Define \(U : L^2(\mathbb{R}) \to L^2(\mathbb{R})\) by
\[Uf(x) = f(x - 1).\]
Show that if \(f \in L^2\) satisfies \(Uf = \lambda f\), for some \(\lambda \in \mathbb{R}\) (i.e., \(f\) is an eigenvector of \(U\)) then \(f\) must be the zero element, i.e., \(f = 0\) almost everywhere.
Solution
1. A nonzero solution would have \(|\lambda|=1\).
Suppose \[ Uf=\lambda f, \qquad\text{that is,}\qquad f(x-1)=\lambda f(x) \quad\text{a.e.} \] The change of variable \(y=x-1\) gives \(\|Uf\|_2^2=\int_{\mathbb R}|f(x-1)|^2\,dx=\|f\|_2^2\). Thus \(U\) is an isometry on \(L^2(\mathbb R)\), so \[ \|f\|_2=\|Uf\|_2=|\lambda|\,\|f\|_2. \] Thus either \(f=0\) in \(L^2\), in which case there is nothing to prove, or else \[ |\lambda|=1. \] Since \(\lambda\in\mathbb R\), a nonzero eigenvector could therefore occur only for \(\lambda=1\) or \(\lambda=-1\).
2. Either remaining value forces the squared modulus to have infinite integral unless \(f=0\).
In either case, \[ |f(x-1)|=|f(x)| \quad\text{a.e.}, \] so \(|f|^2\) is \(1\)-periodic almost everywhere. Let \[ a:=\int_0^1 |f(x)|^2\,dx. \] For each integer \(k\), iteration of the almost-everywhere periodicity gives \(|f(x+k)|^2=|f(x)|^2\) almost everywhere; only finitely many translates of the exceptional null set are needed for that \(k\). Translation invariance therefore gives \[ \int_k^{k+1}|f(x)|^2\,dx=a. \] Hence \[ \|f\|_2^2 =\sum_{k\in\mathbb Z}\int_k^{k+1}|f(x)|^2\,dx =\sum_{k\in\mathbb Z}a. \] Because \(f\in L^2(\mathbb R)\), this sum is finite. Therefore \(a=0\), and consequently \(f=0\) almost everywhere on every interval \([k,k+1]\). Thus \[ f=0\qquad\text{a.e. on }\mathbb R. \]