Let $H$ be a Hilbert space and $U$ be a unitary operator.
Suppose $T$ is self-adjoint.
Show that $T' = U^*TU$ is self-adjoint, and if $f: \sigma(T) \rightarrow \mathbb{C}$ is a continuous function, then $f(T') = U^*f(T)U$.
My attempt:
Since $T$ is self-adjoint, $T= T^*$. Also, since $U$ is unitary $U^*U=I = UU^*$.
So $\langle x ,U^*TU y\rangle = \langle Ux , TU y\rangle = \langle Ux , TU y\rangle = \langle TUx , U y\rangle =\langle U^*TUx , y\rangle$.
Hence $T'$ is self-adjoint.
Now suppose $f$ is continuous. Then we have $f(T') = f (U^*TU)$. From here, I'm not sure what properties of the continuity of $f$ I can use.
Thank you in advance.
Best Answer
$$(T')^*=(U^*TU)^*=U^*T^*U^{**}=U^*TU=T'$$
For any holomorphic function $f$ on $\sigma(T')=\sigma(T)$, \begin{align}f(T')&=\frac{1}{2\pi i}\oint f(z)(z-T')^{-1}\,dz\\ &=\frac{1}{2\pi i}\oint f(z)(z-U^*TU)^{-1}\,dz\\ &=\frac{1}{2\pi i}\oint f(z)U^*(z-T)^{-1}U \,dz\\ &=U^* f(T)U\end{align} since $(z-U^*TU)^{-1}=(U^*(z-T)U)^{-1}=U^*(z-T)^{-1}U$ and the fixed $U$ operators can be taken outside the integral.
Hence, since $C^\omega(\Omega)$ is dense in $C(\Omega)$, it follows that $f(T')=U^*f(T)U$ for continuous functions $f$.