Let $T : L^2(\mathbb R) \to L^2(\mathbb R)$ be a linear operator defined by $$(Tf)(x)=f(x+1).$$ Show that $T$ has no eigenvalues, i.e., there exists no $f \not= 0$ in $L^2(\mathbb R)$ such that $(Tf)(x)=\lambda f(x)$ for any $\lambda \in \mathbb C$.
My work:
Okay, so the $\lambda=0$ case was straightforward for me: If $\lambda=0$, then $Tf(x)=0$, which means $f(x)=0$ because $T(0)=0$ since $T$ is linear.
But I am stuck on the case $\lambda \in \mathbb C \setminus \{0\}$. How can I work with $f(x+1)=\lambda f(x)$ and show that $f(x)=0$?
If $Tf = \lambda f$ for some unit vector $f\in L^2$, then $\|Tf\|=\|f\|$ implies $|\lambda|=1$. Therefore, $$ f(x+n)=\lambda^n f(x),\;\;\; n=1,2,3,\cdots. $$ But that's a problem because it means that $$ \int_{k}^{k+1}|f(x)|^2dx = \int_{l}^{l+1}|f(x)|^2dx,\;\;\; k,l=1,2,3,\cdots. $$ It's a problem because such $f$ could not be in $L^2$ unless $f=0$, thereby contradicting the assumption that $\|f\|=1$. So $T$ has no eigenvalues.