Application of the spectral mapping theorem

Question

Let $T:L^2((0,2)\rightarrow L^2((0,2))$, $(Tx)(t):=\begin{cases} x(t+1), & 0<t<1\\ 0,& \text{elsewhere} \end{cases} $

Show that $T$ is well defined and $\sigma(T)=\sigma_p(T)=\{0\}$

Its easy to see that $(Tx)(t)\in L^2((0,2))$ if $x(t)\in L^2((0,2))$.

And now I have to show the other statement by using the spectral mapping theorem, which says that $\sigma(f(T))=f(\sigma(T))$ with $f\in C(\sigma(T))$.

Can someone give me a tip?

Answer 1

Hint:

Consider the polynomial $f(X)=X^2.$