Ma 5051 — Real Variables and Functional Analysis
Solutions for TakeHome Midterm
Prof. Sawyer — Washington University
Let (
X,
M
, μ
) be a measure space. Recall
R
A
f
(
x
)
dμ
=
R
I
A
(
x
)
f
(
x
)
dμ
for
A
∈ M
and
f
∈
L
+
, where
I
A
(
x
) is the indicator function of
A
.
1.
(a) The problem is to show that the integrand is dominated by an integrable
function.
Since 1

e

x
=
R
x
0
e

y
dy
≤
R
x
0
dy
=
x
, it follows that 1

x
≤
e

x
,
1

(1
/n
)
x
≤
e

x/n
, and
(
1

(1
/n
)
x
)
n
≤
e

x
. Thus the integrand is dominated by
x
k
e

x
, which is integrable on (0
,
∞
). Then by the dominated convergence theorem
lim
n
→∞
Z
n
0
x
k
(
1

n

1
x
)
n
dx
=
Z
∞
0
lim
n
→∞
I
(0
,n
]
(
x
)
x
k
(
1

n

1
x
)
n
dx
=
Z
∞
0
x
k
e

x
dx
=
k
!
(b) The substitution
x
→
x/
√
n
changes the integral to
I
(
n
) =
Z
√
n
0
1 +
x
2
(
1 + (1
/n
)
x
2
)
n
dx
By the binomial theorem (see also the model solutions for Problem 4a of HW5),
1 +
x
2
n
¶
n
≥
1 +
n
x
2
n
+
n
(
n

1)
2
x
2
n
¶
2
≥
1 +
1
3
x
4
for
x >
0 and
n
≥
3. Thus the integrand of
I
(
n
) is dominated by 3(1+
x
2
)
/
(3+
x
4
),
which is integrable on (0
,
∞
). Then by the dominated convergence theorem
lim
n
→∞
I
(
n
) =
Z
∞
0
lim
n
→∞
I
(0
,
√
n
]
(
x
)
1 +
x
2
(
1 + (1
/n
)
x
2
)
n
dx
=
Z
∞
0
1 +
x
2
e
x
2
dx
=
Z
∞
0
e

x
2
dx
+
Z
∞
0
1
2
x
‡
2
xe

x
2
·
dx
=
3
2
Z
∞
0
e

x
2
dx
=
3
4
√
π
2.
The integral is
I
(
a
) =
Z
∞
0
e

y
2
sin(
ay
)
dy
=
Z
∞
0
e

y
2
∞
X
n
=0
(

1)
n
a
2
n
+1
y
2
n
+1
(2
n
+ 1)!
dy
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Ma 5051– Real Variables and Functional Analysis– TakeHome Midterm
. . . . . . . . .
2
The partial sums of the integrand of
I
(
a
) are dominated for fixed
a >
0 by
e

y
2
∞
X
n
=0
a
2
n
+1
y
2
n
+1
(2
n
+ 1)!
≤
e

y
2
∞
X
n
=0
a
n
y
n
n
!
=
e

y
2
e
ay
where the second series is formed from the first by adding in the even terms
a
2
n
y
2
n
/
(2
n
)!.
Since the last expression above is integrable on (0
,
∞
), dominated
convergence allows us to interchange the integral and sum in
I
(
a
) and conclude
I
(
a
) =
∞
X
n
=0
(

1)
n
a
2
n
+1
(2
n
+ 1)!
Z
∞
0
e

y
2
y
2
n
+1
dy
=
∞
X
n
=0
(

1)
n
a
2
n
+1
(2
n
+ 1)!
1
2
Z
∞
0
e

x
x
n
dx
=
∞
X
n
=0
(

1)
n
a
2
n
+1
n
!
2(2
n
+ 1)!
=
a
2
∞
X
n
=0
(

a
2
/
2)
n
(2
n
+ 1)(2
n

1)
. . .
(3)(1)
since 2
n
n
! = (2
n
)(2
n

2)
. . .
2.
One can use complexvariable techniques to show
I
(
a
) =
R
∞
0
e

y
2
sin(
ay
)
dy
=
e

a
2
/
4
R
a/
2
0
e
x
2
dx
, but that is not required.
See the
Appendix
for two derivations of this identity using complexvariable
methods, one based on analytic continuation and one using Cauchy’s theorem.
3.
Let
A
= lim inf
n
→∞
R
f
n
(
x
)
dμ
. Then there exists a sequence
n
k
↑ ∞
such that
lim
k
→∞
R
f
n
k
dμ
=
A
. Since
f
n
k
→
f
in measure, there exists a further subsequence
{
f
n
k
j
}
such that lim
j
→∞
f
n
k
j
(
x
) =
f
(
x
) a.e.
μ
. Since
f
n
(
x
)
≥
0, this implies
Z
f
(
x
)
dμ
≤
lim inf
j
→∞
Z
f
n
k
j
(
x
)
dμ
=
lim
k
→∞
Z
f
n
k
(
x
)
dμ
= lim inf
n
→∞
Z
f
n
(
x
)
dμ
and
R
f
(
x
)
dμ
≤
lim inf
n
→∞
R
f
n
(
x
)
dμ
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 JUNYING
 lim, measure, Lebesgue measure, Functional Analysis– TakeHome, Analysis– TakeHome Midterm

Click to edit the document details