It is well known that for any function $f \in L^p(\mathbb{R}^n)$, we have
$$\lim_{|h|\to 0}\, \lVert f(x-h)-f(x)\rVert_{p} = 0 $$
for $1 \leq p < \infty$. I am trying to find a counterexample to the above holding when $p = \infty$.
Let $\chi_{[0,1]}(x)$ be the indicator function the interval $[0,1]$. I'm trying to compute the value of
$$\lim_{|h|\to 0}\, \lVert\chi_{[0,1]}(x-h)-\chi_{[0,1]}(x)\rVert_{\infty} $$
Now, I'm pretty sure this is non-zero, right? My reasoning is that at $x=0$, no matter how close $h$ gets to zero, we will always be taking the limit of something non-zero (i.e., 1). Is this enough to show that $\chi_{[0,1]}$ doesn't converge
Let $A=[0,1]$. For every $h \neq 0$ with $|h| <1$ we see that $|I_A(x-h)-I_A(x)|=1$ on a non-trivial interval, hence on a set of positive measure. It follows that $\|I_A(x-h)-I_A(x)\|_{\infty} \geq 1$ for every $h \neq 0$.
[For example if $0<h<1$ we can consider the interval $(0,h)$ and if $-1<h<0$ we can consider $(1+h,1)$].