I am trying to understand some of the tools of functional analysis and I came across this exercise:
Given $L^2(\mathbb{R})$ and the family of operators $\{T_a\}_{a \in \mathbb{R}}$ such that:
$(T_af)(x)= f(x)$ if $a < x$
$(T_af)(x)= -f(x)$ if $x < a$
show that $\{T_a\}$ are unitary and self-adjoint, then find eigenvalues and eigenvectors.
As for the first request I noticed that
$(g,T_af) = \int_{\mathbb{R}} g^*T_af dx = \int_{-\infty}^{a} g^* (-1)f dx + \int_{a}^{\infty} g^*f dx$
while
$(T_a g,f) = \int_{\mathbb{R}} T_ag^*f dx = \int_{-\infty}^{a} (-1)g^* f dx + \int_{a}^{\infty} g^*f dx$
which proves self-adjointness. As for unitarity
$T_{a}^{*}T_a f = T_a T_a f = f$
both when $x>a, \; x<a$.
Now the part concerning the eigenvalues and eigenvectors leaves me with some questions. The eigenvalues problem should be the following:
$T_a f(x) = \lambda f(x)$ which admits solutions $\pm 1$ if I am not mistaken. I don't know how to find eigenvectors. They are by definition, functions $\phi$ such that $T_a\phi = \lambda \phi$. Still I don't know how to tackle this problem.
thanks to everyone who is going to participate
You may write $T_a$ as a multiplication operator: $$ T_a f = (-\chi_{(-\infty,a)}+\chi_{(a,\infty)})f $$ It is the difference $T_a=Q_a-P_a$ of two orthogonal projections $$ P_af = \chi_{(-\infty,a)}f,\;\; Q_a f = \chi_{(a,\infty)}f. $$ This projections are mutually orthogonal because $Q_aP_a=P_aQ_a=0$. $T_a$ is self-adjoint because $P_a,Q_a$ are self-adjoint. $T_a^*=T_a$ also follows. $T_a$ is unitary because $T_a^*T_a=T_aT_a^*=I$. $T_a$ has eigenvalues $\pm 1$, and the ranges of $P_a,Q_a$ are the eigenspaces because $$ T_a P_a = -P_a,\;\;\; T_a Q_a = Q_a. $$ These ranges are mutually orthogonal and their direct sum spans the entire space; there are no other eigenvalues.