Spectral decomposition of normal operator

Question

Define $T$ from $L_{2}(R)$ into itself by $T(f)(t)=f(t+1)$. Show that $T$ is normal and finds its spectral decomposition. I've shown that $f$ is normal (in fact it's unitary) but how do I find its spectral decomposition? I know the decomposition is of the form $\int_{{\sigma}(T)} {\lambda}\,dE=T$ so I have presumably have to find the spectrum and the spectral measure. Thanks

Answer 1

$T$ is unitary because (a) $\|Tf\|=\|f\|$ and (b) $T$ is onto. So $T^{\star}f=f(t-1)$ is the inverse of $T$. The spectrum $\sigma(T)$ is a subset of the unit circle in the complex plane. Let $f^{\wedge}$ denote the Fourier transform of $f$. Then $$ (Tf)^{\wedge} =l.i.m.\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t+1)e^{-ist}\,dt =l.i.m. \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-is(t-1)}\,dt =e^{-is}f^{\wedge}(s). $$ The spectral resolution of the identity $P$ for $T$ is supported on the unit circle $C$ in the complex plane. If $E$ is a subset of $C$, let $p_{E}(s)=\chi_{E}(e^{is})$, where $\chi_{E}$ is the characteristic function of $E$. Then $$ P(E)f= (p_{E} f^{\wedge})^{\vee}. $$ Check that $P$ is a spectral measure. For example, $P(C)=I$ because $P(C)f=(f^{\wedge})^{\vee}=f$. Because $p_{E}$ is real, $P(E)^{\star}=P(E)$. Fourier transform properties give $P(E)P(E')=P(E\cap E')$. Notice that $P(E)T=TP(E)$. And, $$ \int_{C}\lambda dP(\lambda)f = Tf, $$ which follows from integration, and follows intuitively from $$TP(\{ e^{i\theta} : |\theta-\theta_{0}|<\delta\})f\approx e^{i\theta_{0}}P(\{ e^{i\theta} : |\theta-\theta_{0}|<\delta\})f. $$