Spectrum of a Multiplication Operator via Fourier Transform

Question

Does the Fourier Transform method to prove the spectrum of the Laplace operator generalize to other multiplication operators?: How to prove the spectrum of the Laplace operator?

As a concrete example: Let $T_a:f(x)\to f(x-a),f\in L^2(\mathbb{R})$. The translation operator, $T_a$ can be expresses as a Fourier multiplier by $e^{-ia\xi}$, $T_a[f]=FT^{-1}[e^{-ia\xi}FT[f]]$.

To get the spectrum of $T_a$ we calculate the resolvent set of $T_a$ (via the resolvent equation) and note that the spectrum is equal to the complement of the resolvent set: $$T_af-\lambda f=g $$

Taking the Fourier Transform:

$$(e^{-ia\xi}-\lambda)\hat{f}=\hat{g} $$ Solving for $\hat{f}$: $$\hat{f}=\frac{\hat{g}}{(e^{-ia\xi}-\lambda)} $$

i believe the last equation determines a solution to the resolvent equation as long as $f\in L^2\implies \hat{f}\in L^2$

Here is where i am stuck- The proof for the spectrum of the laplace operator, linked above, looks at complex $\lambda$ and constructs a $\hat{g}\in L^2$ where the following equation blows up if $\lambda\in (-\infty,0]$, ($|\xi|^2$ as the multiplier for $\Delta$): $$-\frac{1+\lvert\xi\rvert^2}{\lambda+ \lvert\xi\rvert^2}\hat{g}$$

i'd like to know if there is an analogous method to prove the spectrum for the translation operator $T_a$ (and how to properly determine the relevant function $\hat{g}$).

Answer 1

Maybe I'm not giving you a useful answer, but what you are doing is too complicated. The Fourier transform is a unitary $V$ at the level of the Hilbert space, which implies that $T\longmapsto V^{-1}TV$ is a $*$-isomorphism at the level of operators. Which means that the spectrum of $T_a$ is the same as that of the spectrum of the multiplication operator by the function $g(\xi)=e^{-ia\xi}$. It is an easy exercise that the spectrum of a multiplication operator on $L^2$ is the essential range (just the range, in this case, because $g$ is continuous). Thus $\sigma(T_a)=\mathbb T$, the unit circle.

In the particular case of your $T_a$, none of this machinery is needed, though. It is trivial to check that $T_a$ is a surjective isometry, i.e. a unitary. This immediately tells you that $\sigma(T_a)\subset\mathbb T$. Given $\lambda\in\mathbb T$ and $\varepsilon>0$ there exists $h\in L^2(\mathbb R)$ with $\|h\|_2=1$ and $\|Th-\lambda h\|_2<\varepsilon$; this shows that $\lambda\in\sigma(T_a)$.

One easy way to construct $h$ as above is as follows. Fix $n\in\mathbb N$ such that $2n+1>(2/\varepsilon^2|a|)$ and put $$ h=\frac1{\sqrt{(2n+1)|a|}}\sum_{k=-n}^n \lambda^n\,1_{[na,(n+1)a)}. $$ Then $\|h\|_2=1$ and $$\|T_ah-\lambda h\|_2=\sqrt{\frac2{(2n+1)|a|}}<\varepsilon.$$