On nowhere differentiability of functions that just barely fail to be Lipschitz

Question

By Rademacher’s theorem, Lipschitz functions are differentiable almost everywhere. I am wondering how badly this pointwise differentiability fails for functions that “just barely” fail to be Lipschitz.

On $\mathbb R^n$, the space of bounded Lipschitz functions coincides with the Sobolev space $W^{1, \infty}$. Thus, we ask the following:

Question: Does there exist a bounded $f$ on $\mathbb R^n$ that is in $W^{1, p}$ for all $1 \leq p < \infty$ but is not differentiable at any point in $\mathbb R^n$?

Answer 1

As pointed out, this is possible only for $p\le n$. Let $u$ be an unbounded function in $W_0^{1,n}(B_{1/2})$, for instance $$ u(x)=\log |\log |x||-\log \log 2 $$ and let ${q_i}_i$ be a countable dense set. Then the function $$ v(x)=\sum_i 2^{-i} u(x-q_i) $$ is in $W^{1,n}$ and unbounded in the neighborhood of every point, and thus everywhere not differentiable.

Answer 2

No, such a function does not exist. If a bounded function $f$ on $\mathbb{R}^n$ belongs to the Sobolev space $W^{1,p}(\mathbb{R}^n)$ for all $1 \leq p < \infty$, then its weak derivative $\nabla f$ is in $L^p(\mathbb{R}^n)$ for all $p$. This implies that $\nabla f$ is essentially bounded, meaning $\nabla f \in L^\infty(\mathbb{R}^n)$. Therefore, $f$ is Lipschitz continuous, and by Rademacher's theorem, it is differentiable almost everywhere. Consequently, there cannot be a bounded function in all $W^{1,p}$ spaces that is nowhere differentiable.