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On $L^2(-\infty, \infty)$, T is a bounded linear operator and is defined by $$Tf(t)= \begin{cases} f(t), & \text{for } t \geq 0 \ \cr -f(t), & \text{for } t<0, \end{cases}$$

I'm able to prove $T$ is unitary. But then how to determine the spectrum and spectral representation of T?

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  • $\begingroup$ The spectrum of a multiplication operator $f\longmapsto fg$ on $L^2$, where $g\in L^\infty$, is the essential range of $g$. See here. You can verify this directly here. What is $g$? $\endgroup$ – Julien May 7 '13 at 18:37
  • $\begingroup$ If you are not ready to consider multiplication operators in general, simply go to step 2 directly. Let $F=\{f\in L^2\;;\; f=0 \mbox{ a.e on } (-\infty,0)\}$. What is $F^\perp$? How does $T$ act on $F$? On $F^\perp$? This will give you the spectral representation and the spectrum all at once. $\endgroup$ – Julien May 7 '13 at 18:41
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Since $T$ is unitary, its spectrum lies in the unit circle. Since $T$ is selfadjoint, the spectrum lies in the real line. So the only possible elements of the spectrum are $-1,1$ (you can also deduce this from the fact that $T^2=I$).

Both values in the spectrum are eigenvalues, as you can see with the characteristic functions of $[0,\infty)$ and $(-\infty,0)$.

The spectral decomposition of $T$ is $T=P-(I-P)$, where $P$ is the projection $$ Pf(t)=\begin{cases}f(t),&\ t\geq0\\ 0,&\ t<0\end{cases} $$

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