For fixed $r>0$, consider $\mathcal{L}: L^\infty(0,1) \rightarrow L^\infty(0,1)$, with $(\mathcal{L}f)(x) = \|f\|_{L^\infty(B_r(x))}$. If $f$ is Lipschitz continuous, is $\mathcal{L}f$ Lipschitz continuous as well?
Update (answer to comment): I tried to prove it via inverse triangle inequality, but only get, for $x,y \in [0,1]$, that $|f(x)-f(y)| \leq L|x-y| + 2r$, where $L$ is the Lipschitz constant of $f$. Intuitively, it seems at least continuous. I thought one could maybe achieve something in the fashion of $|f(x)-f(y)| \lesssim \mathrm{dist}(B_r(x), B_r(y)) \lesssim |x-y|$. But I am not sure how, do you have any hints?