Math 6210 Homework 5: Linear Operators on Hilbert Space H, Assignments of Quantitative Techniques

Information about linear operators t and s on the hilbert space h = ℓ2(z). How to define and find the adjoints of these operators, and provides examples for bounded and unbounded operators. Students are expected to solve problems related to these concepts.

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Pre 2010

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Math 6210 - Homework 5
Due at 4 PM on 11/10/05
From Rudin: Chapter 5, # 1,2,3,4,5
All of the problems will examine linear operators on the Hilbert space H=`2(Z). Recall that
fHif fis a function
f:Z C
with X
nZ
|f(n)|2<and the inner product of f , g His
(f, g) = X
nZ
f(n)g(n).
We also recall the definition of the adjoint Tof a linear operator T. If Tis bounded then we saw
in class that for any gHthe linear map
f7→ (T f, g )
is bounded. Therefore there is a unique element g0hsuch that (T f, g) = (f , g0) for all fH.
We define Tg=g0.
If Tis unbounded it will in general only be defined on a subspace of Hwhich we denote dom T.
Then given any gHthe map
f7→ (T f, g )
is only defined for fdom T. Furthermore it will not always be bounded. The subset of Hwhere
this map is bounded is the domain of the adjoint, dom T. Then for gdom Tthere is a unique
g0such that (T f, g) = (f , g0) for all fdom T. As in the bounded case we set Tg=g0.
1. Define T:H Hby (T f )(n) = f(n+ 1). Show that Tis bounded and that kTk= 1. Find
the adjoint Tof T.
2. In this problem we will examine an unbounded operator S. Let dom Sbe the set of compactly
supported functions in H. That is fdom Sif f(n)6= 0 for only finitely many nZ. We
then set (Sf )(n) = nf (n).
(a) Show that Sis unbounded.
(b) Show that
dom S=(gHsuch that X
nZ
n2|g(n)|2<).
Here are some hints. If the sum is <apply olders inequality. If the sum is infinite
we need to show that g6∈ dom Sby showing that
f7→ (Sf , g)
1
pf2

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Math 6210 - Homework 5 Due at 4 PM on 11/10/

From Rudin: Chapter 5, # 1,2,3,4,

All of the problems will examine linear operators on the Hilbert space H = `^2 (Z). Recall that f ∈ H if f is a function f : Z −→ C

with

n∈Z

|f (n)|^2 < ∞ and the inner product of f, g ∈ H is

(f, g) =

n∈Z

f (n)g(n).

We also recall the definition of the adjoint T ∗^ of a linear operator T. If T is bounded then we saw in class that for any g ∈ H the linear map

f 7 → (T f, g)

is bounded. Therefore there is a unique element g′^ ∈ h such that (T f, g) = (f, g′) for all f ∈ H. We define T ∗g = g′.

If T is unbounded it will in general only be defined on a subspace of H which we denote dom T. Then given any g ∈ H the map f 7 → (T f, g)

is only defined for f ∈ dom T. Furthermore it will not always be bounded. The subset of H where this map is bounded is the domain of the adjoint, dom T ∗. Then for g ∈ dom T ∗^ there is a unique g′^ such that (T f, g) = (f, g′) for all f ∈ dom T. As in the bounded case we set T ∗g = g′.

  1. Define T : H −→ H by (T f )(n) = f (n + 1). Show that T is bounded and that ‖T ‖ = 1. Find the adjoint T ∗^ of T.
  2. In this problem we will examine an unbounded operator S. Let dom S be the set of compactly supported functions in H. That is f ∈ dom S if f (n) 6 = 0 for only finitely many n ∈ Z. We then set (Sf )(n) = nf (n).

(a) Show that S is unbounded. (b) Show that

dom S∗^ =

g ∈ H such that

n∈Z

n^2 |g(n)|^2 < ∞

Here are some hints. If the sum is < ∞ apply H¨olders inequality. If the sum is infinite we need to show that g 6 ∈ dom S∗^ by showing that

f 7 → (Sf, g)

is not bounded. To do this define fk ∈ dom S with k a positive integer by

fk(n) =

ng(n) if |n| ≤ k 0 if |n| > k

and observe that (Sfk, g) = ‖fk‖^22. Note that ‖fk‖ → ∞ and therefore

|(Sfk, g)| ‖fk‖ 2

(c) Show that (S∗g)(n) = ng(n) for g ∈ dom S∗. In particular S = S∗^ on dom S (which is contained in dom S∗.) However S is not self-adjoint since S and S∗^ have different domains.