MATH 632 Take-Home Final Exam,
due Dec. 18, 2014
Prof. Jonathan Rosenberg
Solutions
1. Let H be a complex Hilbert space and let B : H × H → C be a sesquilinear form (linear in the
first variable when the second variable is fixed, conjugate linear in the second variable when the first
variable is fixed) with the property that there is a constant C > 0 such that |B(ξ, η)| ≤ C kξk kηk.
Show that there is a unique bounded linear map T : H → H such that B(ξ, η) = hT ξ, ηi.
Solution. First fix ξ and consider ϕξ : η 7→ B(ξ, η). This is a linear functional on H (since B
is conjugate linear in the second variable but we took the complex conjugate) and by assumption,
|ϕξ (η)| ≤ C kξk kηk, so ϕξ is bounded with norm ≤ Ckξk. By the Riesz representation theorem,
ϕξ must be given by inner product with a vector ωξ , i.e., B(ξ, η) = hη, ωξ i = hωξ , ηi, where
kωξ k ≤ Ckξk. Now ξ 7→ ωξ is linear, because of the linearity of B in the first variable, so we can
write ωξ = T ξ for some linear map T : H → H. Then inequality kωξ k ≤ Ckξk says that kT k ≤ C,
so T is bounded.
2. For purposes of this problem, you can take a probability measure on [0, 1] to be a linear functional
µ : C([0, 1]) → R (here we work with real scalars) such that µ(1) = 1 and kµk = 1. (a) Show
that these conditions guarantee that µ(f ) ≥ 0 if f ≥ 0 (so that positivity is automatic). Now let the
expected value and variance of such a measure be E(µ) = µ(x) and Var(µ) = µ(x2 ) − E(µ)2 . (Here
x is the identity function [0, 1] → R.) (b) Show that for any 0 ≤ c ≤ 14 , there is a probability measure
µ on [0, 1] with E(µ) = 21 and Var(µ) = c, and that for a probability measure µ with E(µ) = 21 , these
are the only possible values of the variance. (Hint: Hahn-Banach, applied to the linear subspace V of
C([0, 1]) spanned by 1, x, and x2 .)
Solution. (a) Assume f ≥ 0 and let m be the maximum value of f on [0, 1], so that 0 ≤ f ≤ m and
0 ≤ m − f ≤ m. Then kf k = m and km − f k ≤ m, and since µ has norm 1, |µ(f )| ≤ m and
|µ(m − f )| ≤ m. Since µ(m) = m, that means 0 ≤ µ(f ) ≤ m. So µ is positive. (b) Let V be the
span of 1, x, and x2 . If E(µ) = 12 , then µ(1) = 1 and µ(x) = 12 . Let µ(x2 ) = a. Since µ is positive
2
and 0 ≤ x2 ≤ x on [0, 1], we have 0 ≤ a ≤ 21 , and Var(µ) = a − 12 ≤ 12 − 14 = 14 . However if
2
f = x − 12 = x2 − x + 41 , then f ≥ 0, so µ(f ) = a − 12 + 14 ≥ 0, i.e., a ≥ 14 . So Var(µ) ≥ 0.
1
Thus 0 ≤ c ≤ 14 . In the other direction, fix a ∈
1
1
,
4 2 , and define µ on V by
β
+ aγ.
2
This is obviously linear, and because of the choice of a, it’s a convex combination of
β 1
µ1 (α + βx + γx2 ) = α + + γ
2
4
and
β 1
µ2 (α + βx + γx2 ) = α + + γ.
2
2
1
But the formula for µ1 can
be rewritten µ1 (f ) = f 2 and the formula for µ2 can be rewritten
µ2 (f ) = 21 f (0) + f (1) . These are both probability measures, and thus so is µ. By Hahn-Banach,
µ extends to a probability measure on all of C([0, 1]) (or in thiscase you can avoid Hahn-Banach
by
1
1
defining the extension to be a convex combination of f 7→ f 2 and f 7→ 2 f (0) + f (1) .
R∞
3. Recall that the Fourier transform F(f )(ξ) = √12π −∞ f (x)eiξx dx is well-defined and invertible on
R∞
S(R), with inversion formula f (0) = √12π −∞ F(f )(ξ) dξ, and that F extends to a unitary operator on L2 (R). Show that there is a function f ∈ L1 (R) which is continuous at 0 (so that f (0) is
unambiguous) for which the Fourier inversion formula fails (because F(f ) 6∈ L1 (R)). You don’t
necessarily need to construct such an f ; a non-constructive proof is enough.
µ(α + βx + γx2 ) = α +
Solution. The simplest example of such a function is f = χ[−1,1] , the characteristic function of
[−1, 1]. Clearly this is in L1 and is continuous at 0. Its Fourier transform is
Z 1
2 sin ξ
1
1 1 iξ
1 h eiξx i1
(Ff )(ξ) = √
=√
e − e−iξ = √
eiξx dx = √
.
2π −1
2π iξ −1
2π iξ
2πξ
This function is smooth but not in L1 . However the Fourier inversion formula holds in the sense of
principal value integrals:
Z ∞
Z ∞
sin ξ
2
1
2
√ p.v.
p.v.
dξ =
π = 1 = f (0).
(Ff )(ξ) dξ =
2π
2π
2π
−∞ ξ
−∞
RN
One can also give a nonconstructive proof as follows. Let φN (f ) = √12π −N F(f )(ξ) dξ. Unwinding
the definitions and using Fubini, which applies since for f ∈ L1 (R), f (x)eiξx is absolutely integrable
over [−N, N ] × (−∞, ∞), we have
Z N Z ∞
1
φN (f ) =
f (x)eiξx dx dξ
2π −N −∞
Z ∞
Z N
1
=
f (x)
eiξx dξ dx
2π −∞
−N
ixξ ξ=N
Z ∞
Z
e
sin(N x)
1
1 ∞
=
f (x)
dx =
f (x)
dx.
2π −∞
ix ξ=−N
π −∞
x
2
Here sin(N x)/x is interpreted as N at x = 0, and is a C ∞ function on R.
Say for simplicity that f is even and f ≡ 0 on [−ε, ε] for some ε > 0. If this (plus f being in L1 ) were
enough to guarantee that the Fourier inversion formula holds, then we’d have to have φN (f ) → 0,
R∞
x)
i.e., ε f (x) sin(N
dx → 0 as N → ∞, for any f ∈ L1 ([ε, ∞)). Since the dual of L1 is L∞ , by the
x
x)
Uniform Boundedness Principle, that would mean that the L∞ norm of x 7→ sin(N
on [ε, ∞) would
x
∞
−1
have to tend to 0. But it doesn’t; the L norm stays uniformly close to ε .
4. Let H be a Hilbert space and let T : H → H be a self-adjoint bounded linear operator with 0 in its
spectrum. Prove that the range of T is closed if and only if 0 is an isolated point in the spectrum of T .
Solution. First suppose 0 is isolated in the spectrum σ(T ). Thus σ(T ) = {0} ∪ X, where X is
a compact subset of R a positive distance away from 0. Let f be the continuous function on σ(T )
which is 1 on {0} and 0 on X. Then P = f (T ) lies in the unital Banach algebra generated by T and
is the spectral projection onto the kernel of T . Let V = P H, which is a closed subspace of H. On V ,
T is identically 0, whereas on (1 − P )H = V ⊥ , T has spectrum in X, which being positive distance
from 0, implies that T is invertible as a map from V ⊥ to itself. Thus V ⊥ is the range of T , and this is
a closed subspace of H.
For the converse, assume T has closed range and again let V = ker T . Since T is self-adjoint,
ker T = ker T ∗ = range(T )⊥ , and if T has closed range, range(T ) = V ⊥ . Since T is injective on
V ⊥ , the fact that V ⊥ is all of range(T ) means
by the Open Mapping Theorem that T is invertible on
V ⊥ . This means σ(T ) = {0} ∪ σ T |V ⊥ , and since T is invertible on V ⊥ , σ T |V ⊥ is a positive
distance away from 0. This implies 0 is isolated in σ(T ).
5. Let H1 and H2 be Hilbert spaces and let T : H1 → H2 be a (bounded) Fredholm operator. Show
∗
∗
that e−s T T and e−s T T converge in norm to the orthogonal projections onto ker(T ) and ker(T ∗ ),
respectively, as s → ∞.
Solution. Note that T ∗ T is positive self-adjoint on H1 and T T ∗ is positive self-adjoint on H2 . These
operators are both Fredholm, since T and T ∗ are, so they have closed range. Obviously ker T ⊆
ker T ∗ T and ker T ∗ ⊆ ker T T ∗ . But the reverse inclusions also hold; for ξ ∈ ker T ∗ T , kT ξk2 =
hT ξ, T ξi = hT ∗ T ξ, ξi = 0, so ξ ∈ ker T , and similarly ξ ∈ ker T T ∗ implies T ∗ ξ = 0. By the last
problem, T ∗ T and T T ∗ each have a spectral gap, i.e., σ(T ∗ T ) is bounded away from 0 on (ker T )⊥
and σ(T T ∗ ) is bounded away from 0 on (ker T ∗ )⊥ . So there is a ε > 0 such that T ∗ T ≥ ε on
∗
(ker T )⊥ and T T ∗ ≥ ε on (ker T ∗ )⊥ . Then for s > 0, e−s T T is e−s0 = 1 on ker T and is ≤ e−sε
∗
on (ker T )⊥ (though still positive). So as s → ∞, e−s T T converges in norm to the orthogonal
∗
projection onto ker T . Similarly, e−s T T converges in norm to the orthogonal projection onto ker T ∗ .
3