Download Sheet 8 - TUM M7/Analysis

Technische Universität München
Zentrum Mathematik
Prof. Dr. Simone Warzel
Dr. Frank Hofmaier
Sommersemester 2016
Exercise sheet no. 8
Operator Theory (MA 5012)
Exercises (to be discussed on June, 14/15)
29. Let A : D(A) → X, D(A) ⊂ X be a symmetric operator on a Hilbert space
X. Show that the following statements are equivalent:
(i) A is essentially self-adjoint,
(ii) (A − z) D(A) = X for all z ∈ C \ R,
(iii) there is z± ∈ C with Im z+ > 0 and Im z− < 0 and (A − z± ) D(A) = X,
(iv) there is z± ∈ C with Im z+ > 0 and Im z− < 0 and ker(A∗ − z± ) = {0},
(v) σ(A) ⊂ R.
30. Let (M, µ) be a measure space and f : M → C measurable.
a) Prove that the
spectrum of the maximal multiplication Mf coincides
with z ∈ C For all ε > 0 : µ{x ∈ M | |f (x) − z| < ε} > 0 .
b) Give a characterisation of the point spectrum of Mf .
c) Give an example of such an operator which has both point and continuous spectrum.
31. Consider the unilateral shift operator S on ℓ2 (N) defined via (Sa)n+1 = an
for n ∈ N and (Sa)1 = 0.
a) Show that S ∗ S = id and SS ∗ is an orthogonal projection. Determine
this projection.
b) Show that 0 ∈ σr (S).
c) Determine the spectrum of S.
32. Show that the resolvent operators rλ (A) = (A − λ id)−1 , λ ∈ ̺(A) of any
self-adjoint operator A are normal.
33. Let A : D(A) → X, D(A) ⊂ X be a densely defined operator on a Hilbert
space X. Show that:
a) Every normal operator is closed.
b) If A is closed, then the following statements are equivalent.
(i) A is normal,
(ii) A∗ is normal,
(iii) A∗ A = AA∗ .
This exercise sheet is also available online. For more information see
http://www-m7.ma.tum.de/bin/view/Analysis/OperatorTh16
Comments and Solutions
29 Recall that ran(A − z id)
⊥
∗
= ker(A∗ ) = ker(A ) = ran(A − z id)
⊥
holds.
Thus we have ran(A − z id) = ran(A − z id). Morevoer, A is essentially selfadjoint if and only if A is self-adjoint. Hence the equivalences follow almost immediately from the analogous theorem concerning self-adjointness which was proved
in the lectures.
1
30 a) Note that the inverse of λ id − Mf (if existent) is multiplication by λ−f
1
which is a bounded linear operator on L2 (µ) if and only if λ−f
∈ L∞ (µ), as we
have already seen at the very beginning of chapter II in the lectures.
1
∈
/ L∞ (µ) is equivalent to {x ∈ M | |f (x) − λ| < ε} having
Furthermore, λ−f
positive measure for all ε > 0.
b) We have λ ∈ σp (Mf ) of and only if there exists ϕ ∈ L2 (µ) \ {0} such that
f (x)ϕ(x) = λϕ(x) holds for all x up to a µ-nullset or, in other words,
µ{x ∈ M : ϕ(x) 6= 0 and f (x) 6= λ} = 0 .
c) One example can be constructed as follows. Let ν be Lebesgue measure on a
closed disk K ⊂ C, say, and δ point measure at some point w ∈ C. Consider
µ := ν + δ and f : K ∪ {w} → C, f (x) := x. Then
•
1
f −λ
∈ L∞ (µ) ⇐⇒ λ ∈
/ K ∪ {w},
• f (w) is the only eigenvalue of Mf with the multiples of the characteristic
function χ{w} forming the corresponding eigenspace.
31 a) For a, b ∈ ℓ2 (N) we have hSa, bi =
∞
P
an bn+1 .
n=1
It is not difficult to see that S ∗ (b1 , b2 , b3 , . . .) = (b2 , b3 , b4 , . . .), S ∗ S = id and
SS ∗ (b1 , b2 , b3 , . . .) = (0, b2 , b3 , . . .) whence SS ∗ is the orthogonal projection onto
{x ∈ ℓ2 (N) : x1 = 0}.
b) Clearly, ker(0 − S) = ker(S) = {0}.
As ran(0 − S) = {x ∈ ℓ2 (N) : x1 = 0} is not dense in ℓ2 (N), we have 0 ∈ σr (S).
c) Note that S is an isometry for every n whence kSk = 1 and σ(S) ⊂ {z ∈ C :
|z| ≤ 1}.
Assume there exists an eigenvalue λ of S and let x be a corresponding eigenvector.
Since S is an isometry, we get |λ| = 1. Moreover, Sx = λx implies xn = λxn+1 for
all n ∈ N whence |xn+1 | = |xn | holds for all n ∈ N contradicting x ∈ ℓ2 \ {0}.
Pick λ ∈ C such that 0 < |λ| < 1. Then we have y := (λ, λ2 , . . .) ∈ ℓ2 . As we have
just seen, ker(λid − S) = {0}. Moreover, ran(λid − S) is not dense in ℓ2 (N),
as
∞
∞
P
P
(λid − S)x, y = hλx, yi − hSx, yi =
λxn λn −
xn λn+1 = 0
n=1
n=1
⊥
holds for all x ∈ ℓ2 (N) whence y ∈ ran(λid − S) . Therefore, we have shown that
{z ∈ C : 0 < |z| < 1} ⊂ σr (S).
Since σ(S) is compact, we now have σ(S) = {z ∈ C : 0 < |z| ≤ 1}.
Let now |λ| = 1 and assume that λ ∈ σr (S) holds. Then ran(λ id − S) is not
dense in X or, equivalently, ker(λ id − S ∗ ) 6= {0}. For a corresponding eigenvector
x = (xn )n ∈ ℓ2 (N) \ {0} we then have S ∗ x = λx which implies |xi | = |xi+1 | for all
i ∈ N which is impossible.
Thus σc (S) = {z ∈ C : |z| = 1}.
32 Since A is self-adjoint, we have (A − λ id)∗ = A∗ − λ id = A − λ id. For
λ ∈ ̺(A) ∩ R this yields (A − λ id)−1 being the inverse of a self-adjoint operator,
which ist self-adjoint, too.
Note that ran(A − λ id) is dense whenever λ ∈ C \ R ⊂ ̺(A) holds. Using the result
from exercise 25a, we get
∗
−1
(A − λ id)−1 = (A − λ id)∗
= (A − λ id)−1 .
Applying the first resolvent formula, we obtain
(A − λ id)−1 (A − λ id)−1 = (A − λ id)−1 (A − λ id)−1 .
Therefore, (A − λ id)−1 is normal.
33 a) As A∗ is closed, its graph D(A∗ ), k·kA∗
is complete. Since A is normal,
we have kAxk = kA∗ xk for all x ∈ D(A) = D(A∗ ) or, in other words,
the graph
∗
∗
norms of A and A coincide. Thus D(A), k·kA = D(A ), k·kA∗ is complete
whence A is closed.
b) The equivalence (i) ⇐⇒ (ii) follows from A∗∗ = A.
(i)⇒(iii): For arbitrary x, y ∈ D := D(A) = D(A∗ ), using the polarization identity,
we obtain
2 2
2
2
1 hAx, Ayi = 4 A(x + y) − A(x − y) + iA(x − iy) − iA(x + iy)
=
1
4
∗
A (x + y)2 − A∗ (x − y)2 + iA∗ (x − iy)2 − iA∗ (x + iy)2
= hA∗ x, A∗ yi .
Let now x ∈ D(AA∗ ). Then
hAx, Ayi = hA∗ x, A∗ yi = hAA∗ x, yi
for all y ∈ D ,
whence Ax ∈ D(A∗ ) and A∗ Ax = AA∗ x holds, showing that AA∗ ⊆ A∗ A.
Analogously, with A∗ instead of A and, having in mind that A∗∗ = A, we get
A∗ A ⊆ AA∗ .
(iii)⇒(i): First note that for x ∈ D(A∗ A) = D(AA∗ ) we have
kAxk2 = hAx, Axi = hA∗ Ax, xi = hAA∗ x, xi = hA∗ x, A∗ xi = kA∗ xk2 .
Next we show that Graph A D(A∗ A) is dense in Graph(A).
Assume there exists z ∈ D(A) such that (z, Az)⊥(y, Ay) holds for all y ∈ D(A∗ A).
Then
hz, (id + A∗ A)yi = hz, yi + hAz, Ayi = (z, Az), (y, Ay) = 0
yields z = 0 beacuse from exercise 28 we know that ran(id + A∗ A) = X.
Let nor x ∈ D(A). Then there exists a sequence (xn )n in D(A∗ A) such that xn → x
and Axn → Ax. Moreover, from kA(xn − xm )k = kA∗ (xn − xm )k for all n and all
m, we can conclude that (A∗ xn )n is convergent. Since A∗ is closed this means that
x ∈ D(A∗ ) and we have kA∗ xk = kAxk.
Finally we can swap the roles of A and A∗ and eventually get kAxk = kA∗ xk for
all x ∈ D(A) = D(A∗ ), i.e. A is normal.