Eigenvalues for a bounded operator

[Wuberdall]

Homework Statement

Let C be the composition operator on the Hilbert space L_{2}(\mathbb{R}) with the usual inner product. Let f\in L_{2}(\mathbb{R}), then C is defined by

(Cf)(x) = f(2x-1), \hspace{9pt}x\in\mathbb{R}

give a demonstration, which shows that C does not have any eigenvalues.

Homework Equations

C is a unitary operator.

Let \mathcal{F} denote the Fourier Transformation on L_{2}(\mathbb{R}), then
(\mathcal{F}C\mathcal{F}^{\ast}f)(p) = <br /> \frac{\exp\big(-i\tfrac{1}{2}p\big)}{\sqrt{2}}\hspace{1pt}f\big(\tfrac{1}{2}p\big)<br />

The Attempt at a Solution

Direct application of the eigenvalue equation of course yields Schröders equation, that is
f(2x-1) = \lambda f(x)

I don't have a slightest idea on how to proceed from here.Any good suggestions are more than welcome!

 

[RUber]

If C were to have any eigenvalues, they would apply to any choice of f in your space.
Consdider a function that is has the property f(0)=0 and f(-1) $\neq$ 0.
Such a function clearly exists in your space.

 

[Wuberdall]

If C were to have any eigenvalues, they would apply to any choice of f in your space.
Consdider a function that is has the property f(0)=0 and f(-1) $\neq$ 0.
Such a function clearly exists in your space.

Thanks for your reply.
I still don't see how this implies that C doesn't have any eigenvalues ?

I mean, does i not only show that there doesn't exist an eigenfunction having these properties?

 

[Wuberdall]

Is this a valid argument ?
Assume that f:\mathbb{R}\rightarrow\mathbb{R} is non vanishing almost everywhere and \lambda is an eigenvalue of C, then
f(2x-1) = \lambda f(x)
2f&#039;(2x-1) = \lambda f&#039;(x)

For x=1 the first of the equations yields : \hspace{3pt} f(1) = \lambda f(1) \hspace{12pt}\Rightarrow\hspace{12pt} \lambda=1.
While the second of the equations yields : \hspace{3pt} 2f&#039;(1) = \lambda f&#039;(1) \hspace{12pt}\Rightarrow\hspace{12pt} \lambda=2.
This is obvious a contradiction, therefore \lambda can't be a eigenvalue.

Of course f must behave properly around x=1, but this is guaranteed by f \in L_{2}(\mathbb{R}).

 

[Ray Vickson]

If C were to have any eigenvalues, they would apply to any choice of f in your space.
Consdider a function that is has the property f(0)=0 and f(-1) $\neq$ 0.
Such a function clearly exists in your space.

No, the eigenvalue equation $f(2x-1) = \lambda f(x) \; \forall x$ need apply only to an eigenfunction $f = f_{\lambda}$.

 

[RUber]

Thanks, Ray. I must have been operating without caffeine when I posted that.

Wuberdall, your contradiction argument looks reasonable to me.

 

[LCKurtz]

But $f\in \mathbb L^2$ doesn't give you that $f$ is differentiable.

 

[RUber]

LCKurtz makes a good point. The goal should be to contradict your assumptions.
As it stands, 1 is the only potential eigenvalue, as long as f(x) is defined at x=1.
So if f(x) is defined at $x = 1+ \delta$, $f(1+ \delta) = f(1+ 2\delta)=f(1+2^k \delta)$ for any delta and integer k .
You may be able to show that this sort of function is either the zero function (trivial) or not in $L^2$.

 

[Ray Vickson]

Thanks, Ray. I must have been operating without caffeine when I posted that.

Wuberdall, your contradiction argument looks reasonable to me.

More generally: if $f \in C^{\infty}$ near $x = 1$ then we have $2^n f^{(n)}(2x-1) = \lambda f^{(n)}(x), n = 0,1,2,\ldots$, where $f^{(n)}$ denotes the $n$th derivative. Thus $2^n f^{(n)}(1) = \lambda f^{(n)}(1), n = 0,1,2, \ldots$.
(1) If $\lambda = 0$ that would imply that $f^{(n) = 0, n = 0,1,2, \ldots$, meaning that $f$ would be non-analytic (but still $C^{\infty}$. That means that $f$ would not equal its own Taylor series. There are functions like that, and our $f$ would have to be one of them.
(2) If $\lambda \neq 0$ then either (i) $f(0) = 0$; or (ii) $\lambda = 1$. In case (i) we have $2 f'(1) = \lambda f'(1), 4 f''(1) = \lambda f''(1), \ldots$, and this allows $\lambda = 2, f^{(n)}(1) = 0$ for $n \geq 2$ or $\lambda = 4, f^{(n)}(1)=0$ for $n = 1,3,4,\ldots#, etc. Again,$f## would not be analytic

LCKurtz makes a good point. The goal should be to contradict your assumptions.
As it stands, 1 is the only potential eigenvalue, as long as f(x) is defined at x=1.
So if f(x) is defined at $x = 1+ \delta$, $f(1+ \delta) = f(1+ 2\delta)=f(1+2^k \delta)$ for any delta and integer k .
You may be able to show that this sort of function is either the zero function (trivial) or not in $L^2$.

The equation $f(1) = \lambda f(1)$ implies $\lambda = 1$ only in the case $f(1) \neq 0$; otherwise, if $f(1) = 0$ there is no restriction on $\lambda$.