3

We are working on the Hilbert space $H = L^2(\mathbb{R})$ and consider the bounded linear operator $T : H \to H$ defined by $(Tf)(x) = f(x+1) + f(x-1)$. What is the spectrum of $T$?

What I've tried:

  • It's not so hard to show that $T$ is indeed a linear operator and bounded, with $\|Tf\|^2 \leq 4\|f\|^2$ wrt the $L^2$ norm on $\mathbb{R}$. Using for instance the functions ${\bf1}_{[-n,n]}$, I could even show that $\|T\| = 2$.
  • Also not so hard to show that $T$ is self-adjoint, being the sum of obvious self-adjoint $T_1f(x) = f(x+1)$ and $T_2f(x) = f(x-1)$, but this can also be proved by straightforward calculation.

The spectrum is therefore a subset of $[-\|T\|,\|T\|]$. I do not know how to proceed.

Mittens
  • 46,352
Neckverse Herdman
  • 1,746
  • @AnneBauval that is the question I am trying to solve, I thought norm would be easier and could help ... The chief problem is the spectrum for me. As to the second question you are right that is not necessarily the case! – Neckverse Herdman Jan 29 '24 at 15:35
  • The norm seems useful to know so that we only need to focus on the set $[-||T||, ||T||]$ for the spectrum, by self-adjointness – Neckverse Herdman Jan 29 '24 at 15:37
  • @AnneBauval Okay, thank you. That way we can construct a sequence of unit vectors $f_n/(\sqrt{2n})$ and their images converge to 2, confirming that $||T|| = 2$. Thanks. So the spectrum is included in $[-2,2]$. But I still do not really see a way to proceed for the spectrum – Neckverse Herdman Jan 29 '24 at 16:11
  • 2
    $T_1,T_2$ are not self-adjoint, rather $T_1^*=T_2$. – daw Jan 29 '24 at 20:45

1 Answers1

5

Application of the Fourier transform gives

$$(\mathcal{F}Tf)(t)=\big(e^{-2\pi it}+e^{2\pi it}\big)\mathcal{F}f(t)=2\cos(2\pi t)\mathcal{Ff}(t)$$ Consider the multiplication operator $M_\phi:L_2\rightarrow L_2$ given by $f\mapsto \phi f$ where $\phi(t)=2\cos(2\pi t)$. Then $$\mathcal{F}T=M_\phi \mathcal{F}$$

For any $\lambda\in\mathbb{C}$, then $$\mathcal{F}(\lambda -T)=(\lambda\mathcal{F}-\mathcal{F}T)=(\lambda-M_\phi)\mathcal{F}$$ and so $$\lambda-T=\mathcal{F}^{-1}(\lambda - M_\phi)\mathcal{F}$$

Then $\lambda\in\sigma(M_\phi)$ iff $\lambda \in\sigma(T)$. The spectrum of multiplication operators is discussed here.

Mittens
  • 46,352