Continuous function not sobolev

Question

Let $I=(a,b)$ an open bounded interval. It is well known that $W^{1,p}(I)\subset C(I)$. It easy to see that there are $f\in C(I)$ such that $f\notin W^{1,p}(I)$ It is enough to take $I=(0,1)$ and $f(x)=x^{-\alpha}$ for certain $\alpha>0$ depending on $p$.

We are trying to find $f\in C(I)$ bounded such that $f\notin W^{1,p}(I)$. If $f$ has a countable many (even as many as a null set (?)) peaks, surely $f\in W^{1,p}(I)$.

Our first guess is that the Weierstrass function (continuous but non-differentiable everywhere) could work, but we want to know if there is some standard (easier) (counter)-example.

Answer 1

Since any function in $W^{1,p}(I)$ equals the integral of its weak derivative, which is in $L^p(I)$ and hence $L^1(I)$, it is absolutely continuous. Hence any function which is bounded and continuous but not absolutely continuous will be a counterexample. For instance, the Cantor staircase function.

Answer 2

Another example: continuous functions with unbounded derivative at either $0$ or at $1$ such that $f'\notin L^p(I)$. The function $f(x)=x^\alpha\in L^p\quad\forall \alpha>0\quad$, because it is bounded.

$f'=\alpha \frac{1}{x^{1-\alpha}} \Rightarrow \quad$ if $\quad p(1-\alpha)\ge 1\quad\Leftrightarrow \alpha\leq\frac{p-1}{p}$ then $\quad (f')^p\notin L^1(I)\quad$