Functional Analysis: Chapter 9 Solutions
Sarah L. Browne
1. (a) Let T : H → H be unitary. Then T ∗ = T −1 , so T is an isomorphism.
Let x ∈ H. Then
kT xk2 = hT x, T xi = hT ∗ T x, xi = hx, xi = kxk2
so T is an isometry.
Conversely, let T be an isometric isomorphism. Then kT vk2 = kvk2
for all v ∈ H.
Let x, y ∈ H. Then if H is a real Hilbert space,
hx, yi =
1
(kx + yk2 − kx − yk2 )
4
and if H is a complex Hilbert space,
hx, yi =
1
i
(kx + yk2 − kx − yk2 ) + (kx + iyk2 − kx − iyk2 ).
4
4
Hence hT x, T yi = hx, yi. It follows that hT ∗ T x, yi = hx, yi. Since
this identity holds for all x, y ∈ H, we have T ∗ T = I.
Since the map T is an isomorphism, it is invertible. By the above,
we must have T −1 = T ∗ , meaning T is unitary.
(b) The right shift operator, R, on l2 is defined by the formula
R(a1 , a2 , a3 , . . .) = (0, a1 , a2 , a3 , . . .).
Clearly R is an isometry. But R is not surjective, and therefore not
an invertible; it therefore cannot be a unitary.
2. Let T : H → H be self-adjoint. Let v ∈ H. Then
hv, T vi = hT v, vi = hv, T vi
so hv, T vi ∈ R.
Conversely, suppose hv, T vi ∈ R for all v ∈ H. Let x, y ∈ H. We need to
show
hT x, yi = hx, T yi.
Let α ∈ C. Then
hx + αy, T (x + αy)i = hx, T xi + αhy, T xi + αhx, T yi + |α|2 hy, T yi ∈ R.
1
2
Thus
αhT x, yi + αhx, T yi ∈ R.
Set α = 1. Then
=(hT x, yi + hx, T yi) = 0
that is
=(hT x, yi) = =(hx, T yi).
Taking α = i, we similarly see that <(hT x, yi) = <(hx, T yi). Therefore
hT x, yi = hx, T yi
and we are done.
3. (a) Let x ∈ H. Observe
hx, T ∗ T xi = hT x, T xi ≥ 0.
But this is the definition of the operator T ∗ T being positive.
(b) Since A is a positive self-adjoint operator, the Spectrum(A) ⊆
[m(A), M (A)], where
m(A) = inf hv, Avi
kvk=1
M (A) = sup hv, Avi
kvk=1
Now, by positivity, hv, Avi ≥ 0 for all v ∈ V . So m(A) ≥ 0. By the
Cauchy-Schwarz inequality, when kvk = 1, we have hv, Avi ≤ kAk,
meaning M (A) ≤ kAk.
Hence Spectrum(A) ⊆ [0, kAk] as desired.
(c) We have
S ∗ (a1 , a2 , a3 , . . .) = (0, a1 , a2 , a3 , . . .)
so
S ∗ S(a1 , a2 , a3 , . . .) = (0, a2 , a3 , a4 , . . .)
Observe that S ∗ S is positive, and kS ∗ Sk = 1, so by the above
kS ∗ Sk ⊆ [0, 1]. On the other hand, S ∗ S = S ∗ S − 0I is clearly
not invertible, so 0 ∈ Spectrum(S ∗ S).
Further,
(S ∗ S − I)(a1 , a2 , a3 , . . .) = (−a1 , 0, 0, . . .)
is also clearly not invertible, so 1 ∈ Spectrum(S ∗ S).
4. Let T : V → W be compact. Let (xn ) be a bounded sequence in V . Let
B = {xn | n ∈ N}.
Then the image T [B] has compact closure. Thus every sequence in B
has a convergent subsequence. In particular, the sequence (T xn ) has a
convergent subsequence.
3
Conversely, suppose that for every bounded sequence (xn ), the image (T xn ) has a convergent subsequence. Let (yn ) be a sequence in
T [BV (0, 1)]. Then yn = T xn , where (xn ) is bounded. Hence (yn ) has
a convergent subsequence, with limit in the closure T [BV (0, 1)]. It follows
by definition of compactness that the closure T [BV (0, 1)] is compact.
5. Define TN : l2 → l2 by
TN (a1 , a2 , a3 , . . .) = (a1 ,
a2
aN
,...,
, 0, 0, . . .).
2
N
Then
∞
∞
X
1 X
|an |2
1
≤
|an |2 =
k(a1 , a2 , a3 , . . .)k2 .
2
n
N + 1 n=1
N +1
kT (a1 , a2 , a3 , . . .)−TN (a1 , a2 , a3 , . . .)k2 =
n=N +1
Thus kT − TN k ≤ 1/(N + 1), and so kT − TN k → 0 as N → ∞.
Now each map TN has finite rank, and is therefore compact. It follows
that T is compact.
Checking that T is self-adjoint is completely straightforward.
6. Define T : C[0, 1] → C[0, 1] by
(T f )(x) = xf (x).
Then
kT f k2 =
Z
1
1
Z
|f (x)|2 dx = kf k2 .
|xf (x)| dx ≤
0
0
So T is a bounded linear map, and therefore extends to a bounded linear
map on the Hilbert space L2 [0, 1].
Let f, g ∈ C[0, 1]. Then
Z
1
xf (x)g(x) dx = hf, T gi.
hT f, gi =
0
This equation extends by continuity to L2 [0, 1], so the operator T is selfadjoing.
Now, let λ be an eigenvalue of T , and let f be a corresponding eigenvector.
Then, at least for f a continuous function, we have
xf (x) − λf (x) = 0
for all x ∈ [0, 1].
Hence f (x) = 0 or x = λ for any value λ ∈ [0, 1]. Thus f (x) = 0 for all
x ∈ [0, 1] except, possibly, for one value.
It follows that kf k = 0, meaning (even for f not continuous), f corresponds to the point 0 in L2 [0, 1]. Thus, if T f = λf , then f = 0, and the
operator T has no eigenvalues.
4
7. Observe
Z
1
1
Z
hφm,n , φm0 ,n0 i =
em (s)en (t)em0 (s)en0 (t) ds dt = hem , em0 ihen , en0 i.
0
0
It follows that the set (φm,n )m,n∈N is orthonormal.
Let f ∈ L2 ([0, 1] × [0, 1]). Suppose hφm,n , f i = 0 for all m, n ∈ N. Then
1
Z
Z
1
em (s)en (t)f (s, t) ds dt.
0
0
Let
1
Z
em (s)f (s, t) ds.
gm (t) =
0
Then the above tells us that hen , gm i = 0 for all m, n ∈ N.
Since the set (en ) is an orthonormal basis for L2 [0, 1], we know that
{en | n ∈ N}⊥ = {0}
so gm = 0 for all m ∈ N. Hence, for each value of t, the function s 7→ f (s, t)
belongs to the set {em | m ∈ N}⊥ , which means that f = 0.
Hence
{φm,n | m, n ∈ N}⊥ = {0}
so the orthonormal set (φm,n )m,n∈N is in fact a basis for L2 [0, 1].
8. Let
Z
y(s) =
1
k(s, t)f (t) dt.
0
Observe k(0, t) = 0 and k(1, t) = 0 for all t. So y(0) = y(1) = 0. Now, we
can write
Z s
Z 1
y(s) =
t(1 − s)f (t) dt +
s(1 − t)f (t) dt.
0
s
By the fundamental theorem of calculus and the rules for differentiating
under the integral sign, we have
Z s
Z 1
Z s
Z 1
y 0 (s) = s(1−s)f (s)−
tf (t) dt−s(1−s)f (s)+ (1−t)f (t) dt = −
tf (t) dt+ (1−t)f (t) dt.
0
s
0
Differentiating again, by the fundamental theorem of calculus, we have
y 00 (s) = −sf (s) − (1 − s)f (s) = −f (s)
so y 00 (s) + f (s) = 0 as desired.
s
5
Conversely, suppose y 00 (s) + f (s) = 0 and y(0) = y(1) = 0. Using the
above calculation to integrate, we see
Z 1
Z s
(1 − t)f (t) dt + A
tf (t) dt +
y 0 (s) = −
s
0
and
Z
s
Z
1
t(1 − s)f (t) dt +
y(s) =
0
s(1 − t)f (t) dt + As + B
s
where A and B are constants.
Thus
1
Z
y(s) =
k(s, t)f (t) dt + As + B.
0
The conditions y(0) = y(1) = 0 now tell us that A = B = 0, and we are
done.
9. (a) Observe
Z
s
Z
t(1 − s)f (t) dt +
(Af )(s) =
0
1
s(1 − t)f (t) dt.
s
Suppose Af = λf , and λ 6= 0. Then (Af )(0) = (Af )(1) = 0, so
f (0) = f (1) = 0.
Now, let y = Af . Then, differentiating the above, we see
y 00 (s) + f (s) = y 00 (s) +
y(s)
= 0.
λ
This equation has general solution
y(s) = A cos(s/λ) + B sin(s/λ).
But we know that y(0) = y(1) = 0. Hence we must have A = 0, and
1/λ = kπ where k ∈ Z, k 6= 0. It follows that the set of non-zero
eigenvalues is
1
{
| k ∈ Z\{0}}
kπ
and the eigenvalue 1/kπ has a corresponding eigenvector sin(kπs).
(b) Observe that A is a compact self-adjoint operator. Hence
kAk = sup{|λ| | λ ∈ Spectrum(A)} =
1
.
π