a
b
|f 0 (t)|dt ≤
√
Z
b−a
a
b
1/2
|f 0 (t)|2 dt
.
8
CLAY SHONKWILER
Proof. By the Fundamental Theorem of Calculus,
Z b
Z x
Z x
0
0
|f 0 (t)|dt.
|f (t)|dt ≤
f (t)dt| ≤
|f (x)| = |
a
c
c
Also, by Holder,
b
Z
2
|f (t)|dt ≤
a
1/2 Z
b
Z
0
b
1 dt
a
1/2
Z b
1/2
√
0
2
|f (t)|
= b−a
|f (t)| dt
.
0
a
a
(b) If
Rb
a
f (t)dt = 0, show that the above inequality still holds.
Proof. Let F (x) =
Rx
f (t)dt. Then F (a) = 0 and
a
b
Z
F (b) =
f (t)dt = 0,
a
so, by the Mean Value Theorem, there exists c ∈ [a, b] such that 0 = F 0 (c) =
f (c). Now, by (a),
b
Z
0
|f (t)|dt ≤
|f (x)| ≤
√
b
Z
b−a
0
2
|f (t)| dt
a
1/2
.
a
(c) Use the result in part (b) to show that for any smooth f
Z
b
Z
0
[|f (t)| + |f (t)|]dt ≤ β
||f ||unif ≤ α
b
1/2
[|f (t)| + |f (t)| ]dt
,
0
2
2
a
a
where the constants α and β depend only on the region of integration.
Rb
1
Proof. Let g = f − f , where f = b−a
a f (t)dt is the average of f in the
region. Then,
Rb
Rb
Rb
1
g(t)
=
f
(t)
−
f
(t)dt
dt
b−a a
a
a
Rb
R
R
b b
1
= a f (t)dt − b−a
(t)dtdt
a a fR
Rb
b
1
= a f (t)dt − b−a (b − a) a f (t)dt
= 0.
Hence we can use the result in part (b) to conclude that
Z
||g||unif ≤
a
b
|g 0 (t)|dt ≤
√
Z
b−a
a
b
1/2
|g 0 (t)|2 dt
.
ANALYSIS HW 3
9
Now,
||f ||unif = ||g + f ||unif
≤ ||g||unif + |f |
Rb 0
Rb
1
≤ a |g
(t)|dt + b−a
a |f(t)|dt
Rb
1
0
= a |f (t) + b−a |f (t)| dt
Rb
≤ a (α|f 0 (t)| + α|f (t)|) dt
Rb
= α a [|f 0 (t)| + |f (t)|]dt.
10
Define the normed linear space L1,1 (0, 1) as the completion of the C 1 [0, 1]
functions in the norm
Z 1
||f ||L1,1 :=
(|f 0 (t)| + |f (t)|)dt.
0
√
x in this space?
q Why?
Answer: Yes. Let fk (x) = x + k1 . Then each fk ∈ C 1 [0, 1]. Let > 0.
(a) Is f (x) =
Then, if N1 >
1
,
2
then n > N1 implies
r
r
√
√
1 √
1 √
1
|fn (x) − x| = | x + − x| ≤ | x +
− x| = √ < .
n
n
n
√
Hence, fk converges to x = f (x) in [0, 1], which means that fn0 → f 0 .
Therefore, fn and fn0 are Cauchy sequences, which is to say that there exists
N2 ∈ N such that, if n, m > N2 , |fn (x) − fm (x)| < /2 and there exists
0 (x)| < /2 for all x ∈ [0, 1]. Let
N3 ∈ N such that, if n, m > N3 , |fn0 (x) − fm
N = max{N2 , N3 }. Now, then, if n > N ,
Z 1
Z 1
0
0
||fn −fm ||L1,1 =
(|fn (t)−fm (t)|+|fn (t)−fm (t)|)dt <
(/2+/2)dt = .
0
0
So fk is cauchy in the L1,1 norm. Its limit must be f (x) =
√
x ∈ L1,1 (0, 1).
√
x, so f (x) =
♣
(b) Show that all the functions in this space are continuous.
Proof. Fix x ∈ [0, 1]. Let f ∈ L1,1 (0, 1) and let fn be a Cauchy sequence of
C 1 functions that converge to f . Let > 0. Then there exists N ∈ N such
that n > N implies
Z 1
Z 1
/3 > ||f − fn ||L1,1 =
(|f 0 (t) − fn0 (t)|dt +
|f (t) − fn (t)|dt
0
or
Z
1
|f (t) − fn (t)|dt < /3
0
0
10
CLAY SHONKWILER
which is to say that |f (t) − fn (t)| < for all t ∈ [0, 1]. Now, since each fj ∈
C 1 [0, 1], there exists δ > 0 such that |x−y| < δ implies |fj (x)−fj (y)| < /3.
Now, then, if n > N and |x − y| < δ,
|f (x) − f (y)| = |(f (x) − fn (x)) + (fn (x) − fn (y)) + (fn (y) − f (y))|
≤ |f (x) − fn (x)| + |fn (x) − fn (y)| + |fn (y) − f (y)|
< /3 + /3 + /3
= .
Hence, f is continuous at x. Since our choice of x was arbitrary, we see that
f is continuous.
(c) Define the normed linear space L1,2 (0, 1) as the completion of the
functions in the norm
C 1 [0, 1]
Z
||f ||L1,2 :=
1
1/2
(|f (t)| + |f (t)| )dt
.
0
2
2
0
Show that L1,2 (0, 1) is a proper subspace of L1,1 (0, 1).
Proof. If f ∈ L1,2 (0, 1), then, since the smooth functions are dense in the
C 1 functions, we can approximate f by a sequence hk of smooth functions.
We see by part 9(c) above that
||hk ||L1,1 ≤
β
||hk ||L1,2 ,
α
so hk ∈ L1,1 (0, 1). Hence, f ∈ L1,1 (0, 1),
√ so L1,2 (0, 1) ⊂ L1,1 (0, 1). On the
other hand, we saw above that f (x) = x ∈ L1,1 (0, 1). However, ||f ||L1,2 is
not defined, as
Z 1
Z 1
1
1
x2
0
2
|f (t)| + |f (t)| dt =
+ x dt = log x + |10
4x
4
2
0
0
√
and log(0) is undefined. Hence, f (x) = x ∈
/ L1,2 (0, 1), so L1,2 (0, 1) (
L1,1 (0, 1).
11
Let f : [0, 1] → R be a continuous function.
R1
(a) Show that limλ→∞ 0 f (x)sin(λx)dx = 0.
Proof. Let > 0. Since the smooth functions are dense in the continuous
functions, we can choose a smooth function h such that ||f − h||unif < /3.
Since [0, 1] is compact, h and h0 are bounded by M and M 0 , respectively.
ANALYSIS HW 3
11
0
3M
Now, let N1 > 3M
and N2 > . Let N = max{N1 , N2 }. If λ > N , then
R1
R1
R1
0 f (x) sin(λx)dx = R0 (f − h)Rsin(λx)dx + 0 h sin(λx)dx
1
1
< 0 3 dx + 0 |h(x) sin(λx)|dx
i1 R
h
1
|
+ 0 |h0 (x) cos(λx)
|dx
= |h(x) − cos(λx)
λ
λ
0
R
0
1
≤ /3 + Mλ + M
λ 0 dx
< /3 + /3 + /3
= .
In other words,
Z
lim
λ→∞ 0
1
f (x) sin(λx)dx = 0.
R1
(b) Compute limλ→∞ 0 | sin(λx)|dx.
(c) If φ : R → R is continuous with period P , show that
Z 1
Z 1
lim
f (x)φ(λx)dx = φ
f (x)dx,
λ→∞ 0
1
P
0
RP
where φ =
0 φ(t)dt is the average of φ over one period.
(d) What can you say about the validity of all of parts (a) and (c) if one
only knows that f ∈ L1 (0, 1)?
Answer: Part (a) will still hold valid, as the smooth functions are dense
in L1 (0, 1), so we could make the same approximation that we did for any
f ∈ L1 (0, 1).
♣
12
If p, q, r > 1 satisfy
Z
1
p
+ 1q + 1r = 1, prove the three-term Holder inequality
|f (x)g(x)h(x)|dx ≤ ||f ||p ||g||q ||h||r
for all continuous functions f (x), g(x), h(x) that are zero outside of a compact set.
Proof. Since g ◦ h has compact support, the two-term Holder inequality tells
us that
Z
1/α
Z
α
|f (x)g(x)h(x)| ≤ ||f ||p
|g(x)h(x)| dx
for
1
α
=
1
q
+ 1r . Now,
r
q
q+r
+
=
= 1,
q+r q+r
q+r
12
CLAY SHONKWILER
so, again by the two-term Holder inequality,
Z
q
r Z
Z
q+r
q+r
q+r
q+r
α
α r
α q
|g(x)h(x)| dx ≤
dx
(|g(x)| )
dx
(|h(x)| )
.
Now,
α
q+r
qr q + r
qr
=
=
=q
r
q+r r
r
α
q+r
qr q + r
qr
=
=
= r.
q
q+r q
q
and
Therefore,
R
h
q 1
i r hR
i q+r
α
q+r
R
q+r
q+r
1/α
|g(x)h(x)|α dx
≤
(|h(x)|α ) q dx
(|g(x)|α ) r dx
R
1 R
1
=
|g(x)|q dx q
|h(x)|r dx r
= ||g||q ||h||r .
Hence,
Z
Z
|f (x)g(x)h(x)| ≤ ||f ||p
|g(x)h(x)|α dx
1/α
≤ ||f ||p ||g||q ||h||r .
DRL 3E3A, University of Pennsylvania
E-mail address: [email protected]