Homework 4 Spring 2014 Solutions

Prévia do material em texto

Math 426: Homework 4
Mary Radcliffe
due 2 May 2014
In Bartle:
5C. If f ∈ L(X,F , µ) and g is a F-measurable real-valued function such that
f(x) = g(x) almost everywhere, then g ∈ L(X,F , µ) and ∫ f dµ = ∫ g dµ.
Proof. Let h(x) = |f(x)− g(x)|. Then h ∈ M+(X,F), and h = 0 almost
everywhere. Therefore, h is integrable and
∫
h dµ = 0. But this implies
that f−g is integrable, and thus g is integrable. Moreover, ∣∣∫ (f − g) dµ∣∣ ≤∫
h dµ = 0, and thus
∫
(f − g) dµ = 0. The result immediately follows by
linearity of integration.
5E. If f ∈ L and g is a bounded, measurable function, then fg ∈ L.
Proof. Let M be such that |g| ≤M . Note that by linearity of integration,
we have that Mf ∈ L, and |fg| ≤ |Mf | for all x. Then by Corollary 5.4,
fg is integrable.
5I. If f is a complex-valued function on X such that Re f and Im f belong to
L(X,F , µ), we say that f is integrable and define ∫ f dµ = ∫ Re f dµ +
i
∫
Im f dµ. Let f be a complex-valued measurable function. Show that
f is integrable if and only if |f | is integrable, in which case ∣∣∫ f dµ∣∣ ≤∫ |f | dµ.
Proof. Let us write f = f1 + if2, so that f1 = Re f and f2 = Im f . We
have that f is measurable if and only if f1 and f2 are measurable, and
|f | = ((f1)2 + (f2)2)1/2. Thus, f is measurable implies that |f | is also
measurable.
Notice, if a, b ∈ R, then √a2 + b2 = √(a+ b)2 − 2ab ≤ √(a+ b)2 =
|a + b| ≤ |a| + |b|. Therefore, if f is integrable, then |f | is a measurable
function such that |f | ≤ |f1| + |f2|, and thus by Corollary 5.4, |f | is
integrable.
Conversely, if |f | is integrable, note that |f1| < |f | and |f2| < |f |, so we
have f1 and f2 are real-valued integrable functions by Corollary 5.4. But
then by definition f is also integrable.
For the inequality, write
∫
f dµ = reiθ, so that
∣∣∫ f dµ∣∣ = r. Let g(x) =
e−iθf(x), so that g(x) = g1(x) + ig2(x), and
∫
g dµ =
∫
e−iθf(x) dµ =
e−iθ
∫
f(x) dµ = r. But then
∫
g2 dµ = 0, so g2 = 0 almost everywhere.
Moreover, |f | = |e−iθ||f | = |g| = |g1| almost everywhere.
Therefore, we have∣∣∣∣∫ f dµ∣∣∣∣ = r = ∫ g1 dµ = ∣∣∣∣∫ g1 dµ∣∣∣∣ ≤ ∫ |g1| dµ = ∫ |f | dµ,
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since g1 is real valued and we may use Theorem 5.3.
5P. Let fn ∈ L(X,F , µ), and suppose that {fn} converges to a function f .
Show that if lim
∫ |fn − f | dµ = 0, then ∫ |f | dµ = lim ∫ |fn| dµ.
Proof. Note that as |fn − f | is integrable for n sufficiently large, we have
fn − f is integrable, and thus by linearity f is integrable. Moreover, for
all n, we have |fn| ≤ |fn−f |+ |f |, so |fn|− |f | ≤ |fn−f |. Integrating and
taking limits on both sides, we have lim
∫ |fn| dµ− ∫ |f | dµ ≤ lim ∫ |fn −
f | dµ = 0, and thus lim ∫ |fn| dµ = ∫ |f | dµ
5Q. If t > 0, then
∫∞
0
e−txdx = 1t . Moreover, if t ≥ a > 0, then e−tx ≤ e−ax.
Use this and Exercise 4M to justify differentiating under the integral sign
and to obtain the formula
∫∞
0
xne−xdx = n!.
Proof. Now, ∫ ∞
0
e−txdx = lim
b→∞
∫ b
0
e−txdx
= lim
b→∞
−1
t
e−tx
∣∣∣∣b
x=0
= lim
b→∞
(
−1
t
e−tb +
1
t
)
=
1
t
,
where the integral is obtained by the fundamental theorem of calculus.
Let f(x, t) = e−tx for t ∈ [ 12 , 32 ]. Then ∂f∂t (x, t) = −xe−tx, so |∂f∂t (x, t)| ≤
xe−
x
2 for all t. Let g(x) = xe−
x
2 . We claim that g is integrable on [0,∞).
This can be seen in several ways. Note that g is measurable. Moreover,
note that if x is sufficiently large (in fact, x > 8 ln 4 is sufficient), we
have that x < ex/4. Define h(x) = g(x) if x ≤ 8 ln 4 and h(x) = e− x4 if
x > 8 ln 4. Then g ≤ h, and h is clearly integrable, and thus by Corollary
5.4, g is also integrable. But then we have, by Theorem 5.9, that
− 1
t2
=
d
dt
(
1
t
)
=
d
dt
(∫ ∞
0
e−txdx
)
=
d
dt
(∫
[0,∞)
e−txdλ(x)
)
=
∫
[0,∞)
∂
∂t
(
e−tx
)
dλ(x)
=
∫
[0,∞)
−xe−txdλ(x)
= −
∫ ∞
0
xe−txdλ(x).
Taking t = 1 and multiplying by −1 yields the result for n = 1.
Now, proceed by induction on n. Suppose it is known that
∫∞
0
xn−1e−txdx =
(n−1)!
tn . Let fn(x, t) = x
n−1e−tx for t ∈ [ 12 , 32 ]. As above, we have
∂fn
∂t (x, t) = −xne−tx, so |∂fn∂t (x, t)| ≤ xne−
x
2 . Again, for x sufficiently
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large, we have that xn < ex/4, so as above, we have that xne−
x
2 is inte-
grable on [0,∞). Then as above, we have
− n!
tn+1
=
d
dt
(
(n− 1)!
tn
)
=
d
dt
(∫ ∞
0
xn−1e−txdx
)
=
∫
[0,∞)
∂
∂t
(xn−1e−tx)dλ(x)
= −
∫ ∞
0
xne−txdx.
Taking t = 1 and multiplying by −1 yields the result for n.
5R. Suppose that f is defined on X × [a, b] to R and that the function x 7→
f(x, t) is F-measurable for each t ∈ [a, b]. Suppose that for some t0, t1 ∈
[a, b], the function x 7→ f(x, t0) is integrable on X, that ∂f∂t (x, t1) exists,
and that there exists an integrable function g onX such that
∣∣∣ f(x,t)−f(x,t1)t−t1 ∣∣∣ ≤
g(x) for x ∈ X and t ∈ [a, b], t 6= t1. Then[
d
dt
∫
f(x, t) dµ(x)
]
t=t1
=
∫
∂f
∂t
(x, t1) dµ(x).
Proof. Choose tn to be any sequence with tn → t1, tn 6= t1 (we may
start at n = 2 to avoid confusion). Put hn(x) =
f(x,tn)−f(x,t1)
tn−t1 , so by
hypothesis |hn(x)| ≤ g(x) for all x, and hn(x) → ∂f∂t (x, t1). By the
dominated convergence theorem, then, we have limn→∞
∫
X
hn(x)dµ(x) =∫
X
∂f
∂t (x, t1)dµ(x). On the other hand,
lim
n→∞
∫
X
hn(x)dµ(x) = lim
n→∞
∫
f(x, tn)− f(x, t1)
tn − t1 dµ(x)
= lim
n→∞
1
tn − t1
(∫
f(x, tn)dµ(x)−
∫
f(x, t1)dµ(x)
)
=
[
d
dt
∫
f(x, t)dµ(x)
]
t=t1
,
as desired.
5T. Let f be a F-measurable function on X to R. For n ∈ Z+, let {fn}
be the sequence of truncates of f . If f is integrable with respect to µ,
then
∫
f dµ = lim
∫
fn dµ. Conversely, if sup
∫ |fn| dµ < ∞, then f is
integrable.
Proof. Note that for all n, we have |fn| ≤ f , and thus if f is integrable,
the DCT implies that
∫
fn dµ→
∫
f dµ.
For the converse, note that |fn| is a monotonically increasing sequence
of nonnegative functions with limit |f |, and thus the MCT implies that∫ |fn| dµ → ∫ |f | dµ. Thus, by hypothesis, ∫ |f | dµ < ∞, and therefore
|f | and thus f are integrable.
4O. Fatou’s Lemma has an extension to a case where the fn take on negative
values. Let h ∈ M+(X,F), and suppose that ∫ h dµ < ∞. If {fn} is a
sequence in M(X,F) and −h ≤ fn then
∫
lim inf fn dµ ≤ lim inf
∫
fn dµ.
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Proof. Note that fn + h ≥ 0. Thus we have∫
lim inf fn dµ+
∫
h dµ =
∫
(lim inf fn + lim inf h) dµ
=
∫
(lim inf(fn + h)) dµ
≤ lim inf
∫
(fn + h) dµ
= lim inf
(∫
fn dµ+
∫
h dµ
)
= lim inf
∫
fn dµ+
∫
h dµ.
Subtracting
∫
h dµ yields the result.
Also, complete the following:
1. Compute the following limits. Justify each computational step using con-
vergence theorems and/or calculus.
(a) limn→∞
∫∞
0
n
1+n2x2 dx.
(b) limn→∞
∫∞
0
(
1 + xn
)−n
log(2 + cos( xn ))dx
(c) limn→∞
∫ n
−n f
(
1 + xn2
)
g(x)dx, where g : R → R is (Lebesgue) inte-
grable and f : R → R is bounded, measurable, and continuous at
1.
Solution. (a) We consider this as two integrals, over [0, 1] and [1,∞).
Note that on [0, 1], we have
lim
n→∞
∫ 1
0
n
1 + n2x2
dx = lim
n→∞ [arctan(nx)]
1
0
= lim
n→∞ arctan(n) =
pi
2
.
For the otherpart, notice that n1+n2x2 ≤ nn2x2 ≤ 1nx2 ≤ 1x2 , and
moreover,
∫∞
1
1
x2 = 1 < ∞, so 1x2 ∈ L([1,∞),L, λ). Therefore, by
the dominated convergence theorem, we have
lim
n→∞
∫ ∞
1
n
1 + n2x2
dx = lim
n→∞
∫
[1,∞)
n
1 + n2x2
dλ(x)
=
∫
[1,∞)
lim
n→∞
(
n
1 + n2x2
)
dλ(x)
=
∫
[1,∞)
0dλ(x) = 0.
Therefore, we obtain limn→∞
∫∞
0
n
1+n2x2 dx =
pi
2 .
(b) Note that (1 + xn )
−n ≤ e−x, and as cos( xn ) ≤ 1, we have log(2 +
cos( xn )) ≤ log 3. Thus,
(
1 + xn
)−n
log(2+cos( xn )) ≤ (log 3)e−x, which
is clearly integrable. Therefore, by the dominated convergence theo-
rem, we have
lim
n→∞
∫ ∞
0
(
1 +
x
n
)−n
log
(
2 + cos
(x
n
))
dx =
∫ ∞
0
lim
n→∞
((
1 +
x
n
)−n
log
(
2 + cos
(x
n
)))
dx
=
∫ ∞
0
e−x log(3)dx
= log(3)
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(c) Let us rewrite the integral as
lim
n→∞
∫ n
−n
f
(
1 +
x
n2
)
g(x)dx = lim
n→∞
∫
R
f
(
1 +
x
n2
)
χ[−n,n]g(x) dλ(x).
LetM be such that |f(x)| < M for all x. Then ∣∣f (1 + xn2 )χ[−n,n]g(x)∣∣ ≤
M |g(x)|, and since g ∈ L, we also have |g| ∈ L and thus M |g| ∈ L.
Therefore, by the dominated convergence theorem, we have
lim
n→∞
∫ n
−n
f
(
1 +
x
n2
)
g(x)dx = lim
n→∞
∫
R
f
(
1 +
x
n2
)
χ[−n,n]g(x) dλ(x)
=
∫
R
lim
n→∞
(
f
(
1 +
x
n2
)
χ[−n,n]g(x)
)
dλ(x)
=
∫
R
f(1)χRg(x)dλ(x)
= f(1)
∫
R
g(x)dλ(x),
where the limit exists because f is continuous at 1.
2. Let f(t) =
∫ t
0
e−x
2
dx. We will use DCT to evaluate limt→∞ f(t), even
without being able to evaluate the indefinite integral.
(a) Put h(t) = f(t)2 and g(t) =
∫ 1
0
e−t
2(1+x2)
1+x2 dx. Show that h
′(t) =
−g′(t).
(b) Show that h(t) + g(t) = pi4 for all t.
(c) Use the previous parts to conclude that lim
t→∞ f(t) =
∫ ∞
0
e−x
2
dx =
√
pi
2
.
Solution. (a) Note that h′(t) = 2f(t)f ′(t) = 2e−t
2
f(t), by the funda-
mental theorem of calculus.
For g′(t), let f(x, t) = e
−t2(1+x2)
1+x2 , and note that
∂
∂t
[
e−t
2(1+x2)
1+x2
]
=
−2te−t2(1+x2) = −2te−t2e−t2x2 . Put η(t) = 2te−t2 . Note that η′(t) =
(2 − 4t2)e−t2 , and thus by the first derivative test, η has a global
maximum of
√
2/e at t =
√
1/2, and a global minimum of −√2/e
at t = −√12. Therefore,
∣∣∣−2te−t2 ∣∣∣ ≤ √2/e =: C for all t. Thus,∣∣∣∂f∂t ∣∣∣ ≤ Ce−t2x2 ≤ C, which is clearly integrable on [0, 1]. Therefore,
by Theorem 5.9,
d
dt
[∫ 1
0
e−t
2(1+x2)
1 + x2
dx
]
=
∫ 1
0
∂
∂t
(
e−t
2(1+x2)
1 + x2
)
dx
=
∫ 1
0
−2te−t2(1+x2)dx
= −2te−t2
∫ 1
0
e−t
2x2dx
= −2e−t2
∫ t
0
e−u
2
du using the substitution u = tx
= −2e−t2f(t) = −h′(t),
as desired.
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(b) As (h + g)′ = 0 for all t, we have that h + g is constant. Note that
h(0) = 0. Moreover, g(0) =
∫ 1
0
1
1+x2 dx = arctan(1) =
pi
4 . Therefore,
(h+ g)(0) = pi4 , and as h+ g is constant, the result follows.
(c) Note that |f(x, t)| ≤ 11+x2 for all x, t, and thus by the dominated
convergence theorem, we have
lim
t→∞ g(t) =
∫ 1
0
lim
t→∞
e−t
2(1+x2)
1 + x2
dx
=
∫ 1
0
0 dx = 0.
But then limt→∞ h(t) = pi4 , and thus limt→∞ f(t) =
√
pi/4 =
√
pi
2 .
Extras:
5A. If f ∈ L(X,F , µ) and a > 0, show that the set {x ∈ X | |f(x)| > a}
has finite measure. In addition, the set {x ∈ X | f(x) 6= 0} has σ-finite
measure.
Proof. As |f | ∈ L, we have that ∫ |f | dµ = A < ∞. Let Ea = {x ∈
X | |f(x)| > a}. Then we have |f | = |f |χEa + |f |χEca ≥ aχEa + |f |χEca ,
and thus
A =
∫
|f | dµ ≥ aµ(Ea) +
∫
Eca
|f | dµ ≥ aµ(Ea).
Therefore, µ(Ea) ≤ A/a.
Moreover, {x ∈ X | f(x) 6= 0} = ∪n∈Z+En, and thus the set has σ-finite
measure.
5B. If f is a F-measurable real-valued function and if f(x) = 0 for µ-almost
all x ∈ X, then f ∈ L(X,F , µ) and ∫ f dµ = 0.
Proof. Note that |f | ∈ M+ is µ-almost 0, and thus ∫ |f | dµ = 0. This
implies that |f | ∈ L, and thus f ∈ L. Moreover, ∣∣∫ f dµ∣∣ ≤ ∫ |f | dµ = 0,
and thus
∫
f dµ = 0.
5D. If f ∈ L(X,F , µ) and � > 0, then there exists a measurable simple function
ϕ such that
∫ |f − ϕ| dµ < �.
Proof. As f+ and f− are M+, we have measurable simple functions ϕ+
and ϕ− such that ϕ+ ≤ f+, ϕ− ≤ f−, and ∫ (f+ − ϕ+) dµ < �2 , ∫ (f− −
ϕ−) dµ < �2 . Let ϕ = ϕ
+ − ϕ−. Then we have ϕ is simple, and∫
|f−ϕ| dµ =
∫
|f+−ϕ+−(f−−ϕ−)| dµ ≤
∫ (|f+ − ϕ+|+ |(f− − ϕ−)|) dµ < �.
5F. If f belongs to L, it does not follow that f2 belongs to L.
Proof. Let f = 1√
x
on [0, 1]. Then we have
∫
[0,1]
f dλ =
∫ 1
0
1√
x
dx =
2
√
x|10 = 2. However, f2 = 1x does not have a finite integral on [0, 1], so
f2 /∈ L.
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5G. Suppose that f ∈ L(X,F , µ), and that its indefinite integral is λ(E) =∫
E
f dµ for E ∈ F . Show that λ(E) ≥ 0 for all E ∈ F if and only if
f(x) ≥ 0 for almost all x ∈ X. Moreover, λ(E) = 0 for all E if and only
if f(x) = 0 for almost all x ∈ X.
Proof. Write f = f+ − f−. Suppose λ(E) >≥ 0 for all E ∈ F . Let
� > 0, and put E� = {x ∈ X | f(x) < −�}. As f is measurable, E� ∈ F .
Moreover,
∫
E
f dµ ≤ −�µ(E�) ≥ 0, and thus µ(E�) = 0. But then {x ∈
X | f(x) < 0} = ∪n∈Z+E1/n has measure 0, and therefore f ≥ 0 µ-almost
everywhere.
On the other hand, if f ≥ 0 µ-almost everywhere, then for all E ∈ F ,
fχE ≥ 0 µ-almost everywhere, and thus by 5B,
∫
fχE dµ ≥ 0 for all
E ∈ F .
For the second piece, we repeat the above argument with E� = {x ∈
X | |f(x)| > �}.
5H. Suppose that f1, f2 ∈ L(X,F , µ), and let λ1, λ2 be their indefinite inte-
grals. Show that λ1(E) = λ2(E) for all E ∈ F if and only if f1(x) = f2(x)
for almost all x ∈ X.
Proof. Note that λ1(E) = λ2(E) for all E if and only if (λ1 − λ2)(E) = 0
for all E, if and only if f1 − f2 = 0 µ-almost everywhere (by problem
G).
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