From exam preperation. I consider the operator $T f(x) := f(x − 1), x ∈ \mathbb{R}$. First on the space of all function $f:\mathbb{R}\rightarrow \mathbb{C}$. There I found that any number other than $0$ is an eigenvalue. Now it is claimed in the exercise that $T$ has no eigenvalues when $T : L^2
(\mathbb{R}) → L^
2
(\mathbb{R})$. 

Why is that? How does it change the argument, that the function is in $L^
2(\mathbb{R})$?