Consider the $L^p(X)$ space for $1\leq p < \infty$ over $\mathbb{K}$ (either real or complex). For $k\in\mathbb{N}$, define $S_k:L^p(X)\to L^p(X)$ by $S_kf(x)\equiv f(x + 1/k)$. I am trying to show that $\left(S_k\right)_{k\in\mathbb{N}}$ converges to $I$ strongly, i.e.
$$\forall f \in L^p(X):\lim_{k\to\infty}||S_k(f) - f||_{L^p(X)}=0$$
but not in norm, i.e.
$$\lim_{k\to\infty}\sup_{f\in L^p(X):||f||_{L^p(X)}\leq 1}\{||S_k(f) - f||_{L^p(X)}\}\neq0 $$
I have already proved similarish case for $l^p(X)$ with the mapping $P_nu(k) = \begin{cases}u(k)&:k\leq n\\ 0 &: k > n\end{cases}$.
It is not clear to me how to prove the two limits in the case of $S_k$ due to the fact that $S_k$ acts upon the argument of a function, shifting it by $1/k$, instead of cutting a sequence after a certain index. I thought about using Lebesgue dominated convergence theorem to move the limit inside the $L^p$ norm's integral, but elements of $L^p(X)$ need not be continuous and hence one cannot move the limit inside an function. Namely, computations shown below are not valied in my opinion:
$$\lim_{k\to\infty}\int_X\left|u(x + 1/k) - u(x)\right|^pdx = \int_X\lim_{k\to\infty}\left|u(x + 1/k) - u(x)\right|^pdx \Longleftrightarrow$$
$$\int_X\left|u(\lim_{k\to\infty}x + 1/k) - u(x)\right|^pdx = \int_X\left|u(x) - u(x)\right|^pdx = 0$$
What integration/limit tools should I use in this case?
