Eigenvalues of a linear operator from $\mathbb R\to\mathbb C$ My problem is as follows (full problem):

Let $V$ be the vector space of continuous functions $f:\mathbb R\to\mathbb C$ with period $1$, i.e.
$$f(t)=f(t+1),\ \forall t\in\mathbb R$$
Let $V$ have the inner product
$$\langle f\mid g\rangle=\int_0^1\overline{f(t)}g(t)\mathop{dt}.$$
And finally, let $L_a:V\to V$ be the operator defined as $L_a(f)(t)=f(t+a)$, $a\in\mathbb R$.
c) For which values of $a$ is $L_a$ self-adjoint?
d) Give one example of value for $a$ such that $L_a$ has exactly three eigenvalues.

I am having trouble with d) and e). Since $L$ is periodic with period $1$, I have realised that for any integer value of $a$, the only eigenvalue is $1$ for all functions $f$. I can't figure out how to even begin with finding values that give three/infinite eigenvalues.
I thought of finding a matrix representation of $L$ to find the characteristic polynomial, but since $V$ is infinite-dimensional, that is not possible right? Would appreciate any help.
 A: For any $a$, the operator $L_a$ is a unitary. This is easily seen from the fact that $L_a$ is invertible (since $L_a^{-1}=L_{-a}$) and an isometry:
$$
\|f\|^2=\int_0^1|f(t)|^2\,dt=\int_0^1|f(t+a)|^2\,dt=\|L_af\|^2. 
$$
So $L_a^*L_a=L_aL_a^*=I$. In particular $L_a^*=L_a^{-1}$. So for $L_a$ to be selfadjoint we need $L_a=L_{-a}$. This only happens when $2a\in\mathbb Z$.
The equality $L_a^{-1}=L_a$ also tells us that any element of $\lambda$ of the spectrum of $L_a$ (eigenvalue or not) satisfies $|\lambda|=1$.
When $a=m/n$, if $L_a\,h=\lambda h$, then $\lambda^n=1$. So any eigenvalue of $L_a$ has to be a root of unity. What is not immediately clear is that all roots of unity are eigenvalues.
Noticing that $|\lambda|=1$ means that $\lambda=e^{2\pi i\,b}$ for some $b$, it is easy to produce examples of eigenvalues/eigenvectors. Namely, we have
$$
e^{2\pi i (a + t)}=e^{2\pi i a}\,e^{2\pi i t},
$$
so $e^{2\pi i a}$ is always an eigenvalue for $L_a$ with eigenvector $g(t)=e^{2\pi i t}$. Using $g_n(t)=g(t)^n=e^{2\pi i nt}$, we now have
$$
L_ag_n(t)=e^{2\pi i n n(a+ t)}=e^{2\pi i n a}\,g_n(t)=(e^{2\pi i a})^n\,g_n(t). 
$$
So $e^{2\pi i n a}$ is always an eigenvalue for $L_a$. If $a$ is rational, this will produce finitely many eigenvalues; concretely, if $a=1/3$ then $L_a$ has exactly three eigenvalues. When $a$ is irrational, all powers will be distinct, and so $L_a$ has infinitely many eigenvalues.
Using that
the set $\{na\pmod 1:\ n\in\mathbb Z\}$ is dense in $[0,1]$ when $a$ is irrational, we can also fully characterize the spectrum. Assuming that rational numbers are expressed in reduced form,
$$
\sigma(L_a)=\begin{cases} e^{2\pi i k/n},\ k=1,\ldots, n,&\ \text{ if } a=m/n\\[0.3cm]
\mathbb T,&\ \text{ if $a$ is irrational}
\end{cases}
$$
