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.
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.