Analysis Qualifying Exam Solutions
Thomas Goller
September 4, 2011
Contents
1 Spring 2011
1.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2
4
2 Fall 2010
2.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . .
6
6
8
3 Spring 2010
10
3.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 13
4 Fall 2009
15
4.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
4.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 17
5 Spring 2009
20
5.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
5.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 21
6 Fall 2008
24
6.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
7 Spring 2008
27
7.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
1
Chapter 1
Spring 2011
1.1
Real Analysis
A1. (a) `1 (Z) is separable. A countable set whose finite linear combinations
are dense is {en }n∈Z , where en has a 1 in the nth position and is 0 everywhere
PN
else. If x ∈ `1 (Z), then the sums
k=−N xk ek approximate x arbitrarily well
P
in the norm as N → ∞ since k |xk | < ∞.
(b) `∞ (Z) is not separable. This is proved by finding an uncountable number of disjoint open sets, whence every dense set must have an element in each
interval and thus must be uncountable. Recalling that `∞ (Z) consists of all
bounded sequences, consider all sequences containing just 0’s and 1’s. These
map surjectively onto the real numbers by considering binary expansions, hence
are uncountable. Any two distinct sequences of 0’s and 1’s have distance 1 from
each other in the norm, so for any r < 12 , taking a ball of radius r around each
of these sequences gives an uncountable number of disjoint open sets.
A2. (a) If
f is measurable, then so is |f |. We need to show that S(a) :=
|f |−1 (−∞, a) is measurable for each a ∈ R. The case a ≤ 0 is trivial since
then S(a) = ∅ is measurable. Note that when a > 0, we have
|f |−1 (−∞, a) = f −1 (−a, a) .
Now, f measurable
implies f −1 (−∞, a) is measurable for each a ∈ R, so each
f −1 [a, ∞) is measurable by taking complements. Therefore each f −1 (a, ∞)
is measurable since
so f −1
∞
[
1
f −1 (a, ∞) =
f −1 [a + , ∞) ,
n
n=1
(−a, a) = f −1 (−∞, a) ∩ f −1 (−a, ∞) is measurable, as desired.
2
(b) |f | measurable does not imply that f is measurable, as long as X
contains non-measurable sets. For if N is a non-measurable set, define
(
−1 x ∈ N ;
f (x) =
1
otherwise.
Then |f | ≡ 1 is measurable, but f −1 (−∞, 0) = N is not measurable.
A3. Haven’t reviewed Fourier series.
R
A4. f ∈ L1 implies |f | < ∞, and f uniformly continuous means that
for any > 0, there is a δ > 0 such that |f (x) − f (y)| < whenever |x − y| <
δ. Suppose for contradiction that x is a sequence with |xk | → ∞ for which
f (xk ) 6→ 0. Then there is an > 0 such that for each N > 0, there exists n ≥ N
such that |f (xn )| ≥ . Thus x has a subsequence {xnk } such that |f (xnk )| ≥ for each k ≥ 1.
We may assume the xnk are distinct, deleting repeats if necessary. Then
by uniform continuity, choose δ > 0 such that |f (x) − f (y)| < 2 whenever
|x − y| < δ. Also, make δ small enough so that |xnk − xnj | ≥ δ whenever k 6= j.
Then taking a ball of radius 2δ around each xk and letting E be the union of
these balls, we see that
Z
Z
∞
X
δ · = ∞,
|f | ≥
|f | ≥
2
E
k=1
1
contradicting the fact that f ∈ L .
A5. Without (1), consider the sequence
1
1
{fn } = χ[0,1] , χ[0,2] , χ[0,4] , . . . ,
2
4
which converges pointwise to the function f ≡ 0 for all x. Then (1) fails and
Z
Z
lim
|fn − f | = lim
|fn | = 1 6= 0.
n→∞
n→∞
With (1), we can prove the result. Set gn = inf{fn , fn+1 , . . . }, which is
measurable
since the fn are measurable. Then 0 ≤ gn ≤ fn and Rgn % fR, so
R
|gn − f | → 0 by the dominated convergence theorem, whence also gn → f .
Now we use (1) and the fact that |fn − gn | = fn − gn to deduce that
Z
Z
Z
Z
Z
Z
|fn − f | ≤ |fn − gn | + |gn − f | = fn − gn + |gn − f | → 0.
3
1.2
Complex Analysis
B6. g is holomorphic
P∞ on U . Let z0 ∈ U . Since f is holomorphic at z0 ∈ U , we
can write f (z̄) = n=0 an (z − z0 )n for z near z0 . Thus
g(z) := f (z) =
∞
X
an (z − z0 )n
n=0
is a power series expansion for g near z0 .
B7. Since |f (z)| ≤ C|z|k for z outside DR , all the poles of f are inside DR ,
and there can be only finitely many such poles since DR is compact. Let f˜(z)
be the sum of the principal parts of f at each of these poles. Then g := f − f˜
is an entire function that still satisfies a bound of the form |g(z)| ≤ C 0 |z|k for
|z| ≥ R since the principal parts are bounded away from their poles.
We claim that g is a polynomial of degree ≤ k, which will imply that f is a
rational function. Since g is entire, it suffices to prove that g (j) (0) = 0 for all
j > k (look at the power series expansion). For each R0 ≥ R > 0, the Cauchy
inequalities imply
(k + 1)! 0 0 k
C R → 0,
|g (k+1) (0)| ≤
R0 k+1
as R0 → ∞, giving the desired result.
B8.P (i) =⇒ (ii): Suppose a is a zero of f of order l ≥ k. Let f (z) =
(z−a)l n≥0 an (z−a)n be the power series expansion of f near a, so that a0 6= 0.
Let
P C = |a0 | and choose 0 < < 1 sufficiently small such that B (a) ⊆ U and
| n≥1 an n | ≤ |a0 |. Then for all z ∈ B (a),
|f (z)| ≤ |z − a|l (|a0 | + |a0 |) ≤ 2|a0 ||z − a|k .
P
(ii) =⇒ (i): Suppose (ii) holds and let f (z) = n≥0 an (z − a)n be the
power series expansion for f near a. Then for all z ∈ B (a) with z 6= a,
f (z) (z − a)k ≤ C
implies a0 = a1 = · · · = ak−1 = 0.
1
B9. The entire function sin z := 2i
(eiz − e−iz ) has zeroes at πk, k ∈ Z,
an essential singularity at ∞, and no other zeroes or singularities. The function
1
sin z1 therefore has zeroes at πk
and an essential singularity at 0, with no other
zeroes or singularities.
4
3
−3
5
−5
Since sin z = z − z3! + z5! − · · · we also have sin z1 = z −1 − z3! + z5! − · · · .
Thus the residue at 0 of the product of these two series, namely the coefficient
of z −1 of the Laurent series, is 0.
Is this the right answer?
B10. Set
f (z) =
z2
zeiz
.
+ 2z + 10
The denominator factors into (z + 3i + 1)(z − 3i + 1), so there are poles at
3i − 1 and −3i − 1. We will integrate f over the upper-half-plane semicircle
consisting of γ, the real-axis interval [−R, R], and the semicircle γR of radius
R. The imaginary part of the integral over γ gives the real integral we want to
evaluate. Since we are taking R large, we pick up the residue at 3i − 1, which is
lim (z − 3i + 1)
z→3i−1
(3i − 1)e−3−i
zeiz
=
(z + 3i + 1)(z − 3i + 1)
6i
1
= 3 (3 cos 1 + sin 1 + i(cos 1 − 3 sin 1)).
6e
What a mess! The integral over γR vanishes thanks to the 1/z decay of f
near the real axis, together with the better decay of f due to eiz away from the
real axis.
5
Chapter 2
Fall 2010
2.1
Real Analysis
A1. Yes, the fact that f −1 (r, ∞) is measurable for each r ∈Q implies that
f is measurable. Taking complements, we see that f −1 (−∞, r] is measurable
for each r ∈ Q. Now for any s ∈ R, let rk be an increasing sequence of rational
numbers < s such that rk % s. Then
∞
[
f −1 (−∞, s) =
f −1 ((−∞, rk ]) ,
k=1
so f is measurable.
A2.
Suppose 1 ≤ p ≤ q ≤ ∞ such that p1 + 1q = 1, where we allow
p = 1, q = ∞. Since f, fn ∈ Lp and g, gn ∈ Lq , we also have f − fn ∈ Lp and
g − gn ∈ Lq . Hölder’s inequality implies fn gn and f g are in L1 . Note that
|f g| − |fn gn | = |f g| − |fn g| + |fn g| − |fn gn | ≤ |g||f − fn | + |fn ||g − gn |.
Thus using Hölder’s inequality,
Z
Z
Z
|f g| − |fn gn | ≤ |g||f − fn | + |fn ||g − gn |
= kg(f − fn )k1 + kfn (g − gn )k1
≤ kgkq kf − fn kp + kfn kp kg − gn kq .
Since kgkq and kfn kp are bounded but kf − fn kp → 0 and kg − gn kq → 0, we
see that kf g − fn gn k1 → 0, which gives the desired result since
Z
Z
Z
|f g| − |fn gn | ≤ |f g| − |fn gn | → 0.
6
(Maybe I’m missing something, but I don’t think the case p = 1, q = ∞ is
any different than p > 1.)
Tn
/ C denote the bounded linear functional defined by
A3.
Let H
Tn (y) = (y, xn ). Then the countable collection {Tn } is pointwise bounded by
assumption, hence the norms are bounded by the uniform boundedness property.
So there exists M ≥ 0 such that
|(y, xn )| ≤ M kyk
for all y ∈ H.
In particular, choosing y = xn gives kxn k ≤ M , establishing the desired result.
A4. Suppose f is measurable such that
R
|f |p < ∞ and
R
|f |r < ∞. Set
E = {x ∈ X : |f (x)| ≥ 1},
which is measurable. Then
Z
|f |q =
X
Z
|f |q +
E
Z
≤
|f |q
X−E
|f |p +
E
Z
|f |r
X−E
Z
≤
Z
p
Z
|f | +
X
|f |r
X
< ∞,
so f ∈ Lq as desired.
A5. I will only sketch the proof, which is quite long (see Stein 175-177 for
details). If x ∈ M , then take y = x. Otherwise, set d := inf y∈M kx − yk. Since
M is closed and x ∈
/ M , d > 0. Now choose a sequence {yn } in M such that
kx − yn k → d, which we will prove is Cauchy, and whose limit in M (since M
is closed) will be the desired unique element y ∈ M closest to x. For
kx − yk ≤ kx − yn k + kyn − yk → d
as n → ∞,
and uniqueness is shown by first proving that x − y is orthogonal to M (take
perturbations of y in M and look at the norm of the difference with x) and then
applying the Pythagorean theorem to x − y and y − y 0 , where y 0 ∈ M is another
element minimizing distance to x.
To show the sequence is Cauchy, we use the parallelogram law
kA + Bk2 + kA − Bk2 = 2 kAk2 + kBk2
for all A, B ∈ H
with A = x − yn and B = x − ym .
7
2.2
Complex Analysis
B6. We write
2
iz
iz
eiz + e−iz
= e 2 − e− 2
,
2
so the zeroes of f are at z = 2π · k for k ∈ Z and are each of order 2 since
f (z) = 1 − cos z = 1 −
ieiz
eiz − 1
= lim
=1
z→0 1
z→0
z
lim
by L’Hospital’s rule (or apply L’Hospital to limz→2πk (1 − cos z)/(z − 2πk)2 and
note that you get a constant).
B7. (i) We have
iz
iz
1 z+1
e
− e− z+1 .
2i
Clearly the only singularity of f is at z = −1, and we claim that the singularity
is essential. Consider the sequence {xn } := {−1 + ni }. Then xn → −1 and
1 −n+i
e n
n−i |f (xn )| = e
−e
≥ 4 → ∞.
2i
f (z) =
On the other hand, let {yn } := {−1 + n1 }, which also satisfies yn → −1, but
1
i(−n+1)
−i(−n+1) |f (yn )| = e
−e
≤ 1.
2i
So z = −1 is an essential singularity of f .
(ii) Recall that
sin z = z −
and
cos z = 1 −
z3
z5
+
− ···
3!
5!
z2
z4
+
− ··· ,
2!
4!
as well as the identity
sin(a + b) = sin a cos b + cos a sin b,
which holds for all a, b ∈ C (since the proof relies on the properties of exponentials). So
z
1
sin
= sin 1 −
z+1
z+1
1
1
= sin 1 cos
− cos 1 sin
z+1
z+1
1
1
1
= sin 1 1 −
+
·
·
·
−
cos
1
−
+
·
·
·
2!(z + 1)2
z + 1 3!(z + 1)3
8
is the Laurent series for f (z).
(iii) The residue of f at z = −1 is the coefficient of
series, namely − cos 1.
1
z+1
in the Laurent
B8. We integrate
f (z) =
zeiz
(z + 3i − 1)(z − 3i − 1)
on a semicircle of radius R in the upper half plane, picking up the residue at
z = 3i + 1. The real part of the integral along the real axis goes to the desired
integral as R → ∞, while the integral along the curved portion goes to 0 by the
decay of eiz away from the real axis. The residue is
res3i+1 f =
lim (z − 3i − 1)f (z) =
z→3i+1
1
(3 cos 1 + sin 1 + i(− cos 1 + 3 sin 1)).
6e3
The real part of 2πi · res3i+1 f , which is the value of the desired integral, is
therefore 3eπ3 (cos 1 − 3 sin 1).
B9. (i) If f is entire and |f (z)| ≤ C|z|n for all z ∈ C, then f is a polynomial
of degree ≤ n. To see this, it suffices to prove that f (k) (0) = 0 for all k > n
(look at the power series). This in turn can be proved by the Cauchy inequality:
for any R > 0 we have
(k)!
|f (k) (0)| ≤ k CRn → 0
R
as R → ∞ when k > n.
(ii) The dimension is n + 1 since a basis is given by {1, z, z 2 , . . . , z n }.
B10. The polynomials f (z) := 2z 5 − z 3 + 3z 2 − z and g(z) = 8 are both
entire and have no zeroes on the circle |z| = 1. Indeed, |f (z)| ≤ 2 + 1 + 3 + 1 = 7
on this circle, while |g(z)| = 8 on the circle. Thus by Rouche’s theorem, the
number of zeroes of f + g inside the circle is the same as the number of zeroes
of g inside the circle, namely zero. Thus f + g has all 5 of its zeroes (counted
with multiplicities) outside the circle.
9
Chapter 3
Spring 2010
3.1
Real Analysis
1. (a) Since the sequence is Cauchy, we can pick a subsequence {fnk } such
that
1
kfnk − fnk+1 kp < k
2
for each k ≥ 1. We claim that this subsequence converges pointwise almost
everywhere. To prove this, we consider the series
f := fn1 +
∞
X
(fnk+1 − fnk )
k=1
and
g := |fn1 | +
∞
X
|fnk+1 − fnk |
k=1
with partial sums SK (f ) and SK (g), respectively. Then the triangle inequality
implies
K
X
kSK (g)kp ≤ kfn1 kp +
kfnk+1 − fnk kp < kfn1 kp + 1,
k=1
p
p
and we have |SK (g)| % |g| , so the monotone convergence theorem implies
Z
Z
|g|p = lim
|SK (g)|p < ∞.
K→∞
So g ∈ Lp ([0, 1]), and since |f | ≤ g, we have f ∈ Lp ([0, 1]) as well. In particular,
the series defining f converges (is less than infinity) almost everywhere. By
construction, SK−1 (f ) = fnk+1 , so fnk (x) → f (x) for almost every x. (See also
Stein 159-60.)
10
(b) Consider the sequence {χ[0,1] , χ[0, 12 ] , χ[ 21 ,1] , χ[0, 14 ] , . . . }. The sequence is
Cauchy and converges to the 0 function in the norm. But the pointwise limit
fails to exist anywhere since every x ∈ [0, 1] occurs in infinitely many intervals
of the form [ 2mn , m+1
2n ].
2. (a) The key thing is to show that if {fn } → 0 in Lp , then the same is
true in Lq (continuity). Here we must use the fact that Lp ⊂ Lq ; it is not true
in general (e.g. look at R). I don’t know how to do this.
(b) Suppose for contradiction that inf E 0 ∈M µ(E 0 ) = 0, so that we can choose
E1 , E2 , . . . in M0 with µ(En ) → 0. Then set fn = µ(En )−1/p χEn , which is sim
1/q
=
ple and hence measurable, so that kfn kp = 1, but kfn kq = µ(En )1−q/p
µ(En )− for some > 0. But this means that kfn kq → ∞, contradicting the
boundedness from (a).
3. Note that πV is the map that sends x ∈ H to the unique closest element
in V in the norm of H. One shows that x − πV (x) ⊥ V .
(a) For any x ∈ H, x − πV (x) ⊥ V , so by the Pythagorean theorem
kxk2 = k(x − πV (x)) + πV (x)k2 = kx − πV (x)k2 + kπV (x)k2 .
Thus kπV (x)k ≤ kxk, whence kπV k ≤ 1. It follows that kπV k = 1 since
kπV (x)k = kxk for nonzero x ∈ V .
πV is the identity on V since the unique closest element in V of an element
in V is itself. Thus πV is idempotent.
We need to show that (πV x, y) = (x, πV y) for all x, y ∈ H. This follows
easily from
0 = (x − πV x, πV y) = (x, πV y) − (πV x, πV y)
and likewise
0 = (πV x, y − πV y) = (πV x, y) − (πV x, πV y).
(b) Set V := im(P ), which is a subspace since it is the image of a linear
operator, and which we claim is closed. For if {xn } is a Cauchy sequence in V ,
with limit y ∈ H, then P idempotent and continuous implies {xn } = {P xn } →
P y, whence P y = y proves that y ∈ V . So V is closed.
Using the fact that P and πV are self-adjoint,
(P x, y) = (x, P y) = (x, πV P y) = (πV x, P y) = (P πV x, y) = (πV x, y)
for all x, y ∈ H. Thus ((P − πV )x, y) = 0 for all x, y, so (P − πV )x is perpendicular to all of H, hence 0, and this holds for each x, whence P = πV .
11
(Did we not need the assumption that the operator norm is 1?)
Λ
4. (a) The map `1
/ (`∞ )∗ defined by
Λ(ξ)(η1 , η2 , . . . ) =
X
ξi ηi
i
is well defined since
|Λ(ξ)(η)| ≤
X
|ξi ||ηi | ≤ kηk∞
i
X
|ξi | = kηk∞ kξk1 < ∞.
i
This also shows that the operator norm kΛ(ξ)k ≤ kξk1 , which is in fact equality
since if η = (ξ1 /|ξ1 |, ξ2 /|ξ2 |, . . . ), which has kηk∞ = 1, then
X
X |Λ(ξ)(η)| = ξi ξi /|ξi | = |ξi | = kξk1 .
i
i
So Λ is norm-preserving. It is clearly injective since if ξ 6= 0, then some ξi 6= 0,
hence the corresponding η with a 1 in the ith position and 0’s everywhere else
satisfies Λ(ξ)(η) = ξi 6= 0, hence Λ(ξ) 6= 0.
For the failure of surjectivity, I need to show that (`∞ )∗ contains functionals
that vanish on all η such that ηn → 0 as n → ∞. But I don’t know how to find
such a functional. Perhaps as some sort of limit? Heeeeeelp!!!
(b) We need
1
p
+
1
q
= 1, but I’m not sure about other conditions.
5. Suppose for contradiction that there is a set E such that the subset
Σ := {x ∈ E : f n (x) ∈
/ E for all n ≥ 1} has positive. We will prove that
Σ, f (Σ), . . . , f n (Σ)
are almost disjoint (their pairwise intersections have measure 0) for each n ≥ 0,
which will contradict the fact that µ(X) < ∞ since then the fact that f is
measure preserving will imply that
µ(Σ ∪ f (Σ) ∪ · · · ∪ f n (Σ)) = (n + 1)µ(Σ) → ∞.
(Note that if µ(X) = ∞, for instance X = R with the Lebesgue measure, then
f can be a shift by a constant and the claim fails.)
We use induction. When n = 0, the claim is trivial. Now suppose for
induction that for some k ≥ 0,
Σ, f (Σ), . . . , f k (Σ)
are almost disjoint. Since f is measure preserving,
f (Σ), f 2 (Σ), . . . , f k+1 (Σ)
12
must then also be almost disjoint. But Σ has empty intersection with each
f j (Σ) by the definition of Σ, whence
Σ, f (Σ), f 2 (Σ), . . . , f k+1 (Σ)
are almost disjoint.
3.2
Complex Analysis
6. (a) The singularities of f inside the circle of radius 2 are simple poles at
z = ±1. The corresponding residues are
res1 (f ) = lim (z − 1) · f = 1
z→1
and
res−1 (f ) = lim (z + 1) · f = 1.
z→−1
Thus
R
γ
f = 2πi(1 + 1) = 4πi.
(b) In the annulus 1 < |z| < 3, we have
1
1/z
1
=
=
1−
z+1
1 + 1/z
z
1
1/z
1
=
=
1+
z−1
1 − 1/z
z
1/3
1
1
=
=
1−
z+3
1 + z/3
3
1
1
+ 2 − ···
z
z
1
1
+ 2 + ···
z
z
z
z2
+
− ···
3
9
;
;
.
Expanding f by partial fractions, we get
f (z) =
(z 2
1
6z + 2
1
2
=
+
−
,
− 1)(z + 3)
z+1 z−1 z+3
so the coefficient of z1 of the Laurent series of f is 1 + 1 = 2, which agrees with
the sum of the residues in (a). To check the integral, note that every term of
the Laurent series has a primitive (and hence its integral vanishes) except the
one corresponding to z1 , and integrating z1 over any circle centered at the origin
gives 2πi. Multiplying by the coefficient, namely the residue, yields 4πi.
7. Skip.
13
8. (a) If f is analytic and injective on U , then f 0 is nonzero on U . Assume
for contradiction that f 0 (z0 ) = 0. Let the power series for f about z0 be
f (z) = a0 + a1 (z − z0 ) + a2 (z − z0 )2 + · · · .
Then a1 = 0 since the derivative vanishes, so
f (z) − a0 = ak (z − z0 )k + G(z),
where ak 6= 0, k ≥ 2, and G vanishes to order at least k + 1 at z0 . Now for small
w write
f (z) − a0 − w = F (z) + G(z),
where F (z) := ak (z − z0 )k − w.
The idea is that we can choose a small enough circle C around z0 and w small
enough so that |F (z)| > |G(z)| on C and F has k roots inside C. Then by
Rouche’s theorem, F (z)+G(z) has k roots inside C, contradicting the injectivity
of f .
One way to pick C and w is the following. By the vanishing of G, choose
> 0 so small that |G(z)| < |a2k | k on the circle C (z0 ). Then choose 0 < δ < 21 k
and set w = δak , so that
1
|F (z)| = |ak ||(z − z0 )k − δ| ≥ |ak | k > |G(z)|
2
on C (z0 ). Then F has its zeroes inside C (z0 ) since δ 1/k < , so we are done.
(b) f 0 never 0 on U does not ensure that f is injective on U . Let U = C and
f (z) = ez . Then f 0 (z) = ez has no roots, but e0 = e2πi , so f is not injective.
Another example is f (z) = z 2 , on the punctured disc centered at the origin.
9. There is a mistake: the upper half disc should be defined as all z with
|z| < 2 and Im z ≥ 0. First scale this map by 12 , then map to the upper half
plane by z 7→ −(z + z1 ), and finally map to the unit disc with z 7→ i−z
i+z . All
automorphisms of the unit disc are rotations composed with Blaschke factors.
−1
Chasing i, we see that i 7→ 2i 7→ 23 i 7→ −1
5 . So to map 5 to 0, we use the
α−z
−1
Blaschke factor ψα (z) = 1−ᾱz , with α = 5 . Thereafter we can compose with
arbitrary rotations since they fix the origin.
10. Skip.
14
Chapter 4
Fall 2009
4.1
Real Analysis
1. All the fn , f are integrable since they are bounded by f1 , which is integrable.
By assumption
Z
Z
Z
Z
|fn − f | = (fn − f ) = fn − f → 0.
Set
Eδ = {x ∈ X : fn (x) − f (x) ≥ δ for all n},
T
which is measurable since it is the intersection n (fn − f )−1 ([δ, ∞]). Let E be
the setSof all x ∈ X for which fn (x) 6→ f (x), which is measurable since it is the
union k E1/k . Then
Z
Z
|fn − f | ≥
|fn − f | ≥
E1/k
m(E1/k )
k
for all n, so m(E1/k ) = 0. Thus m(E) = 0 as well.
If we don’t assume f1 is integrable, then all bets are off. For instance, if
X
is
a set
R
R with infinite measure (e.g. R), fn ≡ 2 for all n, and f ≡ 1, then
fn → f since both sides are infinite, but fn 6→ f for all x.
2. For boundedness, simply note that
Z ∞
|fˆ(x)| ≤
|f (y)| dy = kf k1 < ∞.
−∞
For continuity, note that f (x)e−2πixy → f (x)e−2πix0 y for all y ∈ R as x → x0
by the continuity of the exponential, so fˆ(x) → fˆ(x0 ) by the dominated convergence theorem.
15
f
/ C sending y 7→ limn→∞ (xn , y) = limn→∞ (y, xn )
3. The function H
is well-defined and linear. We will show that f is bounded/continuous. Then
by the Riesz representation theorem, there exists x ∈ H such that f (y) = (y, x)
for all y ∈ H. Thus
lim (xn , y) = f (y) = (y, x) = (x, y)
n→∞
gives the desired result.
To prove boundedness, define the collection of bounded linear functionals
Tn
/ C by
H
Tn y = (y, xn ).
By assumption, supn |Tn y| < ∞ for each y, so by the uniform boundedness
principle, supn kTn k < ∞. Thus there is some M ≥ 0 such that |Tn y| ≤ M kyk
for all y ∈ H, whence
|f (y)| = lim (y, xn ) ≤ lim |(y, xn )| ≤ M kyk
n→∞
n→∞
proves that f is bounded.
4. Since f ∈ L∞ (X), |f | ≤ kf k∞ a.e. Thus
Z
Z
Z
f g ≤ |f g| ≤ kf k∞ |g|
by monotonicity and the fact that we can ignore a set of measure 0 when integrating. This proves that T is bounded and that kT k ≤ kf k∞ . Next we will
prove that kT k ≥ kf k∞ . We need to show that given > 0, there is a g such
that
Z
Z
f g > (kf k − ) |g|.
∞
By definition of kf k∞ , the set
Eδ := {x ∈ X : |f (x)| > kf k∞ − δ},
δ>0
has positive measure, and it is measurable since f is measurable. Set g(x) :=
χE (x)f (x)/|f (x)|, which is in L1 (X) since m(Eδ ) ≤ m(X) < ∞. Then
Z
Z
Z
f g =
|f
|
>
m(E
)(kf
k
−
δ)
=
(kf
k
−
δ)
|g|,
δ
∞
∞
Eδ
so taking δ = gives the desired result. Note that the strict inequality depends
on the fact that m(Eδ ) > 0.
16
5. The claim is false. Consider the sequence of intervals
1
1
1
1 2
2
1
{In } := [0, 1], 0,
, , 1 , 0,
, ,
, , 1 , 0,
,...
2
2
3
3 3
3
4
and let {χp
In } be the corresponding sequence of characteristic functions. Then
kχIn k2 = µ(In ) → 0 but the sequence {χIn (x)} has infinitely many 0’s and
1’s for each x, and therefore does not converge.
4.2
Complex Analysis
6. We prove Schwarz’s lemma.
Lemma 1 (Schwarz’s lemma). Let ∆
0, where ∆ is the open unit disc. Then
f
/ ∆ be holomorphic such that f (0) =
(i) |f (z)| ≤ |z| for all z ∈ ∆;
(ii) if equality holds in (i) for some z0 6= 0, then f is a rotation;
(iii) |f 0 (0)| ≤ 1, and if equality holds then f is a rotation.
Proof. (i): Since f (0) = 0, f (z)/z is holomorphic in ∆. If |z| = r < 1, we have
f (z) 1
z ≤ r,
and by maximum modulus this 1r bound holds for all |z| ≤ r. Letting r → 1
shows that 1 is an upper bound for all z ∈ ∆.
(ii): By assumption |f (z)| = |z|, so by maximum modulus f is constant.
Thus f (z) = cz, where |c| = 1, namely f is a rotation.
(iii): Write g(z) := f (z)/z. Then
f (z) − f (0)
= f 0 (0),
z→0
z
and we already showed in (i) that g is bounded by 1 in absolute value. If
|g(0)| = 1, then by (ii) we see that f is a rotation.
g(0) = lim
7. We will compute the Laurent series of f (z) = (1 + z 2 )e1/z to find the
residue of f at the essential singularity z = 0. Since
ez = 1 + z +
z2
z3
+
+ ··· ,
2!
3!
17
we have
e1/z = 1 +
1
1
1
+
+ ··· .
+
z
2!z 2
3!z 3
Thus
1
1 1
1 1
1
+ ) + (1 + ) + (1 + ) + z + z 2 ,
2! 4! z 2
3! z
2!
so by the residue theorem the integral of f on the ccw. unit circle is
f (z) = · · · + (
2πi · (1 +
7πi
1
)=
.
3!
3
8. We begin by proving that f is also holomorphic on the real axis. By
Morera’s theorem, it is enough to check that the integral of g over every triangle
is 0. If the triangle T has a side on the real axis, then slightly shifting it ensures
that T lies entirely in one of the open half planes, hence the integral over an
arbitrarily close contour is 0. Continuity then ensures the integral over T is 0.
Similarly, if the triangle crosses the real axis, then splitting the triangle into
smaller triangles and shifting them slightly gives the result. So f is entire.
Now define a new function
(
f (z) Im(z) ≥ 0;
g(z) :=
f (z̄) otherwise.
Then g is holomorphic on the open set Im(z) < 0 since for z near z0 in the lower
half plane we can use the power series expansion for f in the upper half plane
to write
X
f (z̄) =
an (z̄ − z̄0 )n ,
whence
g(z) = f (z̄) =
X
ān (z − z0 )n
is a power series expansion for g. So g is holomorphic except possibly on the
real axis, and continuous on the real axis since f is continuous and real-valued
there. Replacing f by g in the above argument thus shows that g is entire.
Since f and g agree on the entire upper half plane, they must agree everywhere
by the uniqueness of analytic continuation.
9. Skip.
10. Let F (z) = z 4 and G(z) = −6z − 3. Then for all |z| = 2, we have
|F (z)| = 16 > 15 = 6 · 2 + 3 ≥ |G(z)|,
18
whence by Rouche’s theorem the functions F and F + G have the same number
of solutions within the unit circle, namely all 4.
19
Chapter 5
Spring 2009
5.1
Real Analysis
1. There are typos in the problem, which I assume is meant to be the following. Let {xn } be a sequence of measurable functions with 0 ≤ fn ≤ 1 for each
R1
n such that 0 fn → 0. Then is it necessarily true that fn → 0 a.e.?
No, this is false. Consider the sequence of intervals
1
1
1 2
2
1
1
, , 1 , 0,
, ,
, , 1 , 0,
,...
{In } := [0, 1], 0,
2
2
3
3 3
3
4
and let χIn be the corresponding characteristic functions. Then the assumptions
hold but χIn → 0 nowhere.
2. If f ∈ L∞ , then for each p ≥ 1,
Z
Z
kf kpp = |f |p ≤ kf kp∞ ≤ kf kp∞
since µ(X) = 1. Thus f ∈ Lp for each p.
For any > 0, we wish to show that there is an N such that for all p ≥ N ,
kf k∞ − kf kp < . Choose δ such that > δ > 0 and set Eδ := {x ∈ X : |f (x)| >
kf k∞ − δ}, which has positive measure. Then
Z
kf kpp ≥
|f |p ≥ (kf k∞ − δ)p µ(Eδ ),
Eδ
which implies
kf kp ≥ (kf k∞ − δ)µ(Eδ )1/p .
Since µ(Eδ ) has positive measure ≤ 1, µ(Eδ )1/p → 1 as p → ∞, whence we get
the desired result since δ < .
20
3. First, we may assume T is injective. (Assuming the axiom of choice,
we can extend a basis of ker(T ) to a basis for X, obtaining a decomposition
X = ker(T ) ⊕ X 0 . T |X 0 is a compact bijective linear map from X 0 to Y .)
Suppose for contradiction that Y is infinite dimensional. Let y1 , y2 , . . . be a
normal basis of Y . Since T is bijective, there exist unique x1 , x2 , . . . such that
T xn = yn . Then {xn } cannot be bounded since if it were, compactness of
T would ensure the existence of a convergent subsequence {T xnk }, which is
impossible. So we may assume |xn | → ∞
This basis method does not seem so good. I can’t even prove that the yk s
don’t converge!
Help!
/ C by Tn y = hy, vn i for
4.
Define bounded linear functionals H
all y ∈ H. By assumption, this collection of linear functionals is bounded for
each y ∈ H, hence bounded in the norm by the uniform boundedness principle.
Thus there exists M ≥ 0 such that |Tn y| ≤ M kyk for all n, y. In particular,
kvn k2 = |Tn vn | ≤ M kvn k implies kvn k ≤ M for all n.
Tn
5. Skip.
5.2
Complex Analysis
6. Since f (z) sin z → 1 as z → 0 and sin z = z − z 3 /3! + · · · has a zero of order
1 at 0, the limit implies that f has a pole of order 1 at 0, with residue 1. Thus
the residue theorem shows that the integral of f on the ccw. unit circle is 2πi.
7.
Since f is bounded by B outside DR , all its poles must be in DR ,
which is compact, so there can be only finitely many poles, at a1 , . . . , an . Then
g(z) = f (z)(z − a1 ) · · · (z − an ) is entire and satisfies
|g(z)| ≤ C|z|n
for some constant C ≥ 0. We claim that this implies that g is a polynomial of
degree ≤ n, for which it suffices to show that g (k) (0) = 0 for all k > n (look at
the power series). This is proved by using the Cauchy inequalities:
|g (k) (0)| ≤
k!
CRn → 0
Rk
21
as R → ∞ when k > n. Thus
f (z) =
g(z)
(z − a1 ) · · · (z − an )
expresses f as a rational function.
R
8. Fix a point z0 ∈ Ω. Define F (z) = γ f (z) dz, where γ is any curve consisting of line segments parallel to the axes, starting at z0 and ending at z. This
is well-defined since our assumption ensures that the definition is independent
of the curve chosen. We now prove that F is holomorphic, with derivative f .
For h so small that z + h ∈ Ω, note that
Z
F (z + h) − F (z) = f (w) dw,
ξ
where ξ is any curve from z to z + h composed of line segments parallel to the
axes. Actually, we insist that ξ is of minimal length (as direct as possible from
z to z + h, to ensure so that we can use the continuity of f to bound part of
the integral, as follows. Since f is continuous, write f (w) = f (z) + ψ(w), where
ψ(w) → 0 as w → z. Thus
Z
Z
Z
f (w) dw = f (z) dw + ψ(w) dw.
ξ
Since
R
ξ
ξ
ξ
dw = h and
Z
ψ(w) ≤ |h| sup |ψ(w)|,
w∈ξ
ξ
with the supremum tending to 0 as h → 0, we see that
F (z + h) − F (z)
= f (z),
h→0
h
lim
as desired.
9. As stated, the problem is false since, for instance, f (z) = z + z 2 has
f (0) = 0, f 0 (0) = 1, but f is not the identity map. Clearly we are meant to
assume that f maps Q to Q. Let z0 ∈ Q denote the point at which f (z0 ) = z0
and f 0 (z0 ) = 1. I want to use an argument similar to the Schwarz lemma but I
don’t know how. Riemann mapping theorem to map Q to the unit disc? But
then the derivative may change...
10. The idea is to rewrite the integral as a complex integral that reduces
to the desired integral when we use the parametrization z = eiθ . Let γ be the
22
ccw. unit circle and consider the integral
Z
Z
1
1
1
2i
dz =
dz
1
z−
i γ z a+ z
i γ 2aiz + z 2 − 1
2i
Z
1
√
√
=2
dz.
2 − 1))(z + i(a +
(z
+
i(a
−
a
a2 − 1))
γ
From the left side, we see that taking z = eiθ and dz = ieiθ dθ gives the desired
integral. On
√ the other hand, the right side shows that the integrand has poles
at −i(a ± a2 − 1). The one corresponding
to (+) clearly does not lie in the
√
2 − 1) always does for a > 1, namely
unit circle, but we
claim
that
−i(a
−
a
√
that f (a) := a − a2 − 1 < 1. For this it suffices to argue that f (1) = 1, and
that f 0 (a) < 0 for a > 1, hence f is strictly decreasing for a > 1. This is easy:
2a
a
f 0 (a) = 1 − √
=1− √
< 1 − 1 = 0.
2 a2 − 1
a2 − 1
√ Now we proceed to use the residue theorem. The residue at the pole −i(a −
a2 − 1) is
lim√
z→−i(a−
i
2
2
√
= ,
=
2
−2ai
a
a − 1)
a2 −1) z − i(a +
so the residue theorem shows that the integral is
i
2π
2πi( ) = − .
a
a
23
Chapter 6
Fall 2008
6.1
Real Analysis
1. (a) Ta is clearly linear since multiplication in R distributes over addition.
Ta x ∈ `∞ since
n
∞
X
X
kTa xk∞ = sup |(Ta x)n | = sup ai xi ≤
|ai xi | ≤ kxk∞ kak1 < ∞,
i=1
i=1
which also shows that kTa k ≤ kak1 .
(b) To prove equality, note first that if a = 0, then Ta = 0 and the result
is trivial. So suppose a 6= 0 and let x be the sequence defined by
(
an /|an | when an 6= 0;
xn =
0
when an = 0.
(xn is the sign of an or 0). Then kxn k∞ = 1 since a 6= 0, and
n
n
X
X
sup ai xi = sup |ai | = kak1 = kak1 kxn k∞ .
i=1
i=1
2. (a) Suppose f ∈ L2 . Let E be those points x ∈ X for which |f (x)| ≤ 1.
Then
Z
Z
Z
Z
|f | =
|f | +
|f | ≤ µ(E) +
|f |2 ≤ µ(E) + kf k22 < ∞,
E
X−E
X−E
so f ∈ L1 .
24
Since |f | is measurable and non-negative, there is a sequence of simple functions {ϕn } such
thatR ϕn % |f |. Thus the
convergence theorem guarR
R monotone
R
antees that ϕn → |f |, and likewise ϕ2n → |f |2 . If we can prove that
p
kϕk1 ≤ µ(X)kϕk2
holds for all non-negative simple functions ϕ, then we get the inequality for all
f ∈ L2 since
p
p
kϕn k1 ≤ µ(X)kϕn k2 ≤ µ(X)kf k2
by monotonicity, and the inequality is preserved upon taking the limit of the
left side.
PN
So let ϕ be a simple function, with canonical form ϕn = k=1 ak χEk , where
the Ek are disjoint finite measurable sets and the ak are distinct positive real
numbers. Then
!2
Z 2
N
N
X
X
X
a2k µ(Ek )2 +
ϕ =
ak µ(Ek )
=
aj ak µ(Ej )µ(Ek ). (∗)
k=1
k=1
1≤j<k≤N
On the other hand,
!Z
! N
!
N
N
X
X
X
2
2
µ(Ek )
ϕ =
µ(Ek )
ak µ(Ek )
k=1
k=1
=
N
X
k=1
a2k µ(Ek )2 +
k=1
X
(a2j + a2k )µ(Ej )µ(Ek ).
(∗∗)
1≤j<k≤N
Since aj ak ≤ a2j +a2k for non-negative real numbers aj , ak , we see that (∗) ≤ (∗∗).
Thus
v
uN
Z
uX
p
µ(Ek ) · kϕn k2 ≤ µ(X)kϕn k2 ,
kϕn k1 = ϕn ≤ t
k=1
as desired.
p
To see that kik ≥ µ(X), simply take f ≡ 1.
(b) Suppose for contradiction that there is a sequence {En } of measurable sets such that µ(En ) → 0. Taking a subsequence if necessary, we may
assume µ(En ) ≤ 41n . Then define a sequence of simple functions {ϕn } by
ϕn = √ 1 χEn . Then
µ(En )
N
N Z
N p
∞ p
∞
X X
X
X
X
1
ϕn ≤
ϕn =
µ(En ) ≤
µ(En ) ≤
≤ 1.
n
2
n=1
n=1
n=1
n=1
n=1
1
P∞
Taking the limit of the left side, we see that f := n=1 ϕn is well-defined and
in L1 . But
N Z
N
X
X
2
kf k2 ≥
ϕ2n =
1=N
n=1
n=1
25
for each N , whence f ∈
/ L2 , contradicting the assumption.
3.
For only if, suppose f = zg for some z ∈ C∗ . If h ∈ H such that
hg, hi = 0, then
hf, hi = hzg, hi = zhg, hi = 0,
so there is no h with the stated properties.
For if, suppose that for every h ∈ H orthogonal to g, hf, hi 6= 1. Then in
fact hf, hi = 0 is necessary since otherwise we could scale h appropriately. Let
S be the subspace spanned by g, which is closed, and consider H = S ⊕ S ⊥ .
Choose λ such that f = λg + h, where h ∈ S ⊥ . Then h ⊥ g and therefore h ⊥ f
imply the equalities
1
λhg, gi = hf, gi = hf, f i,
λ̄
whence kf k = |λ|kgk. But the Pythagorean theorem gives
kf k2 = |λ|2 kgk2 + khk2 ,
whence h = 0, so f = λg as desired.
4. Define g(x, y) = f (x)χ[y,1]×[0,1] (x, y), which is non-negative and measurable since f (x) is measurable on [0, 1] × [0, 1] (see Stein 85). Note that g is
really just
(
f (x) if x ≥ y;
g(x, y) =
0
otherwise.
By Fubini’s theorem, we obtain
!
Z
Z
Z
f (x)χ[y,1]×[0,1] (x, y) dy dx =
[0,1]
[0,1]
f (x)χ[y,1]×[0,1] (x, y) dx
[0,1]
[0,1]
in the extended sense, where the left side yields
Z
xf (x) dx
[0,1]
and the right side yields
Z
!
Z
f (x) dx
[0,1]
[0,y]
This proves the claim.
5. Skip.
26
!
Z
dy.
dy,
Chapter 7
Spring 2008
7.1
Real Analysis
1. Set fn := f χEn , where
En := {x ∈ X : f (x) ≤ n}.
R
R
Then
R
Rfn % f , so fn → f . Thus given > 0, we can choose N > 0 such that
f − fn < /2 whenever n ≥ N . Now if we choose δ > 0 such that δ < /2N ,
then for any measurable E ⊆ X with µ(E) < δ, we compute
Z
Z
Z
Z
n
≤ ,
f=
(f − fn ) +
fn ≤ (f − fn ) + n · µ(E) < +
2 2N
E
E
E
which proves the result.
2. Since K is continuous on a compact subset of R2 , K is bounded, namely
K ∈ L∞ ([0, 1] × [0, 1]), with |K| having supremum kKk∞ . Then
Z
Z
|(T f )(x)| = K(x, y)f (y) dy ≤
|K(x, y)||f (y)| dy ≤ kKk∞ kf k1 ,
[0,1]
[0,1]
so T f is in fact uniformly bounded, hence square integrable on [0, 1], a set with
finite measure. Since in fact [0, 1] has measure 1, the above inequality implies
that
Z
Z
kT f k1 :=
|T f | ≤
kKk∞ kf k1 ≤ kKk∞ kf k1 ,
[0,1]
[0,1]
whence kT k ≤ kKk∞ .
27
3. We can write T = ker T ⊕ (ker T )⊥ , where T |(ker T )⊥ gives a bijection
with im(T ). Then T |(ker T )⊥ is bounded since linear transformations between
finite-dimensional Hilbert spaces are bounded, whence T has the same bound.
Bounded linear transformations map closed sets to closed sets since if xk →
x, then T xk → T x. This holds because
kT x − T xk k = kT (x − xk )k ≤ kT kkx − xk .k
Now, every bounded closed set of a finite-dimensional normed space is compact. For this it suffices to show that every bounded sequence in a finitedimensional vector space has a convergent subsequence. Choosing a normal basis {e1 , . . . , en }, we have ka1 e1 + · · · + an en k ≤ |a1 | + · · · + |an |, so convergence
is analogous to convergence in Cn , which is guaranteed by Bolzano-Weierstrass.
Then if T is bounded and {xk } is bounded, {T xk } is bounded, hence has a
convergent subsequence, so T is compact.
28