Let $\tau_h$ denote the the translation of $f$ by $h$, i.e $\tau_h f(x) = f(x+h)$ . Let $f \in L^\infty$, then $f$ is not continuous under $h$, i.e $\tau_h f$ does not converge to $f$ in the $\infty$ norm as $h$ tends to $0$.
My proof: Consider the function given by $2x\sin \left(\frac{1}{x}\right) - \cos (1/x)$ at $x \neq 0$, and $0$ at $x = 0$, restricted to the interval $[-1,1]$. Then, we see that this function is bounded everywhere, but the translation operator is not continuous. Indeed, no matter how small we take $h$, we have at $x = 0$ that $\|f_h - f\|_\infty = \|f_h\|_\infty$ oscillates rapidly between $1$, $-1$, and thus we cannot have convergence under the supremum norm, as it will always bre greater than or equal to $2 -\epsilon$ for some point.