Suppose we want to find the eigenvalues of the operator $T_y$ defined in $B(L^p(\mathbb{R}))$ which is defined as
$$ (T_y f)(x) = f(x - y) $$
The eigenvalues equation is $$ T_y f - \lambda f = 0 \Leftrightarrow (T_y f)(x) - \lambda f(x) = 0 \;\; \forall x \in \mathbb{R}\Leftrightarrow f(x - y) - \lambda f(x) = 0 \;\; \forall x \in \mathbb{R} $$ I've seen arguments to take the norm of $f(x- y)$ and $\lambda f(x)$ to show that the only eigenvalue is $\lambda = 0$.
I was wondering if somewhat the density of $C_c(\mathbb{R})$ in $L^p(\mathbb{R})$ can be used to prove the same result.
Indeed if $f \in C_c(\mathbb{R})$ we have
$$ f(x - y) - \lambda f(x) = 0 \;\; \forall x \in \mathbb{R} $$
Assuming without loss of generality $y > 0$ let $x = \min \overline{\left\{x : f(x - y) = 0\right\}}$ and let $\epsilon > 0$ be chosen such that $f(x + \epsilon - y) = 0$ but $f(x + \epsilon) \neq = 0$ from such choice we end up having
$$ \lambda f(x + \epsilon) = 0 \iff \lambda = 0 $$
Somewhat, but not sure how, I think I can extend this to all the $L^p(\mathbb{R})$.
A similar argument still works. Assume that $T_yf=\mu f$ for some $f\neq 0$. Fix some interval $[b,b+y]$ such that $\delta:=\int_b^{b+y} |f(x)|^pdx>0$. As $T_yf=\mu f$, we see that, for any $n\in \mathbb{Z}$, we have that $\int_{b+ny}^{b+(n+1)y}|f(x)|^pdx=|\mu|^{np}\int_{b}^{b+y}|f(x)|^pdx=|\mu|^{np}\delta$ by induction.
Thus we have that $\|f\|_p^p=\sum_{n=-\infty}^{\infty}|\mu|^{np}\delta.$ This is clearly infinite unless $\mu=0$.