REAL ANALYSIS QUALIFYING EXAMS
MINH KHA
Abstract. Here are some of my own solutions of recent qualifying exams of Real
Analysis in TAMU. For three exams Jan 2013, August 2012 and January 2012, I
type all full solutions. For previous exams before 2012, I type solutions of some
selected problems. Sometimes, there are some comments and similar exercises after
some problems. This is written for my Real Analysis Qualifying Exam Preparation
Course in Summer 2013 at TAMU.
1. January 2013
Problem 1: Let f be a Lebesgue integrable, real valued function on (0, 1) and for
x ∈ (0, 1) define
Z 1
g(x) =
t−1 f (t)dt
(1)
x
Z 1
Z 1
Show that g is Lebesgue integrable and
g(x)dx =
f (x)dx.
0
0
Proof. Since f (t)/t is measurable, we could approximate it by a sequence of simple
functions on (0, 1). Hence, g is Lebesgue measurable. Applying Fubini theorem for
t−1 |f (t)|,
Z
1
Z
1
Z
|g(x)|dx ≤
0
1
Z
−1
|t f (t)|dtdx =
0
x
1
Z
−1
t |f (t)|
0
t
dxdt = kf k1 < ∞
(2)
0
Thus g is Lebesgue integrable.
The identity (1) is proved by Fubini theorem again since g is integrable now.
Problem 2: Let fn ∈ C[0, 1]. Show that fn → 0 weakly if and only if the sequence
(kfn k)∞
n=1 is bounded and fn converges pointwise to 0.
Proof. fn → 0 weakly iff for any nonnegative Borel measure µ on [0, 1], then
Z 1
fn dµ → 0
(3)
0
. Moreover, fn converges pointwise to 0 iff for any x0 ∈ [0, 1], (3) is satisfied for
µ = δx 0 .
Z 1
1
∗
∞
Assume (3). Since L [0, 1] = L [0, 1], kfn k = sup |
fn gdx|. If we view
kgk1 ≤1
0
Z 1
1
each fn as a linear functional on L [0, 1] and since sup |
fn gdx| < ∞ for each
n
Date: 18:14 o’clock, 9 August 2013.
1
0
2
KHA
function g ∈ L1 [0, 1] (by (3)), the sequence (kfn k)∞
n=1 is bounded by Banach Steinhaus
theorem.
For the converse, i.e assume the sequence (kfn k)∞
n=1 is bounded and fn converges
pointwise to 0, by Lebesgue dominated convergence, we get (3).
Problem 3: Let (X, µ) be a measure space with 0 < µ(X) ≤ 1 and let f : X → R
be measurable. State the definition for kf kp for p ∈ [1, ∞]. Show that kf kp is a
monotone increasing function of p ∈ [1, ∞] and limp→∞ kf kp = kf k∞ .
Proof. By Hölder inequality, for p ≤ q, we get kf kp ≤ kf kq k1kr where r = pq/(q − p).
Since µ(X) ≤ 1, k1kr ≤ 1. Thus, kf kp ≤ kf kq . If kf k∞ < ∞ then limp→∞ kf kp
exists and is bounded above by kf k∞ . For any > 0, we pick a subsetZ U ⊂ X such
that µ(U ) > 0 and f (x) ≥ kf k∞ − for any x ∈ U . Therefore, kf kpp ≥
(kf k∞ − )p
U
or kf kp ≥ µ(U )1/p (kf k∞ − ). By letting p → ∞ first and then let → 0, we get
limp→∞ kf kp ≥ kf k∞ which yields the identity we wish to prove. If kf k∞ = ∞ then
we just repeat the above proof to show that limp→∞ kf kp = ∞.
Problem 4: (a) Is there a signed Borel measure µ on [0, 1] such that
Z
0
1
p (0) =
p(x)dµ(x)
(4)
0
for all real polynomials p of degree at most 19?
(b) Is there a signed Borel measure µ on [0, 1] such that
Z
0
p (0) =
1
p(x)dµ(x)
(5)
0
for all real polynomials p?
Proof. (a) Write p as
19
X
ai xi , where each ai ∈ R. Then if (4) is satisfied for some µ,
i=0
we get a1 =
19
X
Z
ai
1
Z
i
x dµ(x) for every tuple (a0 , . . . , a19 ). This means
0
i=0
Z
0 if i 6= 1 and
1
xi dµ(x) =
0
1
xdµ(x) = 1. To solve this, we think about the form of discrete
0
measures µ =
k
X
aj δxj where aj ∈ R, xj ∈ [0, 1]. Then
j=1
k
X
j=1
aj xij = δi,1
(6)
REAL ANALYSIS QUALIFYING EXAM 18:14 o’clock, 9 August 2013
for every i = 0, . . . , 19. For example, let xj

1 1
1 x1
A=
 ... ...
= 2−j , j = 0, . . . , 19.

··· 1
· · · x19 
. 
..
. .. 
···
1 x19
1
3
(7)
x19
19
Q
This is a Vandermonde matrix and hence its determinant det(A) = 0≤i<j≤19 (xj −xi )
is nonzero. So (6) A(a0 , . . . , a19 )T = (0, 1, 0, . . . , 0)T is solvable.
(b) Similarly, Rsuppose µ is a signed Borel measure such that (5) is hold. Then
1
since µ([0, 1]) = 0 dµ = 0, we get µ+ ([0, 1]) = µ− ([0, 1]) and by definition of signed
Borel measure, at least one of them is finite and thus
Z both of them are finite. Hence
1
|µ|([0, 1]) < ∞. This yields that the mapping p →
p(x)dµ(x) where the domain
0
are real polynomials can be extended continuously onto C[0, 1]]. However, if we take
fn (x) = 0 on [1/n, 1] and fn (0) = 1 such that fn is smooth on [0, 1] then we must have
the inequality |fn0 (0)| ≤ kµkkfn k by taking a C 1 -approximation (by real polynomials)
for fn . But this is clearly a contradiction since limn→∞ |fn0 (0)| = ∞.
Problem 5: Let F be the set of all real-valued functions on [0,1] of the form
1
f (t) = Qn
j=1 (t − cj )
(8)
for natural numbers n and for real numbers cj ∈
/ [0, 1]. Prove or disprove: for
all continuous, real-valued functions g and h on [0,1] such that g(t) < h(t) for all
t ∈ [0, 1], there is a function a ∈ spanF such that g(t) < a(t) < h(t) for all t ∈ [0, 1].
Proof. Consider the subset A = spanF in C[0, 1]. From the definition of F, A is a
1
subalgebra. Furthermore, given distinct points x 6= y in [0, 1], the function t → t+1
in the family F separates x and y. By Stone-Weierstrass theorem (e.g see theorem
4.45 in [2]), the closure of A is either all C[0, 1] or {f ∈ C[0, 1] : f (x0 ) = 0} for
some x0 ∈ [0, 1]. However, given any x0 ∈ [0, 1], the evaluation of the element
1
∈ F at x0 is nonzero. Thus, A is dense in C[0, 1]. It’s clear that the subset
t+1
K = {a(t) : g(t) < a(t) < h(t), t ∈ [0, 1]} of C[0, 1] is open (i.e given any a0 ∈ K, the
open ball B(a0 , ) ⊂ K, where 0 < < 12 minx∈[0,1] {(a0 (x) − g(x)), (h(x) − a0 (x))}).
Thus, given any such pair (g(t), h(t)), we can find an element a ∈ A ∩ K.
Problem 6: Let k : [0, 1] × [0, 1] → R be continuous and let 1 < p < ∞. For
f ∈ Lp [0, 1], let T f be the function on [0, 1] defined by
Z 1
(T f )(x) =
k(x, y)f (y)dy
(9)
0
Show that T f is a continuous function on [0, 1] and that the image under T of the
unit ball in Lp [0, 1] has compact closure in C[0, 1].
Proof. Let q = p/(p − 1). By Hölder inequality, we have
Z 1
|(T f )(x) − (T f )(y)| ≤
|k(x, z) − k(y, z)|.|f (z)|dz ≤ kk(x, .) − k(y, .)kq .kf kp (10)
0
4
KHA
Since k is uniformly continuous on the unit square, when x ≈ y, kk(x, .) − k(y, .)kq
is small. Thus, T f is a continuous function on [0, 1]. For the later part, the family
F = {T f : kf kp ≤ 1} ⊂ C[0, 1] is equicontinuous by (10). Moreover, it is also
pointwise bounded because |T f (x)| ≤ maxy∈[0,1] |k(x, y)|.kf kp ≤ maxy∈[0,1] |k(x, y)|
for each x ∈ [0, 1]. By Ascoli-Arzela theorem (see [2]), F is totally bounded and its
closure is compact in C[0, 1].
Problem 7: (a) Define a total variation of a function f : [0, 1] → R and absolute
continuity of f . (b) Suppose f : [0, 1] → R is absolutely continuous and define
g ∈ C[0, 1] by
Z 1
f (xy)dy
(11)
g(x) =
0
Show that g is absolutely continuous.
Proof. To prove g is absolutely continuous, we need to show that ∀ > 0, there exists
a δ > 0 such that for any finite set of disjoint subintervals (a1 , b1 ), . . . , (an , bn )
n
X
|bi − ai | < δ ⇒
i=1
n
X
|g(bi ) − g(ai )| < (12)
i=1
Because f is absolutely continuous, for such an > 0 we can find a suitable δ > 0 such
that the above condition (12) satisfies for f instead of g. When we fix any y ∈ [0, 1]]
and consider P
the corresponding finite set of disjoint
P intervals (a1 y, b1 y), . . . , (an y, bn y),
we still have ni=1 |bi y − ai y| < δy ≤ δ and thus ni=1 |f (bi y) − f (ai y)| < . Therefore
Z 1X
Z 1
n
n
X
|g(bi ) − g(ai )| ≤
|f (bi y) − f (ai y)|dy < dy = (13)
i=1
0
i=1
0
Problem 8: (a) State the definition of absolute continuity, ν µ, for positive
measures ν and µ, and state the Radon-Nikodym theorem. (b) Suppose that we have
ν1 µ1 and ν2 µ2 for positive measures νi and µi on measurable spaces (Xi , Mi )
(i = 1, 2). Show that we have ν1 × ν2 µ1 × µ2 and
d(ν1 × ν2 )
dν1
dν2
(x, y) =
(x)
(y)
d(µ1 × µ2 )
dµ1
dµ2
(14)
Proof. (a) ν µ if and only if when E ∈ M satisfies µ(E) = 0, then ν(E) = 0.
The Radon-Nikodym theorem states that for σ-finite positive measure µ and σ-finite
(signed) measure ν on (X, M), there exists σ-finite measures λ and ρ such that
ν = λ + ρ and λ ⊥ µ and ρ µ. Moreover there is uniquely (µ-a.e) an extended
µ-integrable function f : X → R such that dρ = f dµ.
(b) Assume E ∈ M1 ⊗ M2 such that (µ1 × µ2 )(E) = 0. Define sections Ex = {y ∈
X2 , (x, y) ∈ E} and E y = {x ∈ X1 , (x, y) ∈ E} then Ex ∈ M2 and ZE y ∈ M1 for all
x ∈ X1 , y ∈ X2 . By Fubini-Tonelli, we can write 0 = µ1 × µ2 (E) =
µ1 (E y )dµ2 (y).
Since µ1 , µ2 are positive measures, µ1 (E y ) = 0 for µ2 -a.e y ∈ X2 . Thus, ν1 (E y ) = 0
for µ2 -a.e y ∈ X2 by assumption. Note that the set {y ∈ X2 : ν1 (E y ) > 0} is
REAL ANALYSIS QUALIFYING EXAM 18:14 o’clock, 9 August 2013
5
measurable by Theorem 2.36, p.66 in [2]. Since µ2 {y ∈ X2 : ν1 (E y ) > 0} = 0, we
y
must have ν2 {y
Z ∈ X2 : ν1 (E ) > 0} = 0 by the asbolute continuity. Therefore,
ν1 (E y )dν2 (y) = 0. This shows that ν1 × ν2 µ1 × µ2 .
Z
d(ν1 × ν2 )
By Radon-Nikodym theorem, we can write ν1 × ν2 (E) =
(x, y)d(µ1 ×
E d(µ1 ×Zµ2 )
µ2 ) for E ∈ M1 ⊗ M2 . Again by Fubini-Tonelli, ν1 × ν2 (E) =
ν2 (Ex )dν1 (x).
Z
Moreover, by Proposition (3.9) (a) in [2], since νi µi we have
ν2 (Ex )dν1 (x) =
Z
Z
Z Z
dν1
dν2
dν1
ν2 (Ex )
(x)dµ1 = ν2 (Ex )dν1 (x) = (
(y)dµ2 )
(x)dµ1 .
dµ1
dµ1
Ex dµ
Z Z
Z2
dν2
dν2
dν1
dν1
Furthermore, (
(y)dµ2 )
(x)dµ1 =
(y)
(x)d(µ1 × µ2 ). The
dµ1
dµ1
Ex dµ2
E dµ2
uniqueness of the Radon-Nikodym derivative yields the indetity we want to prove. ν1 × ν2 (E) =
Problem 9: (a) Let E be a nonzero Banach space and show that for every x ∈ E
there is φ ∈ E ∗ such that kφk = 1 and |φ(x)| = kxk.
(b) Let E and F be Banach spaces, let π : E → F be a bounded linear map and
let π ∗ : F ∗ → E ∗ be the induced map on dual spaces. Show that kπk = kπ ∗ k.
Proof. (a) Use Hahn-Banach theorem for the linear functional T (λx) = λkxk . . .
(b) Explicitly, π ∗ (y ∗ )(x) = y ∗ (π(x)) with y ∗ ∈ F ∗ , x ∈ E. Hence, kπ ∗ (y ∗ )(x)k ≤
ky ∗ k.kπk.kxk, so kπ ∗ (y ∗ )k ≤ ky ∗ k.kπk or kπ ∗ k ≤ kπk. For the reverse, use (a), for
each x ∈ E such that kxk ≤ 1 we could find y ∗ ∈ F ∗ such that y ∗ (π(x)) = kπ(x)k
and ky ∗ k = 1. Then kπ ∗ k ≥ kπ ∗ (y ∗ )k ≥ kπ ∗ (y ∗ )(x)k = kπ(x)k. Taking supremum
over the unit ball of E, we get kπ ∗ k ≥ kπk.
Problem 10: Let X be a real Banach space and suppose C is a closed subset of X
such that
(i) x1 + x2 ∈ C for all x1 , x2 ∈ C.
(ii) λx ∈ C for all x ∈ C and λ > 0.
(iii) For all x ∈ X there exists x1 , x2 ∈ C such that x = x1 − x2 .
Prove that for some M > 0, the unit ball of X is contained in the closure of
{x1 − x2 | xi ∈ C, kxi k ≤ M, i = 1, 2}
(15)
Deduce that every x ∈ X can be written x = x1 − x2 with xi ∈ C and kxi k ≤ 2M kxk
for i = 1, 2.
Proof. By (iii), X = ∪M >0 CM , where each CM is the closure of {x1 − x2 | xi ∈
C, kxi k ≤ M, i = 1, 2}. By Baire category theorem, there exists an M > 0 such that
int(CM ) is non-empty. Hence we can find x0 ∈ X and r > 0 such that the open
ball B(x0 , r) is contained in CM . For any x in the unit ball of X, rx + x0 ∈ CM .
The difference set CM − CM is clearly contained in C2M by (i). Thus, rx ∈ C2M and
by (ii) we have x ∈ C2M/r and this shows the first assertion. Hence, there exists an
M > 0 such that for any x ∈ X, x ∈ CM kxk by (ii). First, we find z1 , y1 ∈ C such that
kz1 k, ky1 k ≤ M kxk and kx − (z1 − y1 )k ≤ 2−1 kxk. Inductively, we can find sequences
6
KHA
Pn
−n+1
−n
{zn }, {yn } in C such
that
kz
k,
ky
k
≤
2
M
kxk
and
kx−
kxk.
n
n
i=1 (zi −yi )k ≤ 2
P∞
P∞
Thus the series i=1 zi , i=1 yi converge to elements
Px1 , x2−jrespectively. Since C is
closed and (i), x1 , x2 ∈ C. Moreover, kxi k ≤ M kxk ∞
= 2M kxk for i = 1, 2.
j=0 2
Pn
Also, x = limn→∞ ( i=1 (zi − yi )) = x1 − x2 .
Comment: The above proof is inspired from the standard proof of the open mapping
theorem in functional analysis. Indeed, by modifying constants, the final assertion of
Problem 10 is still true for any K > M instead of 2M .
2. August 2012
Problem 2: Fix two measure spaces (X, M, µ), (Y, N , ν) with µ(X), ν(Y ) > 0. Let
f : X → C, g : Y → C be measurable. Suppose f (x) = g(y)(µ ⊗ ν) − a.e. Show that
there is a constant a ∈ C such that f (x) = a, µ − a.e and g(y) = a, ν − a.e
Proof. Let E = {(x, y) ∈ X × Y : f (x) = g(y)}. By problem 51 (a) p.68 in [2], it
follows that the functions m(x, y) = f (x), n(x, y) = g(y) are µ⊗ν-measurable. Thus,
E = {(x, y) : m(x, y) = n(x, y)} is also measurable. By assumption, (µ ⊗ ν)(E c ) = 0.
It is clear that (µ ⊗ ν){(x, y) : f (x) = a, g(y) 6= a} = 0. By Fubini-Tonelli theorem,
this means µ{x ∈ X : f (x) = a}.ν{y ∈ Y : g(y) 6= a} = 0. If we could show that
there exists at least one a ∈ C such that µ{x ∈ X : f (x) = a} > 0 then ν{y ∈ Y :
g(y) 6= a} = 0 or g(y) = a for ν-a.e and hence, ν{y ∈ Y : g(y) = a} = ν(Y ) > 0.
Similarly, µ{x ∈ X : f (x) = 6= a}.ν{y ∈ Y : g(y) = a} = 0 or this yields f (x) = a
for µ-a.e
So suppose for contradiction, µ{x ∈ X : f (x) = a} = 0 for all a ∈ C. By Fubini
again:
Z
0 < (µ ⊗ ν)(X × Y ) = (µ ⊗ ν)(E) =
χ{(x,y)∈X×Y :f (x)=g(y)} dµ(x)dν(y)
X×Y
Z Z
= ( χ{x∈X:f (x)=g(y)} dµ(x))dν(y) = 0.
Y
X
Problem 3: Let f : R3 → R be a Borel measurable
function. Suppose for every
R
ball B, f is Lebesgue integrable on B and B f (x)dx = 0. What can you deduce
about f ? Justify your answer carefully.
Proof. By the Lebesgue Differentiation Theorem ([2]), since f is locally integrable on
R3 , for almost every x0 ∈ R3
Z
1
lim
f (x)dx = f (x0 )
(16)
r→0 |Br (x0 )| B (x )
r 0
and hence by our assumption, f (x0 ) = 0 for a.e x0 . In other words, f = 0-a.e.
Problem 4: Let X be a locally compact Hausdorff space. Denote by C0 (X) the
space of complex-valued continuous functions on X which vanish at infinity, and by
Cc (X) the subset of compactly supported functions. Use an appropriate version of
the Stone-Weierstrass theorem to prove that Cc (X) is dense in C0 (X).
REAL ANALYSIS QUALIFYING EXAM 18:14 o’clock, 9 August 2013
7
Proof. For convenience, we recall the complex version of the Stone-Weierstrass theorem for noncompact LCH space X (Theorems (4.51) and (4.52) in [2]): If A is a
closed subalgebra of C0 (X, C) which separates points and is closed under complex
conjugation, then either A = C0 (X, C) or A = {f ∈ C0 (X, C) : f (x0 ) = 0} for some
x0 ∈ X.
It is clear that given two compactly supported functions f, g ∈ Cc (X), its product
f g has compact support too. Hence, Cc (X) is a subalgebra of C0 (X). Furthermore,
given two distinct points x 6= y in X, first we pick a precompact open subset U such
that {x} ⊂ U ⊂ Ū ⊂ X −{y}. By Urysohn lemma (the locally compact version (4.32)
in [2]), we can always find a continuous function f : X → [0, 1] such that f (x) = 1
and f = 0 outside a compact subset of the open subset U . So f ∈ Cc (X) and
f (y) = 0. In other words, Cc (X) separates points of X. Obviously, Cc (X) is closed
under complex conjugation since given any f ∈ Cc (X), the complex conjugation of
f has compact support too.
Finally, Cc (X) * {f ∈ C0 (X, C) : f (x0 ) = 0} for any x0 ∈ X. Hence the closure
of Cc (X) is C0 (X) by Stone-Weierstrass theorem.
Exercise: Given a locally compact group G, for each f ∈ C0 (G), define f˜ ∈
Cb (G × G) by mapping each pair (p, q) ∈ G × G to f (pq). Prove that the ideal
generated by {f˜g : f ∈ C0 (G), g ∈ C0 (G × G)} is dense in C0 (G × G).
Problem 5: Give an example of each of the following. Justify your answers. (a) A
nowhere dense subset of R of positive Lebesgue measure. (b) A closed, convex subset
of a Banach space with multiple points of minimal norm.
Proof. (a) The typical example is the Fat Cantor set (or the Smith-Volterra-Cantor
set, for more information, see [5]), i.e the complement in [0, 1] of the following set
[
a
1
a
1
A=
( n − 2n+1 , n + 2n+1 ).
(17)
2
2
2
2
n
n
a,n∈N;gcd(a,2 )=1;a∈[0,2 ]
P∞ −2n−1 n
Hence m(A) = n=0 2
.2 = 1/2. So the measure of the Fat Cantor set is at
least 1/2.
Moreover, it is nowhere dense. Otherwise, since the dyadic rational numbers are
dense in [0, 1], as the Fat Cantor set is closed, it must have empty interior since it
contains no dyadic rational.
(b) Consider the Banach space X = L1 [0, 1], relative to the Lebesgue Zmeasure, and
the closed, convex subset in X is the set of all functions C = {f ∈ X :
f (t)dt =
[0,1]
Z
1}. Then the minimal norm of elements in C is 1 since kf k1 ≥ |
f (x)dx|. Each
[0,1]
member of the infinite family of functions {aχ[1/2,1] + (2 − a)χ[0,1/2] }0≤a≤2 in C has
norm 1.
Comment: For things related to question (b), the Day-James theorem states that
every closed and convex subset in a normed space X has a unique point of minimal
norm iff X is strictly convex and reflexive (hence, X is complete). Note that X is
8
KHA
strictly convex iff x 6= 0, y 6= 0, kx + yk = kxk + kyk together implies x = cy for some
constant c > 0.
Exercise: Instead of question (b), can you construct an example of a closed and
convex subset of a Banach space X with no point of minimal norm?
Problem 6:
1
Let S = {f ∈ L∞ (R) : |f (x)| ≤ 2
a.e}. Which of the following statements are
x +1
true? Prove your answers.
(a) The closure of S is compact in the norm topology.
(b) S is closed in the norm topology.
(c) The closure of S is compact in the weak-* topology.
1
χ[n−1 ,(n+1)−1 ] (x), n ∈ N. Thus, fn ∈ S.
+1
However, no subsequence of {fn } is a Cauchy sequence in L∞ (R).
(b) Yes. Suppose fn ∈ S such that fn → f in L∞ (R). f ∈ S because |f (x)| ≤
|fn (x)| + kfn − f k∞ , ∀x ∈ R, n ∈ N.
(c) Yes. Indeed, the unit ball of L∞ (R) is weak-* compact by Banach-Alaoglu
theorem. Note that (x2 + 1)−1 ≤ 1 for all x, S is a subset of the unit ball of L∞ (R).
Hence, its closure in the weak-* topology should be weak-* compact.
Proof. (a) No. For example, let fn (x) =
x2
Problem 7: Let T be a bounded operator on a Hilbert space H. Prove that
kT ∗ T k = kT k2 . State the results you are using.
Proof. We want to prove kT ∗ T k ≥ kT k2 since the reverse side is obvious. Note that
for a bounded operator on a Hilbert space, we have kT k2 = sup{|hT x, T xi| : kxk ≤
1} = sup{|hT ∗ T x, xi| : kxk ≤ 1} ≤ kT ∗ T k.
Problem 8: (a) Let g be an integrable function onZ [0, 1]. Does there exist a bounded
1
f g = kgk1 kf k∞ ? Give a con-
measurable function f such that kf k∞ 6= 0 and
0
struction or a counterexample.
(b) Let g be a bounded measurable Zfunction on [0, 1]. Does there exist an integrable
1
function f such that kf k1 6= 0 and
f g = kgk∞ kf k1 ? Give a construction or a
0
counterexample.
Proof. (a) Yes. Given such g, let f be the following function
|g(x)|
, g(x) 6= 0
g(x)
= 1,
g(x) = 0
f (x) =
Z
Then f is a bounded (kf k∞ = 1) measurable function and
1
Z
0
1
|g| =
fg =
0
kgk1 kf k∞ .
(b) Not always. For example, let g(x) = x for all x ∈ [0, 1]. Suppose we could find
such a function f .
REAL ANALYSIS QUALIFYING EXAM 18:14 o’clock, 9 August 2013
Z
Z
f (x)xdx |≤ (1 − 1/n)
Thus kf k1 =|
[0,1]
Z
|f (x)|dx +
[0,1−1/n]
9
|f (x)|dx or
[1−1/n,1]
Z
|f (x)|dx = 0. Let n → ∞, kf k1 = 0 by the monotone converging theorem.
[0,1−1/n]
Problem 9: Let F : R → C be a bounded continuous function, µ the Lebesgue
measure, and f, g ∈ L1 (µ). Let
Z
Z
f˜(x) = F (xy)f (y)dµ(y), g̃(x) = F (xy)g(y)dµ(y)
(18)
Show that f˜ and g̃ are bounded continuous functions which satisfy
Z
Z
f g̃dµ = f˜gdµ
(19)
Proof. It’s obvious that kf˜k∞ ≤ kF k∞ kf k1 and kg̃k∞ ≤ kF k∞ kgk1 . Hence, they are
bounded functions.
Z
By the dominated converging theorem, lim
|f (x)|dµ = kf k1 . Hence for
n→∞ [−n,n]
Z
|f (x)|dµ is arbitrarily small. For any > 0, we find n
n large enough,
R\[−n,n]
Z
˜
˜
large enough so that |f (x) − f (y)| ≤ |F (xz) − F (yz)|.|f (z)|dµ ≤ sup |F (xz) −
z∈[−n,n]
F (yz)|kf k1 + 2kF k∞ . By the uniform continuity of F on the compact interval
[−n, n], f˜ is continuous (and g̃ too).
Since (f g̃)x is integrable for almost everywhere x ∈ R, Fubini-Tonelli gives
Z
Z Z
Z
Z
f g̃dµ =
f (x)F (xy)g(y)dµ(y)dµ(x) = g(y) f (x)F (xy)dµ(x)dµ(y)
Z
= g f˜dµ
Problem 10: Let µ, {µn : n ∈ N} be finite Borel measures on [0, 1]. µn → µ vaguely
∗
if it converges in the
Z weak-* topologyZ on M [0, 1] = (C[0, 1]) . µn → µ in moments if
xk dµn (x) →
for each k ∈ 0 ∪ N,
[0,1]
xk dµ(x). Show that µn → µ vaguely if and
[0,1]
only if µn → µ in moments.
Proof. The vague convergence obviously implies the convergence Zin moments. Assume µn → µ in moments, we need to show that for any f ∈ C[0, 1]:
f (x)dµn (x) →
[0,1]
Z
f (x)dµ(x). By Weierstrass theorem, let {Pn } be the sequence of polynomials
[0,1]
that converges uniformly to f on [0, 1].
10
KHA
Z
Z
|
Z
f (x)dµ(x) −
Z
f (x)dµn (x)| ≤ |
Pm (x)dµ(x) −
[0,1]
[0,1]
[0,1]
Z
Z
Z
f (x)dµ(x) −
+|
[0,1]
Pm (x)dµ(x)| + |
[0,1]
Z
f (x)dµn (x) −
[0,1]
Pm (x)dµn (x)|
[0,1]
Z
Z
Pm (x)dµ(x) −
=|
Pm (x)dµn (x)|
[0,1]
[0,1]
Pm (x)dµn (x)| + kf − Pm k∞ (µ([0, 1]) + µn ([0, 1]))
[0,1]
We have µn [0, 1] → µ[0, 1] by the convergence in moment (k = 0). Hence, there
is some M > 0 such that (µ([0, 1]) + µn ([0, 1])) ≤ M, ∀n. ∀ > 0, we find m > 0
such
Z that kf − Pm k∞Z≤ /2M and then for such m, choose n large enough such that
|
Pm (x)dµ(x) −
Pm (x)dµn (x) |< /2.
[0,1]
[0,1]
Z
Z
This means |
f (x)dµ(x) −
f (x)dµn (x)| < , and hence gives us the con[0,1]
[0,1]
clusion.
3. January 2012
Problem 1: Let A be the subset of [0, 1] consisting of numbers whose decimal
expansions contain no sevens. Show that A is Lebesgue measurable, and find its
measure. Why does non-uniqueness of decimal expansions not cause any problems?
Proof. Let Ai , i ∈ N be the subset of [0, 1] consisting of numbers
whose first i digits
T
in decimal expansions are not 7. Then An+1 ⊂ An , A = n∈N An . It’s not hard to
see that each An is a finite union of 2n Borel intervals in [0, 1], and hence they are
Lebesgue measurable. Thus A is Lebesgue measurable. The remaining work is to
compute the Lebesgue measure of A, i.e m(A).
It suffices to compute each m(An ) since m(A) = limn→∞ m(An ). Recall the definition of each An , an element in An+1 is also in An and its n + 1 digit is not
7. Let Ain , P
i = 0, . . . , 9 be the subset of An such that its first digit is i. Then
m(An+1 ) = i6=7 m(Ain+1 ). For each i, under the linear transform fi : x → 10x − i,
Ain+1 is mapped bijectively onto An . By Lebesgue theory, we must have m(Ain+1 ) =
9
9n+1
m(An )/10. So m(An+1 ) = m(An ) = . . . = n+1 . Thus m(A) = 0.
10
10
Problem 2: Let the functions fa be defined by
(
1
xa cos( ) x > 0
fa (x) =
x
0
x=0
Find all values a ≥ 0 such that
(a) fa is continuous
(b) fa is bounded of variations on [0, 1]
(c) fa is absolutely continuous on [0, 1]
REAL ANALYSIS QUALIFYING EXAM 18:14 o’clock, 9 August 2013
11
1
Proof. (a) xa cos( ) ≤ |x|a . Hence {a ∈ R : a > 0} is the answer.
x
(b) When x > 0, fa0 (x) = axa−1 cos(x−1 ) + xa−2 sin(x−1 ). By mean value theorem:
|fa (x + h) − fa (x)| ≤ h(axa−1 + xa−2 ) . hxa−2
(20)
Consider the case a > 1 (Another way to prove fa is of bounded variation is to use
the proof of (c), in fact fa is even absolutely continuous in this case)
For any partition P = {0 = x0 < x1 < . . . < xn = 1}, the variation Tfa (P) is
bounded by
Z
n−1
n−1 Z
X
X
a−2
a−2
(xj+1 − xj )xj ≤
x dx =
xa−2 dx < ∞
j=0
j=0
[xj ,xj+1 ]
[0,1]
We will prove if 0 < a ≤ 1 then fa is of unbounded variation on [0, 1].
To see this, for each m ∈ N, consider the following partition
1
1
1
Pm = {0,
,
, . . . , , 1}
π2m π(2m − 1)
π
The values of fa at these points of this partition are
1
−1
−1
fa (Pm ) = {0,
,
,...,
, cos(1)}
a
a
(π2m) (π(2m − 1))
(π)a
For this partition, the variation of fa is:
1
−1
1
−1
1
−1
Tfa (Pm ) = |
− 0| + |
−
| + ... + | a −
| + |cos(1) −
|
a
a
a
a
(π2m)
(π(2m − 1))
(π2m)
(π)
(π2)
(π)a
2m
X
1
≈(
) → ∞.
a
(πj)
j=1
So the answer is {a ∈ R : a > 1}.
(c)fa is absolutely continuous if and only if fa0 is integrable over [0, 1] by the
Lebesgue fundamental calculus theorem (e.g see theorem 3.11 in [3]). When a > 1,
a trivial calculation shows that fa0 is integrable on [0, 1] and fa0 (0) = 0. Of course, if
a ≤ 1, by (b) fa is not of bounded variation, and hence it could not be absolutely
continuous.
Problem 3: Let F denote the family of functions on [0, 1] of the form
f (x) =
∞
X
an sin(nx)
(21)
n=1
−3
where an are real and |an | ≤ n . State a general theorem and use that theorem to
prove that any sequence in F has a subsequence that converges uniformly on [0, 1].
Proof. Since |an | ≤ n−3 , F ⊂ C[0, 1]. We use Ascoli-Arzela theorem for this problem.
|
∞
X
n=0
an sin(nx)| ≤
∞
X
n=0
n−3 < ∞, ∀x
(22)
12
KHA
So the family F is uniformly bounded. By mean value theorem, ∀ > 0, for any f as
in (21), x 6= y:
∞
X
|f (x) − f (y)| ≤
|an ||sin(nx) − sin(ny)|
n=1
=
∞
X
2n−3 |cos((nx + ny)/2)|.|sin((nx − ny)/2)|
n=1
≤ |x − y|
∞
X
n−2
n=1
Thus, F is equicontinuous. We are done.
Problem 4: A basic theorem in textbooks. We skip this.
Problem 5: Suppose A is a bounded linear operator on a Hilbert space H with
the property that kp(A)k ≤ C sup{|p(z)| : z ∈ C, |z| = 1} for all polynomials p with
complex coefficients, and a fixed constant C. Show that to each pair x, y ∈ H, there
corresponds a complex Borel measure µ on the circle S 1 such that
Z
n
hA x, yi =
z n dµ(z), n = 0, 1, 2, . . .
(23)
S1
Proof. Consider the following linear mapping
Tx,y : P(S 1 ) → C
P 7→ hP (A)x, yi
where P(S 1 ) is the space of polynomials on S 1 . By assumption, |hP (A)x, yi| ≤
Ckpk∞ kxk.kyk. Hence for pair x, y ∈ H, Tx,y could be extended continuously on
C(S 1 ). Applying the Riesz Representation Theorem for Tx,y , there exists a complex
Borel measure µ on the circle such that (23) is hold.
Problem 6: Let φ be the linear functional
Z
1
φ(f ) = f (0) −
f (t)dt
(24)
−1
(a) Compute the norm of φ as the linear functional on C[−1, 1] with the uniform
norm.
(b) Compute the norm of φ as the linear functional on LC[−1, 1], i.e C[−1, 1] with
the L1 norm.
Proof. (a)
Z
1
|φ(f )| ≤ |f (0)| +
|f (t)|dt ≤ 3kf k∞ ⇒ kφk ≤ 3
−1
For showing kφk = 3, we choose a sequence of piecewise linear functions fn whose
−1 1 −1 1 1 1 1 1
values at −1,
− ,
+ , − , + , 1 are 1, 1, −1, −1, 1, 1 respectively. Then
2 Zn 2
n 2 n 2 n
1
kfn k = 1 = −f (0),
fn (t)dt ≈ 2.
−1
REAL ANALYSIS QUALIFYING EXAM 18:14 o’clock, 9 August 2013
13
(b) f can be chosen so that |f (0)| is much larger than kf k1 , e.g ∀n ∈ N let f
be the linear piecewise function whose values at −1, −1/n, 0, 1/n, 1 are 0, 0, 1, 0, 0
respectively. Hence kφk = ∞.
Exercise: For question 6(a), explain in details why the norm kφk could not be attained its maximum 3 in C[−1, 1].
Problem 7: Let X be a normed space, and A ⊂ X a subset. Show that A is
bounded (as a set) if and only if it is weakly bounded (that is, f (A) ⊂ C is bounded
for each f ∈ X ∗ ).
Proof. It is enough to prove if A is weakly bounded then A is bounded. Indeed,
∀a ∈ A, a = sup{|f (a)| : f ∈ X ∗ , kf k ≤ 1} by Hahn-Banach theorem. On X ∗∗ ,
we consider the family A = {a∗∗ : a ∈ A}, where a∗∗ (f ) = f (a), ∀f ∈ X ∗ . By
the Uniform Boundedness Principle, A is normed bounded iff for each f ∈ X ∗ ,
{a∗∗ (f ) : a ∈ A} is bounded. This latter condition is just the boundedness of
{f (a) : a ∈ A} = f (A).
Problem 8: Let X be a topological vector space. (a) Define what this means.
(b) Let A ⊂ X be compact and B ⊂ X be closed. Show that A + B ⊂ X is closed.
(c) Give an example indicating that the condition A closed is insufficient for the
conclusion.
√
√
Proof. (c) A = Z, B = 2Z then A + B = Z + 2Z √
is not closed. Our claim for
this is that A + B would be dense in R and certainly 3 ∈
/ A + B. The density of
this set comes from the Dirichlet Approximation Theorem: For any real number α
and any positive integer N , there exists integers p and q such that 1 ≤ q ≤ N and
|p − αq| ≤ (N + 1)−1 .
Another example is that A = {n : n = 1, 2, . . .}, B = {−n + 1/n : n = 2, 3, . . .}
then A + B contains {1/n : n ≥ 2} but 0 ∈
/ A + B.
Problem 9: Let (X, M, µ) be a finite measure space. Let f, fn ∈ L3 (X, dµ) for
n ∈ N be functions such that fn → f µ − a.e and |fn | ≤ M, ∀n. Let g ∈ L3/2 (X, dµ).
Show that
Z
Z
lim
fn gdµ =
f gdµ
(25)
n→∞
X
X
Proof. This is just a simple application of the Lebesgue Dominated Convergence.
Note that ∀n ∈ N, |fn g| ≤ M |g|. Since g ∈ L3/2 (X, dµ) and µ(X) < ∞, we must have
g ∈ L1 (X, dµ) by Hölder inequality: kgk1 ≤ µ(X)1/3 kgk3/2 . The rest is routine. Problem 10: Let (X, M, µ) be a σ-finite measure space, and fn : X → R a sequence
of measurable functions on it. Suppose fn → 0 in L2 , L4 .
(a) Does fn → 0 in L1 ? Give a proof or a counterexample.
(b) Does fn → 0 in L3 ? Give a proof or a counterexample.
(c) Does fn → 0 in L5 ? Give a proof or a counterexample.
Proof. (a) No. Take X = R, µ: Lebesgue measure, fn = n−1 χ[0,n] . Then kfn k1 = 1
while kfn k2 = n−1/2 , kfn k4 = n−3/4 .
14
KHA
(b) Yes. We use the log convexity of Lp -norms (see e.g Lemma 2 in [4]):
1/3
2/3
kfn k3 ≤ kfn k2 kfn k4 .
(c) No. Take X = [0, 1], µ: Lebesgue measure, fn = nχ[0,n−5 ] . Then kfn k5 = 1
while kfn k2 = n−3/2 , kfn k4 = n−1/4 .
4. August 2011
Problem 2
Proof. Let {xn : n ∈ N} be a dense sequence in X. For each integer n, we pick
x∗n ∈ X ∗ such that kx∗n k = 1 and x∗n (xn ) = kxn k. Now we consider the following
linear map
T : X → `∞
x 7→ (x∗n (x))n∈N ,
x∈X
Obviously, kxk ≥ |x∗n (x)|. On the other hand, ∀ > 0, we pick n ∈ N such that
kxn − xk < . Then, |x∗n (x)| = |x∗n (x − xn ) + kxn k| ≥ kxn k − |x∗n (x − xn )| ≥ −2 + kxk.
Let → 0, kxk = supn∈N (|x∗n (x)|), or T is isometric linear.
Suppose for any separable Banach space X, there is always an isometric linear map
T : X → `2 . This is false if we let X = `1 . Indeed, the isometry map T : `1 → `2
would yield that kx+yk21 +kx−yk21 = kT x+T yk22 +kT x−T yk22 = 2(kT xk22 +kT yk22 ) =
2(kxk21 + kyk21 ), ∀x, y ∈ `1 or `1 is a Hilbert space. This is a contradiction since
simply take x = (1, 0, 2−2 , 0, . . . , 0, n−2 , 0, . . .), y = (0, 1, 0, 2−2 , . . . , 0, n−2 , 0, . . .) as
an example.
Problem 3
P
Proof. Let’s denote yn = n−1 nk=1 δxk for each n ∈ N. Then yn converges to 0
in the weak*-topology
on C0 (X)∗ if and only if yn (f ) → 0, ∀f ∈PC0 (X). Since
P
yn (f ) = n−1 nk=1 f (xk ), this is equivalent to the convergence of (n−1 nk=1 f (xk ))n∈N
to 0, for each f ∈ C0 (X). By Cesaro mean theorem, it suffices to prove f (xn ) → 0.
Suppose for contradiction, f (xn ) does not converge to 0. Hence, we can find an > 0
and a subsequence {xnk } such that |f (xnk )| ≥ , ∀k. For such f we find a compact
subset K ⊂ X such that |f (z)| < , ∀z ∈
/ K. This means {xnk } ⊂ K, ∀k. The
compactness of K implies there is a subsequence of {xnk } that converges in X, which
is a contradiction with our assumption.
Problem 4
Proof. Let C = {f ∈ C 1 [0, 1] : f (0) = f 0 (0) = 0}. Our claim is P is dense in
C. Indeed, fix an element f ∈ C. By Weierstrass theorem, we approximate f by
a sequence of polynomials {pn } in C 1 [0, 1]. Hence, pn (0), p0n (0) → 0. Let hn (t) =
pn (t) − pn (0) − tp0n (0) so hn (0) = h0n (0) = 0. Thus, kf − hn kC 1 = kf − hn k∞ + kf 0 −
h0n k∞ ≤ kf − pn k∞ + kf 0 − p0n k∞ + kpn − hn k∞ + kp0n − h0n k∞ ≤ kf − pn kC 1 + kpn (0) +
tp0n (0)k∞ + |p0n (0)|. So, hn ∈ P is a C 1 -approximation of f . By the same proof P
is dense in E = {f ∈ C[0, 1] : f (0) = f 0 (0) = 0}. Consider the case p < ∞. The
claim reduces the problem into another equivalent problem: Show that in this case,
C is dense in (C 1 [0, 1], k.kp ) (because the Lp -norm on [0, 1] is dominated by the C 1
REAL ANALYSIS QUALIFYING EXAM 18:14 o’clock, 9 August 2013
15
norm and C 1 [0, 1] is dense in Lp [0, 1], e.g Proposition 8.17 in [2]). Given any function
f ∈ C 1 [0, 1] and ∀ > 0, let g be the following function
(
f (x)
x > n−1
g(x) =
(n2 f 0 (n−1 ) − 2n3 f (n−1 ))x3 + n(−f 0 (n−1 ) + 3nf (n−1 ))x2 x ≤ n−1
where n ∈ N is chosen later. Note that g(0) = g 0 (0) = 0 and g(1/n) = f (1/n), g 0 (1/n) =
f 0 (1/n). Thus g ∈ C. Moreover,
Z
p
|f (t) − (n2 f 0 (n−1 ) − 2n3 f (n−1 ))t3 − n(−f 0 (n−1 ) + 3nf (n−1 ))t2 |p dt
kf − gkp =
[0,n−1 ]
1
) . n−1 kf kpC 1 < p
p
n
p −p
if we let n > kf kC 1 . In other word, kf − gkp < , and our problem is solved
affirmatively in this case.
Consider p = ∞. Then P is not dense in L∞ since χ[0,1/2] ∈ L∞ [0, 1] could not be
approximated in the sup norm by polynomials in P, otherwise ∀ > 0 we could find
p ∈ P so that kp − χ[0,1/2] k∞ < and this implies supt∈[0,1/2] |1 − p(t)| < . Because p
is continuous and p(0) = 0, there is some 1/2 > δ > 0 such that |p(t)| ≤ , ∀t ∈ [0, δ].
Hence supt∈[0,1/2] |1 − p(t)| > 1 − or 1 < 2, ∀ > 0 (contradiction!)
. n−1 (kf kp∞ + kf 0 kp∞
Comment: Thanks to Zach in class, we get another shorter proof by just applying the Stone-Weierstrass theorem directly to the algebra of polynomials P such that
P (0) = P 0 (0) = 0 to get the density in {f ∈ C[0, 1] : f (0) = 0} and then use a linear
approximation to cut off f around a small neighborhood of 0.
Exercise: Given k ∈ N, find all values p ∈ [1, ∞] such that the set of all polynomials P satisfying P (0) = P (1) (0) = . . . = P (k) (0) = 0 is dense in Lp [0, 1].
Problem 5
Proof. Banach-Alaoglu theorem implies once kxk k ≤ M, ∀k, there exists a subsequence xkn such that it converges in the weak*-topology on `p . Since 1 < p < ∞,
`p is reflexive. Hence, its weak* topology coincides with the weak topology on
`p . So the subsequence xkn converges weakly in `p . Assume its P
weak limit is
x = (x(1), . . . , x(n), . . .) ∈ `p . Then ∀y = (y(1), .., y(n), . . .) ∈ `q , ∞
j=1 (xk (j) −
x(j))y(j) → 0. By substituting y = ej (j ∈ N), x(j) = limk→∞ xk (j) = 0. Hence
x = 0 or there is a subsequence of xn that converges weakly in `p to 0. The above
argument is true for any subsequence of xn , i.e for any subsequence xkn of xk , we
can find a subsubsequence xknj that converges weakly to the same limit 0. By a
simple contradiction argument, we easily conclude that the whole sequence xn must
converge weakly to 0.
The argument fails if the sequence kxk kp is unbounded. One simple example is
to take the sequence xk = k 2/q ek where ek = (0, . . . , 1, 0 . . .). Then
P∞it is clear that
−2/q
q
limk→∞ xk (n) = 0, ∀n ∈ N. However, if we let y = (n
)n∈N ∈ ` , n=1 xk (n)y(n) =
k 2/q y(k) = 1, ∀k ∈ N. So xk does not converge weakly to 0.
16
KHA
Exercise: Is the above statement still true if we consider `1 or `∞ (with the boundedness assumption)? Notice the Schur property of `1 means weak convergence implies
strong convergence.
Problem 6
Proof. (a) Let A = {ft : t ∈ [0, 1]}. Since f vanishes at infinity, ∀ > 0, pick a
compact subset of R such that |f (z)| < , ∀z ∈
/ K. Consider any sequence {ftn } in
A. Since [0, 1] is compact, there is a convergent subsequence of tn . Without loss of
generality, we assume tn → t0 ∈ [0, 1]. ∀x ∈
/ (K −tn )∪(K −t0 ), |f (x+tn )−f (x+t0 )| ≤
2. So kftn − ft0 k = supx∈R |f (x + tn ) − f (x + t0 )| ≤ min{2, supx∈(K−tn )∪(K−t0 ) |f (x +
tn ) − f (x + t0 )|}. Since tn → t0 , (K − tn ) ∪ (K − t0 ), ∀n ∈ N is bounded in some fixed
compact set C = K−t0 +[−M, M ] in R. Hence supx∈(K−tn )∪(K−t0 ) |f (x+tn )−f (x+t0 )|
is small by the uniform continuity of f on C. Thus we can select a subsequence ftnk
such that it converges in the norm topology.
Therefore, A is compact in C0 (R).
(b) Let B = {ft : t ∈ R}. By Riesz representation theorem, the dual of C0 (R)
is M (R) consisting of complex Radon measures on R. We will apply the useful
criterion for proving weak convergence on C0 (R) is Exercise 22 (p.125, Chap 7) in [2]
(or Problem 2, Jan 2013): fn → f weakly in C0 (R) iff supn∈N kfn k < ∞ and fn → f
pointwise. Consider any sequence {ftn : tn ∈ R} in B. kftn k = kf k, ∀n ∈ N so
supn∈N kftn k < ∞. An application of the above criterion yields that the weak closure
B̄ of B contains all {ft : t ∈ {tn : n ∈ N}}. Given any bounded sequence tn , we can
always extract a convergent subsequence tnk and so the corresponding subsequence
ftnk of ftn converges weakly in C0 (R) and its weak limit must belong to B̄. On the
other hand, for any unbounded sequence tn , since f vanishes at infinity, we must have
ftn (x) = f (x + tn ) → 0, ∀x. So ftn always converges weakly to 0.
This means in any case, ftn contains a weakly convergent subsequence. So B is
relatively compact in the weak topology. Also, this gives us the description of the
weak closure of B: B̄ = {ft : t ∈ {tn : n ∈ N}} ∪ {0}.
Comment: We can prove sequentially weak compactness instead of weak compactness using nets because of the Erbelein-Smulian theorem.
Exercise: Can you generalize the above problem on higher dimensional Euclidean
space Rn , or even on any locally compact group G?
Problem 7
Proof. Let An = x ∈ R : ∃rn (x) > 0 ∀x0 , x00 ∈ B x, rn (x) T|f (x00 ) − f (x0 )| < n1 for
each n > 0. Then if f is continuous at x, we have x ∈ Tn∈N An and the converse
is also true. Hence the set of continuity points of f is n∈N An . If we can show
each An is open, then we are done. Indeed, if we consider any x ∈ An , then we see
immediately B(x, rn (x)/2) ⊂ An since for those y ∈ B(x, rn (x)/2), we would have
supz∈B(y,rn (x)/2) f (z) − inf z∈B(y,rn (x)/2) f (z) < 1/n.
REAL ANALYSIS QUALIFYING EXAM 18:14 o’clock, 9 August 2013
17
Exercise: Show that there exists no real-valued function on R such that the set of
its continuity points is exactly Q.
Problem 8
Proof. The latter part is standard. We give an example for the first part of the
n+1
problem. Consider the following nonempty bounded subset of `2 : A = {
en :
n
n ∈ N}
n+1
en −
Then A is nonempty and nonconvex. Moreover, A is closed since k
n
√
m+1
em k2 ≥ 2, ∀n 6= m so any convergent sequence in A must contain an evenm
tually constant subsequence. Finally, A has no element of minimal norm since
n+1
inf k
en k2 = 1 while no element in A has norm equal to 1.
n∈N
n
Problem 9
Proof. fn (t) = eint and apply the Riemann-Lebesgue lemma.
Problem 10b
Proof. For any finite collection of subintervals (ai , bi ), i = 1, . . . , n of [0, 1] and ∀ > 0
n
n
X
X
f (bi ) f (ai )
|f (bi )g(ai ) − f (ai )g(bi )|
|
−
|=
g(bi )
g(ai )
g(ai )g(bi )
i=1
i=1
≤
n
X
|f (bi ) − f (ai )|
g(bi )
i=1
≤ A−1
n
X
+
n
X
f (ai )|g(bi ) − g(ai )|
i=1
g(ai )g(bi )
|f (bi ) − f (ai )| + A−2 kf kC[a,b]
i=1
.
n
X
n
X
|g(bi ) − g(ai )|
i=1
|f (bi ) − f (ai )| + |g(bi ) − g(ai )|
i=1
where A = minx∈[a,b] g(x) > 0. Thus, f /g is absolutely continuous because so are f
and g.
5. January 2011
Problem 1
1/4
Proof. Let gn = |fn − f0 |
, ∀n ∈ N. By assumption,
∞. By monotone convergence theorem,
∞ Z
X
n=1
X
|gn |dµ =
∞ Z
X
|gn |dµ <
n=1 X
Z X
∞
X n=1
∞
X
n−2 <
n=1
|gn |dµ. This means
18
KHA
∞
X
|gn | ∈ L1 (µ). Hence this series function is finite µ-almost everywhere on X. In
n=1
other words, gn or |fn − f0 | converges to 0 µ-almost everywhere.
Problem 2
Proof. It’s well known that K is separable. Let {xn } be dense in K. For each n,
consider the following function dn (s) = d(s, xn ), ∀s ∈ K. Given different x 6= y ∈ K,
we take any subsequence xnk converging to x. Thus d(y, xnk ) → d(y, x) > 0 =
limk→∞ d(x, xnk ). Hence, we can find at least one xn such that d(x, xn ) < d(y, xn ). In
other words, the family {dn } separates points of K. Let A be the algebra generated
by dn over Q. It is clear that A is countable. By Stone-Weierstrass, A is dense in
C(K). Hence C(K) is separable.
Optional reading: Another way to find a separating family is to use the normality
of the compact metric space K to construct for each pair of distinct points xn , xm
and each pair of positive rational numbers ri , rj such that ri + rj < d(xn , xm ) (i.e
r ,r
r ,r
B(xn , ri )∩B(xm , rj ) = ∅), a function fxni ,xjm ∈ C(K) such that fxni ,xjm = 1 on B(xn , ri )
r ,r
and fxni ,xjm = 0 on B(xm , rj ). Then given x 6= y ∈ K, choose xn , xm such that
B(xn , 2d(x, xn )) ∩ B(xm , 2d(y, xm )) = ∅ and hence we could still find suitable ri , rj ∈
r ,r
r ,r
Q so that fxni ,xjm (x) = 1 6= 0 = fxni ,xjm (y). Now we just let A be the algebra generated
ri ,rj
by these fxn ,xm over Q and then apply the Stone-Weierstrass theorem.
Comment: An elegant classification theorem of Miljutin states that if K is an
uncountable compact metric space then C(K) is isormophic to C[0, 1] as Banach
spaces (See [1]).
Exercise: Prove that the converse is true, i.e if C(K) is separable for a compact
Hausdorff space K, then K is metrizable by using Riesz Representation Theorem and
the fact that if X is a separable Banach space then BX ∗ is metrizable.
Problem 3
Proof. This is the Vitali Convergence theorem (see Exercise 15, p.187 in [2]). We
follow the hint in [2]. Let Am = {x ∈ X : |fm (x)| ≥ }.
Z
kfm k1 =
Z
|fm |dµ +
Am
|fm |dµ
X\Am
Since fm is uniformly integrable, for each > 0, we pick a suitable δ > 0. When m
is large enough, µ{x ∈ X : |fm (x)| ≥ } < δ because fn converges in measure to 0.
REAL ANALYSIS QUALIFYING EXAM 18:14 o’clock, 9 August 2013
19
Since
Z
Z
fm dµ| ≥
>|
Am ∩{Re(fm )≥0}
Re(fm )dµ
Am ∩{Re(fm )≥0}
Z
Z
fm dµ| ≥
>|
Am ∩{Re(fm )<0}
−Re(fm )dµ
Am ∩{Re(fm )<0}
Z
Z
fm dµ| ≥
>|
Am ∩{Im(fm )≥0}
Im(fm )dµ
Am ∩Im(fm )≥0}
Z
Z
fm dµ| ≥
>|
Am ∩{Im(fm )<0}
−Im(fm )dµ
Am ∩{Im(fm )<0}
and
Z
Z
Z
|Im(fm )|dµ
|Re(fm )|dµ +
|fm |dµ ≤
Am
Am
Am
we have
Z
|fm |dµ < 4
Am
Obviously by definition of Am
Z
|fm |dµ < µ(X)
X\Am
Summing these two inequalities, we get
kfm k1 ≤ (4 + µ(X))
Hence kfn k1 → 0.
Problem 4
Proof. f ∈ Lp [0, 1](+) means f > 0 almost everywhere on [0, 1]. Consider any sequence {f 1/tn : tn ∈ N}. The only interesting case is when limn tn = ∞. Hence
f 1/tn → 1 almost everywhere. Moreover f 1/tn (x) ≤ f (x) + 1, ∀x ∈ [0, 1] and certainly
f + 1 ∈ Lp [0, 1]. By the Lebesgue dominated convergence theorem, we conclude
kf 1/tn − 1kp → 0. So {f 1/n : n ∈ N} has compact closure in Lp .
This is false in L∞ . Take f = id, then kf 1/n − 1k∞ = 1, ∀n ∈ N. Hence the family
{x1/n } cannot have compact closure in L∞ [0, 1].
Problem 5
Proof. Assume X is a reflexive Banach space. Let a = inf y∈K kyk. Then take a
sequence yn ∈ K such that a = limn→∞ kyn k. Hence the sequence (kyn k)n∈N is
bounded and since X is reflexive, we can extract a subsequence ynk such that it
converges weakly to an element y ∈ X by Banach-Alaoglu theorem. Suppose y ∈
/
K. By Hahn-Banach theorem, we could find a bounded linear functional T on X
such that T (z) = 0, ∀z ∈ K and T (y) > 0 since K is nonempty closed convex.
Since T (ynk ) → T (y), there exists at least k so that T (ynk ) > 0 while ynk ∈ K
(contradiction!). Thus, y must be in K and so a ≤ kyk ≤ lim inf n kyn k = a as we
20
KHA
wish. The uniqueness part when X is Hilbert is standard in any textbook such as
[2], [3].
Comment: After applying the Banach-Alaoglu theorem, we can apply the Mazur
theorem (which states that for any convex subset A in a Banach space X, its weak
and norm closures are the same) instead of Hahn-Banach theorem to prove y ∈ K.
Problem 6
Proof. Suppose X ∗ is separable with a dense sequence {x∗n }. For each n, we pick an
element xn ∈ X such that kxn k = 1 and x∗n (xn ) ≥ kx∗n k/2. Let L0 be the vector
space generated by vectors xn over Q. Obviously, L0 is countable. Let L be the
vector space generated by xn over R then L0 is dense in L. To prove X is separable,
it suffices to show that L is dense in X. Suppose for contradiction, L is not dense in
X. Using Hahn-Banach theorem, we can find a linear functional f ∈ X ∗ such that
kf k = 1 and f (x) = 0, ∀x ∈ L. Given any > 0, we pick n so that kf − x∗n k < .
Hence kx∗n k/2 ≤ |f (xn ) − x∗n (xn )| < ⇒ kf k ≤ kf − x∗n k + 2 ≤ 3. So f = 0.
(contradiction!).
Problem 8
Proof. Since C[0, 1] is dense in L1 [0, 1] and L1 -norm is dominated above by sup-norm,
it suffices to prove the space of all Lipschitz functions on [0, 1] is dense in C[0, 1] by
Stone-Weierstrass. First, notice that for two Lipschitz function f, g, their sum is also
Lipschitz with the Lipschitz constant kf + gkL ≤ kf kL + kgkL . Also, the product f g
satisfies
|f (x)g(x) − f (y)g(y)|
kf gkL = |f (0)|.|g(0)| + sup
y−x
0≤x<y≤1
|f (x) − f (y)|
|g(x) − g(y)|
+ kf k∞
≤ |f (0)|.|g(0)| + kgk∞
y−x
y−x
≤ (1 + kf k∞ + kgk∞ )(1 + kf kL )(1 + kgkL ) < ∞
Hence the space of Lipschitz functions on [0, 1] is a subalgebra (Lipschitz algebra).
It separates points in [0, 1] since given x 6= y ∈ [0, 1], the identity function obviously
separates them. So the Lipschitz algebra is dense in C[0, 1] and hence in L1 [0, 1]. Exercise: Give another proof by using the denseness of Cc∞ [0, 1] in Lp [0, 1], 1 ≤ p <
∞.
Problem 9
Proof. Let J = {x ∈ [0, 1] : g(x) = 1}. Then J is measurable. Take f = χJ we have
µ(J) = kg n f k2 , ∀n hence µ(J) = 0. We claim µ(J) = 0 is the equivalent condition.
Fix any f ∈ L2 [0, 1]. Then g n f → 0 µ-almost everywhere on [0, 1] if µ(J) = 0.
Moreover, ∀n ∈ N, |g n f |2 ≤ |f |2 since |g| ≤ 1 and because f ∈ L2 , by the Lebesgue
Dominated Convergence theorem, we must have kg n f k2 → 0.
Problem 10b
REAL ANALYSIS QUALIFYING EXAM 18:14 o’clock, 9 August 2013
21
Proof. For any g ∈ L2 (µ × µ) and fn ∈ L∞ , by Fubini we get
Z 1 Z 1
Z 1Z 1
fx,n (y)gx (y)dµ(y))dµ(x)
(
fn (x, y)g(x, y)dµ(x)dµ(y) =
0
0
0
0
Let
Z
1
fx,n (y)gx (y)dµ(y)
hn (x) =
0
By our assumption, hn (x) → 0, ∀x ∈ [0, 1] since
Z
Z 1 Z 1
2
(
|gx (y)| dµ(y))dµ(x) < ∞ ⇒
0
0
1
|gx (y)|2 dµ(y) < ∞,
0
for almost every x.
Z
|hn (x)| ≤ sup kfn k∞
n∈N
Z
1
Z
1
|g(x, y)|dµ(y)
0
1
|g(x, y)|dµ(y)dµ(x) ≤ kgkL2 (µ×µ) . Use Lebesgue dominated convergence
and
0
0
theorem to conclude.
6. August 2010
Problem 1
Proof. (a) For each n = 2k + j where 0 ≤ j < 2k , let fn = χ[j/2k ,(j+1)/2k ] . Then
kfn k1 = 1/2k → 0 since k = [log2 (n)]. On the other hand, ∀x ∈ [0, 1], k ∈ N, there
is exactly one 0 ≤ jx,k < 2k such that x ∈ [j/2k , (j + 1)/2k ] and hence for each x
we can always find a subsequence f2k +jx,k (x) = 1. This implies fn cannot converge
to 0 almost everywhere. More generally, we can pick any collection of partitions Pn
of finite subintervals of [0, 1] as long as the sequence of the maximal sizes of these
subintervals approaches
R
Pto 0.
P
P
−k
(b) kfk k1 ≤ 2 ⇒ k kfk k1 < ∞ ⇒ [0,1] k |fk (x)|dx < ∞ ⇒ k |fk (x)| < ∞
for almost every x. This means fk (x) → 0 almost everywhere.
Problem 2
Proof. Suppose for contradiction, there is an α ∈ (0, 1) such that for any interval
∗
I ⊂ [0, 1], m
ThenP
given any family of intervals
In ⊂ [0, 1] such
S (E ∩ I) < αm(I).
P
∗
that E ⊂ n In , we have m∗ (E) ≤
m
(E
∩
I
)
<
α
|I
|.
By taking the
n
n
n
n
infimum over such these families of intervals and the definition of the outer measure
m∗ , we must get a contradiction since m∗ (E) > 0.
Problem 3
Proof. For each T ∈ X ∗ , supn∈N |x∗∗
n (T )| = |T (xn )| < ∞ since T (xn ) → 0. By Banach
Steinhaus theorem, we must have supn∈N kx∗∗
n k < ∞ and this implies the sequence
xn is bounded.
Problem 4
22
KHA
Proof. 4a/ Problem 2 in Jan 2013.
4b/ Without loss of generality, we can assume fn converges weakly in C[0, 1] to 0
then by problem 3, kfn k is a bounded sequence and hence we can apply part (a) to see
fn converge pointwise to 0. Together with the boundedness of kfn k, we can deduce
fn is norm convergent to 0 in L1 [0, 1] by Lebesgue dominated convergence.
Problem 5
Proof.
Z
Z
Z
|f (x)|dx =
E
m({x ∈ E : |f (x)| ≥ λ})dλ ≤ C
[0,∞]
Z
≤C
min{m(E), λ−2 }dλ
[0,∞]
λ−2 dλ + Cm(E)−1/2 m(E) ≈
p
m(E)
[m(E)−1/2 ,∞]
Problem 6
Proof. By Weierstrass
theorem, we find a polynomial q such that kf 0 − qk∞ < /2.
Z x
Now let p(x) =
q(t)dt + f (0) then q is a polynomial and we should have q 0 = p.
0
Z x
Z x
0
p0 (t)dt − f (0)| ≤
f (t)dt + f (0) −
Moreover, ∀x ∈ [0, 1], |f (x) − p(x)| = |
0
xkf 0 − qk∞ < /2. Hence kf − pk∞ + kf 0 − p0 k∞ < .
0
Problem 7
Proof. Define Uk = {x ∈ X : {fn (x)}n≥k is monotone increasing}, Vk = {x ∈ X :
{fn (x)}n≥k is monotone decreasing}. It is clear that these subsets
are closed in X
S
since each fn is continuous on X. By assumption, X = k Uk ∪ Vk . By Baire
category theorem, there exists N ∈ N such that either int(UN ) 6= ∅ or int(VN ) 6= ∅.
Without of loss generality, we suppose int(Uk ) 6= ∅ and we let U = int(UN ) then U
is a non-empty open subset of X. Moreover, ∀x ∈ U , the sequence {fn (x)}n≥N is
monotone.
Problem 8
Proof. To prove T is bounded, it suffices to prove its graph is closed in Lp × Lp .
Consider any sequence (fn , T (fn )) which converges to (f, g) ∈ Lp × Lp . Then let
hn = fn − f we have hn → 0 in Lp . We can extract a subsequence hnk such that
hnk → 0 almost everywhere. By our assumption, the sequence T hnk → 0 almost
everywhere. Moreover, the sequence T hnk = T fnkl − T f → g − T f in Lp and hence
we can extract a subsequence T hnkl → g − T f almost everywhere. Hence g = T f
almost everywhere. So the graph of T is closed.
Comment: It suffices to relax the condition on T in the above problem by the following condition: If fn converges almost everywhere to 0, there is a subsequence of
T fn which converges almost everywhere to 0. Then T is still bounded on Lp [0, 1]. Problem 10
REAL ANALYSIS QUALIFYING EXAM 18:14 o’clock, 9 August 2013
23
Proof. Consider an extreme point f in U . Assume there is x0 ∈ [0, 1] such that
|f (x0 )| < 1 then by continuity of f , we can find α, δ > 0 such that |f (y)| ≤ α <
1, ∀y ∈ (x0 − δ, x0 + δ) ∩ [0, 1]. Choose a nice bump function g on [0, 1] such that its
support is contained in (x0 − δ, x0 + δ) ∩ [0, 1] and 0 ≤ g ≤ 1. It is not hard to see that
f 6= f +(1−α)g, f −(1−α)g ∈ U . But then f = 1/2(f +(1−α)g)+1/2(f −(1−α)g)
(contradiction!). Hence |f | = 1 and by continuity f ∈ {−1, 1}. It is clear that −1, 1
are extreme points of U .
Suppose C[0, 1] is a dual space. Banach-Alaoglu gives us the weak*-compactness
of the closed unit ball U of C[0, 1]. By the Krein-Milman theorem, the weak*closure of the convex hull of extreme points of U is exactly U . Since we know
extreme points are just constant functions −1, 1, its weak*-closure of its convex hull
is contained in the weak* closure of the two dimensional vector subspace generated
by the constant functions −1, 1 and hence coincides with span{−1, 1}. Of course, U
cannot be contained in that two dimensional subspace.
7. January 2009
Problem 2
Proof. First, we assume µ : U → R. For each
Pnsubalgebra U of M, we define SU to be
the set of all simple functions of the form i=1 ci 1Ei where ci ∈ R, Ei ∈ U. Then SU
is a vector subspace of SM .
P
P
Now we define p : SM → R such that p(f ) = sup{ ni=1 |ci |p(Ei ), f = ni=1 ci 1Ei , Ei ∩
Ej = ∅ ∀i 6= j, Ei ∈ M, ci ∈ R}, ∀f ∈ SM . It’s not difficult to check that p satisfies
p(f + g) ≤ p(f ) + p(g), ∀f, g ∈ SM and p(tf ) = tp(f ), ∀t ∈ R+ , f ∈ SM . Note that
the definition the above seminorm p is just an extension of the total variation of the
measure p when you apply to the function f = 1 (see Exercise 21, p.94 in [2] to recall
different equivalence of the total variation of a complex measure). Z
Consider the linear mapping T : SU → R defined by T (f ) =
f dµ, ∀f ∈ SU
X
(finite additive property of µ gives linearity of T ). Then |T (f )| ≤ p(f ), ∀f ∈ SU .
Using a version of Hahn-Banach theorem to get a linear extension of T on SM which
we still denote it by T for convenience, morever this extension T : SM → R satisfies
|T (f )| ≤ p(f ), ∀f ∈ SM . Now we define a finite additive measure ν on M by just
setting ν(E) = T (1E ), ∀E ∈ M. So ν|U = µ, |ν(E)| ≤ p(E), ∀E ∈ M. To check
the countably additive property of ν, we consider
P any countable collection
P of disjoint
measurablePsubsets Ei ∈ M and so 1∪i Ei = i 1Ei . Thus, ν(∪i Ei ) = i ν(Ei ) since
the series i T (Ei ) converges (use |T (f )| ≤ p(f ), ∀f ∈ SM and properties of the
measure p).
For the complex case, we simply repeat the trick in proving the complex version
of Hahn-Banach theorem from the real version.
Exercise: Check the triangle inequality for the seminorm p defined in the above
proof.
24
KHA
8. Sample Exam
Problem 1: Assume K ⊂ C is a compact subset such that it has finite (at most
countable) boundary. Prove that K itself is finite (at most countable, respectively).
Problem 2: ZSuppose µ(X) = 1 and f ∈ Lp for some p < ∞. Prove that
lim kf kp = exp( log|f |dµ).
p→0
Problem 3: Let X be a reflexive Banach space. Assume C is a weakly closed
subset in X. Prove that the norm function k.k on C attains its minimum on C.
Problem 4: Find an example to show that there exists a non-empty closed convex
subset in some Banach space such that the norm function attains no minimum on
that subset.
Problem 5: Use Stone-Weierstrass to prove that the set of all functions of the
type
m
X
φ(x) = a0 +
(bn cosnx + cn sinnx)
n=1
(for some finite m) is dense in the space of all real valued continuous function f on
R such that f is 2π-periodic.
Problem 6: (a) Let X be a compact topological space and {fn } is a sequence of
real-valued continuous functions on X such that fn+1 (x) ≥ fn (x), ∀x ∈ X, n ∈ N.
Assume fn converges pointwise to 0. Show that fn converges uniformly on X to 0.
(b) Assume X 6= ∅ is a locally compact and complete metric space. Let {fn } be a
sequence of real-valued continuous functions on X such that for each x ∈ X, there
exists an integer Nx so that {fn (x)}n≥Nx is either a monotone increasing or decreasing sequence and suppose fn converges to a continuous function f . Prove that there
exists a nonempty interior compact subset K ⊂ X such that fn converges uniformly
to f on K.
Problem 7: Let G ⊂ On be a compact group, where On is the orthogonal group in
Mn (C). For each 1 ≤ i, j ≤ n, define the coordinate function uij in C(G) which maps
a matrix A into its (i, j)-entry. Prove that C(G) is generated by these n2 coordinate
functions.
Selected Solutions
Z Problem 4: ZLet X = C[0, 1] (real-valued functions) and C = {f ∈ C[0, 1] :
f dt =
f dt + 1} then C is closed and convex in X. First we notice
[0,1/2]
[1/2,1]
that for any f ∈ C, one has kf k∞ /2 ≥ 1 − 1/2kf k∞ by the definition of C. So
inf f ∈C kf k∞ ≥ 1.
REAL ANALYSIS QUALIFYING EXAM 18:14 o’clock, 9 August 2013
25
Now we construct for each n ∈ N a piecewise linear function fn ∈ X such that
n−1
fn (t) = 1 whenever 0 ≤ t ≤
, fn (1/2) = 0 and fn (t) = −fn (1 − t), ∀t ∈ [0, 1].
2n + 2
Then a simple calculation yields fn ∈ C. Morever, kfn k∞ = 1 + 1/n. Let n → ∞,
we must have inf f ∈C kf k∞ = 1.
Now suppose there
exists g ∈ C such
Z
Z that kgk∞ = 1. Using definition of C, we
immediately get
(1 − g)dt = 0,
(g + 1)dt = 0. This and the continuity of
[0,1/2]
[1/2,1]
g would imply g = 1 on [0, 1/2] and g = −1 on [1/2, 1] which is a contradiction. So
there cannot be any point of minimal norm in C.
References
[1] Fernando Albiac and Nigel Kalton, Topics in Banach space theory, Graduate Texts in Mathematics, vol. 233, Springer, New York, 2006.
[2] Gerald B. Folland, Real analysis, Second, Pure and Applied Mathematics (New York), John Wiley & Sons Inc., New York, 1999. Modern techniques and their applications, A Wiley-Interscience
Publication.
[3] Elias Stein and Rami Shakarchi, Real analysis, Princeton Lectures in Analysis, III, Princeton
University Press, Princeton, NJ, 2005. Measure theory, integration, and Hilbert spaces.
[4] Terry Tao, http://terrytao.wordpress.com/2009/03/30/245c-notes-1-interpolation-of-lp-spaces/.
[5] Wikipedia, http://en.wikipedia.org/wiki/Fat-Cantor-set.
M.K., Department of Mathematics, Texas A&M University, College Station, TX
77843-3368, USA
E-mail address: kha@math.tamu.edu