Say $\Omega\subset\mathbb R^n$ is a bounded set. Let $p\in[1,\infty)$, and $f\in L^p(\Omega)$. It is well-known that $$ \sup_{|h|\leq\rho}\Vert f(\cdot+h)-f\Vert_{L^p(\Omega)}\longrightarrow0 $$ as $\rho\searrow0$. My question is: what if one brings the supremum inside the integral? More precisely, is it the case that if $f\in L^p(\Omega)$, then $$ \left(\int\limits_{\Omega}\text{ess}\sup_{|h|\leq\rho}|f(x+h)-f(x)|^p\,dx\right)^{\frac1p}\longrightarrow0\quad\text{as}\quad \rho\searrow0\quad?\tag1 $$
It is trivial to show (1) for $\phi\in C_c^\infty(\Omega)$ (or even just uniformly continuous functions on $\Omega$). Through some work (approximate through mollifiers), (1) can be shown for essentially bounded functions on $\Omega$.
I have not thought yet of a counter-example in a more general case, and I'm not sure whether it is even true. My intuition tells me that it should be false, since putting the supremum inside is effectively an $L^\infty$-constraint. If a counter-example exists, it seems like it must be a function which blows up at every scale. In any case, any help is appreciated!
You are right that the result is not true. I give an example that works for any $1 \le p < \infty$.
For $\Omega = (0,1)$ consider the function $f(x) = x^{\frac{-1}{2p}}$. Then $f \in L^{p}(\Omega)$ for $1 \le p < \infty$.
Fix $\rho > 0$. Then $\text{ess} \sup_{|h| \le \rho} |f(x+h) - f(x)| = + \infty$ for all $|x| \le \rho$.
We define $g(\rho) = ( \int_{\Omega} \text{ess} \sup_{|h| \le \rho} |f(x+h) - f(x)|^{p} dx )^{\frac{1}{p}}$. Then we just need to show that $\lim_{\rho \downarrow 0} g(\rho) \neq 0$.
In fact, which show the much stronger result: $g(\rho) \equiv + \infty$. Indeed, $$ g(\rho)^{p} \ge \int_{0}^{\rho} \text{ess} \sup_{|h| \le \rho} |f(x+h) - f(x)|^{p} dx \ge \int_{0}^{\rho} M^{p} dx = \rho M^{p} \qquad \forall M \in \mathbb{R}^{+}. $$ Which shows that $g(\rho)$ is infinity for all $\rho > 0$. In particular, $\lim_{r \downarrow 0} g(\rho) \neq 0 $.