Linear algebra inner products, self adjoint operator,unitary operation

[Karl Karlsson]

Homework Statement

Let $V = \mathbb{C}([0, 1], \mathbb{C})$ be the vector space of continuous functions $f: [0,1] \rightarrow \mathbb{C}$ with inner product:
$$\langle f|g \rangle = \int_0^1 \overline {f(t)}g(t)\;\mathrm{d}t$$

For $h \in V$ , let $L_h: V \rightarrow V$ be the operator defined by $L_h(f)(t)=h(t)f(t)$.
b) Determine the adjoint operator to $L_h$
c) For which functions is the operator $L_h$ self adjoint?
d) For which functions is the operator $L_h$ unitary?
e) Give example of one function h(t) such that the operator $L_h$ is self adjoint and has at least two different eigenvalues

Relevant Equations

$$\langle f|g \rangle = \int_0^1 \overline {f(t)}g(t)\;\mathrm{d}t$$
$L_h(f)(t)=h(t)f(t)$

b)

c and d):
In c) I say that $L_h$ is only self adjoint if the imaginary part of h is 0, is this correct?

e) Here I could only come up with eigenvalues when h is some constant say C, then C is an eigenvalue. But I' can't find twtherwise does b-d above look correct?

Thanks in advance!

 

[PeroK]

It's looks all right. For d) perhaps think about writing $h(t)$ in polar form - that might be neater.

For e), remember these functions are only continuous. You have lots of options for where they are zero and non-zero.

 

[Karl Karlsson]

It's looks all right. For d) perhaps think about writing $h(t)$ in polar form - that might be neater.

Do you mean having $a(t)=r\cdot cos(t)$ and $b(t)=r\cdot sin(t)$ and r=1 ?

For e), remember these functions are only continuous. You have lots of options for where they are zero and non-zero.

Like what? I can't see how it would work for cos(t) $e^t$ t or any other normal continuous function

 

[PeroK]

Do you mean having $a(t)=r\cdot cos(t)$ and $b(t)=r\cdot sin(t)$ and r=1 ?

Like what? I can't see how it would work for cos(t) $e^t$ t or any other normal continuous function

I was thinking $e^{ia(t)}$ for some real function $a(t)$. But, what you have is fine.

Try thinking about continuous functions more generally. Something like $\cos t$ is no good. That would never work. What's "normal"?

 

[Karl Karlsson]

Try thinking about continuous functions more generally. Something like $\cos t$ is no good. That would never work. What's "normal"?

I was just mentioning som elementary continuous functions that came to my mind but none of those seem to work and I can't find one that does.

 

[PeroK]

I was just mentioning som elementary continuous functions that came to my mind but none of those seem to work and I can't find one that does.

What about a continuous step-function?

 

[Karl Karlsson]

What about a continuous step-function?

Ooh, right! You mean if I for example said h(t)=1 in the intervall 0<t<0.5 and h(t)=2 in the intervall 0.5<t<1. One eigenvector for the eigenvalue 1 could be the function f(t) which is $t-0.5$ on the intervall 0<t<0.5 and f(t)=0 on 0.5<t<1 ? And for for the eigenvalue 2 an eigenvector could be g(t) = 0 on 0<t<0.5 and $g(t) =t-0.5$ on 0.5<t<1?

 

[PeroK]

Ooh, right! You mean if I for example said h(t)=1 in the intervall 0<t<0.5 and h(t)=2 in the intervall 0.5<t<1. One eigenvector for the eigenvalue 1 could be the function f(t) which is $t-0.5$ on the intervall 0<t<0.5 and f(t)=0 on 0.5<t<1 ? And for for the eigenvalue 2 an eigenvector could be g(t) = 0 on 0<t<0.5 and $g(t) =t-0.5$ on 0.5<t<1?

That's not a continuous step function. You need to join the steps. The eigenfunctions would be any function that is zero outside one of the steps.

 

[Karl Karlsson]

That's not a continuous step function. You need to join the steps. The eigenfunctions would be any function that is zero outside one of the steps.

Yeah, I realized that afterwards. So i can choose h = 1 on 0 <= t <= 1/3 and h=3t on 1/3 < t < 2/3 and h=2 on 2/3 <=t <= 1. An eigenvector to eigenvalue 1 could be a function that is t-1/3 on the intervall 0<=t<=1/3 and 0 on 1/3<t<=1. An eigenvector to eigenvalue 2 could be a function that is 0 on the interval 0<=t<2/3 and t-2/3 on 2/3<=t<=1. Is this correct?

 

[PeroK]

Yeah, I realized that afterwards. So i can choose h = 1 on 0 <= t <= 1/3 and h=3t on 1/3 < t < 2/3 and h=2 on 2/3 <=t <= 1. An eigenvector to eigenvalue 1 could be a function that is t-1/3 on the intervall 0<=t<=1/3 and 0 on 1/3<t<=1. An eigenvector to eigenvalue 2 could be a function that is 0 on the interval 0<=t<2/3 and t-2/3 on 2/3<=t<=1. Is this correct?

Yes. Note that you have a whole eigenspace of functions that are zero on one of these regions. The set of functions that are zero on $[0, 2/3]$ is a vector subspace - and is the eigenspace of your operator corresponding to eigenvalue $2$.