Real Analysis – September 2011
Part A
1. Let E and F be Lebesgue measurable subsets of R, with E ⊂ F. Suppose that F is Borel
measurable, but suppose that E is not Borel measurable.
Prove or disprove: F − E is uncountable.
True. Proof:
By way of contradiction, assume that F − E is countable, thus F − E is Borel measurable.
Now, notice that Ec = ( F − E) ∪ F c , where both F − E and F c are Borel measurable. So, Ec
is Borel measurable, and thus, E would also be Borel measurable, which is a contradiction.
Thus, F − E must be uncountable.
2. Let f : [0, 1] → [0, 1] be a nondecreasing continuous function satisfying f (0) = 0 and
f (1) = 1. Prove or give a counterexample:
Z 1
0
f 0 ( x ) dx = 1.
False. Counterexample: Cantor function
Let C be the Cantor set.
n That is, let
o
−n , where each a = 0 or 2
C =
x ∈ [0, 1] : x = ∑∞
a
3
n
n =1 n
(
)
m
−
1
∞ 3 [−1
[
3k + 1 3k + 2
=
x ∈ [0, 1] \
,
m
3
3m
m =1 k =0
n
o
=
x ∈ [0, 1] \ “open middle thirds” .
Notice that µ(C ) = 0, since C is totally disconnected, where µ is the Lebesgue measure.
Define the Cantor function
f : [0, 1] → [0, 1] as


∞

an −n


2 ,
if x ∈ Cas defined above
∑
2
n
=
1
f (x) =




 p 2−k , for some p, k ∈ Z, if x ∈ [0, 1] \ C
so that if x ∈ [0, 1] \ C, then f ( x ) is equal to the value of f at the endpoints of the “open
middle third” interval that x is in.
Notice: f (0) = 0 and f (1) = 1, (i.e. f goes from 0 to 1 as x goes from 0 to 1). Also, f is
continuous everywhere.
Now, since µ(C ) = 0 and f 0 ( x ) = 0 for every x 6∈ C, f is differentiable almost everywhere
with f 0 ≡ 0. Thus, f is nondecreasing, also. So f satisfies the hypotheses.
Z 1
Z 1
However,
0
f ( x ) dx =
0 dx = 0 6= 1 = f (1) − f (0).
0
0
1
3. Prove or disprove:
a. Every locally compact metric space is complete.
b. Every locally compact inner product space is finite dimensional.
a. False. Counterexample:
Let E = (0, 1) ⊆ R. Then E is a locally compact metric space since for
every x∈ (0, 1)
x x+1
there exists some interval [ a, b] ⊂ (0, 1) such that x ∈ [ a, b]. Let [ a, b] =
,
.
2
2
However, E is not complete: consider the sequence
∞
1
n n =2
1
1
1
which is Cauchy since − → 0 as m, n → ∞. However, lim = 0 6∈ E. Thus, E is
n→∞ n
m n
not a complete metric place.
b. True. Proof:
By way of contradiction, assume that X is a locally compact inner product space with
infinite dimension. Then,
every inner product space has an orthonormal basis, there
since
∞
exists a countable set u j j=1 of orthonormal basis elements such that
(
0 if j 6= k
u j , uk =
1 if j = k
Now, since X is locally compact, it’s locally compact at 0. So, there is a compact K and
open U such that 0 ∈ U ⊆ K. Also, since U is open, there is some δ > 0 such that
(−δ, δ) ⊆ U.
δ
δ
Now, scale each u j by 2 so that 2 u j − 0 = 2δ u j = 2δ u j = 2δ · 1 = 2δ . Thus,
δ
2 uj
∈ (−δ, δ) for all j. But,
δ
δ 2 u j − 2 uk =
=
=
=
=
n
rD
δ
2 (u j
− uk ) , 2δ (u j − uk )
E
q
u j − uk , u j − uk
q
δ
u j , u j − 2 u j , uk + uk , uk
2
√
δ
2 1−0+1
δ
2
√δ
2
6→ 0 as j, k → ∞.
o
⊆ K is not Cauchy and cannot have a convergent subsequence. But X
is locally compact and K is compact, so every sequence in K should have a convergent
subsequence. This has led us to a contradiction. Thus, X must be finite dimensional.
Thus,
δ
2 uj
2
4. Let f : [0, 1] → R be absolutely continuous, and assume that f 0 ∈ L2 ([0, 1]) and that
f (0) = 0. Show that the following limit exists, and compute its value.
lim x −1/2 f ( x )
x →0+
Since f is absolutely continuous, f 0 exists a.e.
Z x and is integrable on [0, x ] for x ∈ [0, 1] and,
by the Fundamental Theorem of Calculus,
f 0 (t) dt = f ( x ) − f (0) = f ( x ). Also, note
0
Z 1
21
0 2
0
2
f (t) dt
that f ∈ L ([0, 1]) implies that
< ∞.
0
Now,
0
≤
lim x −1/2 f ( x )
x →0+
=
=
≤
=
≤
=
=
=
lim x −1/2 · f ( x )
x →0+
Z x
−1/2 0
lim x
f
(
t
)
dt
x →0+
0
Z x
0 lim x −1/2
f
(
t
)
dt
x →0+
0
on [0, x ]
lim x −1/2 f 0 (t)
+
1
x →0
lim x −1/2 12 · f 0 (t)
+
x →0
2
lim x −1/2
x →0+
Z x
0
lim x −1/2 x1/2
x →0+
lim
x →0+
Z
R
! 21
|1|2 dt
Z
R
By Hölder’s Inequality
Z x
0 2
f
(
t
)
dt
! 12
0
0 2
χ[0,x] f (t) dt
2
χ[0,x] f 0 (t) dt
! 12
! 21
Now, let { xn } be a sequence of positive real numbers in [0, 1] such that xn → 0 as
2
n → ∞. Define gn ( x ) := χ[0,xn ] f 0 ( x ) . Then, for x ∈ [0, 1], gn ( x ) → 0 as n → ∞
2
and gn ( x ) ≤ f 0 ( x ) , which is integrable, so by the Lebesgue Dominated Convergence
Theorem,
Z
21
12
Z
12
Z
21
Z
0 2
=
lim
gn (t) dt
=
lim gn (t) dt
=
0 dt
= 0.
lim
χ[0,x] f (t) dt
x →0+
R
n→∞ R
R n→∞
Thus, lim x −1/2 f ( x ) = 0.
R
x →0+
3
Part B
5. Suppose that f : R → C is locally integrable (i.e., f · χK ∈ L1 (R), for every compact
K), and suppose that g : R → C is continuous.
Prove or disprove:
a. If g has compact support, then f ∗ g is uniformly continuous.
b. If g is bounded and if f has compact support, then f ∗ g is uniformly continuous.
a. False. Counterexample:
Let
f (x) = ex


if x ∈ [−1, 0]
x + 1
g( x ) = − x + 1 if x ∈ (0, 1]


0
otherwise.
and
Then, g is bounded and continuous, and has compact support since g = 0 outside of
[−1, 1]. Also, f is locally integrable. However,
Z
f ( x − y) g(y) dy
f ∗ g (x) =
R
=
=
Z
R
e x−y g(y) dy
Z −1
−∞
= ex
= e
x
e
x −y
0
Z
−1
h
· 0 dy +
−1
Z 1
e−y y dy −
− ye
Z 0
−y
−e
−y
= e x − 2 − e + e −1
0
i0
−1
e
x −y
(y + 1) dy +
e−y y dy +
−
h
Z 1
− ye
−y
Z 1
−1
−e
0
e−y dy
−y
i1
0
e
x −y
(−y + 1) dy +
Z ∞
1
e x−y · 0 dy
+
h
−e
−y
i1 −1
which is not uniformly continuous since e x is not uniformly continuous.
b. True.
Since f has compact support, f = 0 outside some compact interval [− N, N ]. Since f is
locally integrable, f is integrable on [− N, N ], so we have that
Z Z N f =
f = M f < ∞,
for some M f .
R
−N
Since compactly supported functions are dense in L1 (R
Z ), given ε > 0 there exists some
f ( x ) − h( x ) dx < ε.
continuous function h with compact support such that
R
4
Now, since g is bounded, there is some Mg such that | g| ≤ Mg . Then,
f ∗ g ( x1 ) − f ∗ g ( x2 ) =
=
≤
≤
=
=
≤
Z
Z
f ( x1 − y) g(y) dy −
f ( x2 − y) g(y) dy
R
R
Z f
(
x
−
y
)
−
f
(
x
−
y
)
g
(
y
)
dy
2
1
R
Z f ( x1 − y) − f ( x2 − y) g(y) dy
R
Mg
Mg
Mg
Mg
Z R
f ( x1 − y) − f ( x2 − y) dy
Z R
f (v) − f (v + u) (−dv), if u = x2 − x1 and v = x1 − y
Z R
"Z
f (v) − h(v) + h(v) − h(v + u) − h(v + u) − f (v + u) dv
Z h(v) − h(v + u) dv
f (v) − h(v) dv +
R
R
#
Z h(v + u) − f (v + u) dv
+
R
Z R h(v) − h(v + u) dv + Mg ε
<
Mg ε + Mg
<
2Mg ε + Mg (2R)ε
ε 2Mg (1 + R) .
=
−R
err... are we done?
if |u| < δ (since h is continuous)
5
6. Consider a function f : R → R which is periodic with period one and which satisfies
f ( x ) = x for | x | < 1/2.
a. Compute the Fourier series for f .
∞
b. Use your result from item (a) to compute
a. The Fourier series for f is
So, c0 =
Z
1
2
− 12
∑
1
.
n2
n =1
∑
cn e2πinx where cn =
n ∈Z
Z
1
2
− 12
f ( x )e−2πinx dx.
x dx = 0, and for n 6= 0,
cn =
1
2
Z
− 12
=
=
=
xe−2πinx dx
− xe−2πinx
2πin
12
−
Z
x =− 21
e−2πinx
− xe−2πinx
−
2πin
(2πin)2
1
2
− 12
e−2πinx
dx
−2πin
12
x =− 21
eπin
e−πin
eπin
−e−πin
−
−
+
4πin
4πin (2πin)2 (2πin)2
2i sin (πn)
2 cos (πn)
+
−4πin
(2πin)2
(−1)n
=
+ 0.
−2πin
=
So, the Fourier series for f is
(−1)n 2πinx
∑ −2πin e .
n∈Z,n6=0
b. Parseval’s Identity tells us that k f k L2 = k fˆkl 2 . Thus, k f k2L2 = k fˆk2l 2 . Well,
Z 1
(−1)n 2πinx 2
2
2
2
2
k f k L2 =
x dx
and
k fˆkl 2 =
∑ −2πin e − 12
n∈Z,n6=0
=
1
12
=
1
4π 2 n2
n∈Z,n6=0
=
2
∑
∞
∞
Thus,
1
∑ n2
n =1
=
π2
.
6
1
4π 2 n2
n =1
∑
6
7. Let f : R → R be a bounded Lebesgue measurable function, let K be a compact subset
Z
of R, and let
g(t) =
cos(tx ) f ( x ) dx, t ∈ R.
K
Prove or disprove:
a. g has compact support.
b. g is continuously differentiable.
a. False. Counterexample:
π
− sin(πt)
− sin(tx ) Let f ( x ) = χK , where K = [0, π ]. Then g(t) =
=
,
cos(tx ) dx =
t
t
0
0
so, g(t) = 0 for t ∈ Z \ {0}. Thus, the support of g is R \ Z \ {0} , which is not compact.
Z π
b. True.
First, let’s show that g is differentiable:
0
g (t)
=
=
R
R
g(t + h) − g(t)
K cos ( t + h ) x f ( x ) dx − K cos( tx ) f ( x ) dx
lim
= lim
h
h
h →0
h →0
Z
cos (t + h) x − cos(tx ) f ( x )
lim
dx, since K is compact (bounded).
h
h →0 K
Let {hn } be a sequence such that hn
the Mean Value Theorem
cos (t + h ) x − cos(tx ) n
=
hn
≤
→ 0 as n → ∞. Now, fix t ∈ R and x ∈ K. Then, by
for some c ∈ [t, t + hn ]
x sin(cx )
x ≤ sup x = M, since K is compact.
x ∈K
cos (t + h ) x − cos(tx ) n
Now, f ( x ) ≤ M f ( x ), and f ∈ L1 (K ), so, by the Lebesgue
hn
Dominated Convergence Theorem, we have
Z cos (t + h) x − cos(tx ) f ( x )
0
g ( x ) = lim
dx
h
h →0 K
Z cos (t + hn ) x − cos(tx ) f ( x )
= lim
dx
n→∞ K
hn
Z
cos (t + hn ) x − cos(tx )
=
lim
f ( x ) dx
hn
K n→∞
=
Z
K
− x sin(tx ) f ( x ) dx, which exists (and is finite) since K is compact.
Thus, g0 ( x ) exists.
7
Now, to show that g0 ( x ) is continuous.
Let {tn } be a sequence in R
such that tn → t0 as n → ∞. Then, we want to show that
0
0
lim g (tn ) = g (t0 ). Well, − x sin(tn x ) f ( x ) ≤ x f ( x ) and x f ( x ) ∈ L1 (K ). So, by the
n→∞
Lebesgue Dominated Convergence Theorem, we have
lim
n→∞
Z
K
− x sin(tn x ) f ( x ) dx
=
=
Z
lim − x sin(tn x ) f ( x ) dx
K n→∞
Z
K
=
Z
K
− x sin
lim tn x f ( x ) dx, since sin(tx ) is continuous
n→∞
− x sin(t0 x ) f ( x ) dx
=
g 0 ( t0 ).
Thus, g0 (t) is continuous. So, g(t) is continuously differentiable, as required.
8