# Eigenvalues for a bounded operator

1. Jan 20, 2015

### Wuberdall

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$C$ is a unitary operator.

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

3. 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!

2. Jan 20, 2015

### 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.

3. Jan 20, 2015

### Wuberdall

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?

4. Jan 20, 2015

### 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
$f(2x-1) = \lambda f(x)$
$2f'(2x-1) = \lambda f'(x)$

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 : $\hspace{3pt} 2f'(1) = \lambda f'(1) \hspace{12pt}\Rightarrow\hspace{12pt} \lambda=2$.
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 $f \in L_{2}(\mathbb{R})$.

5. Jan 20, 2015

### Ray Vickson

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

6. Jan 21, 2015

### RUber

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

7. Jan 21, 2015

### LCKurtz

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

8. Jan 21, 2015

### 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 $L^2$.

9. Jan 21, 2015

### Ray Vickson

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$would not be analytic The equation$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##.