$u\in W^{1,1}((0,1))$, $f\in C^1(\mathbb R)$, but: $f\circ u\notin W^{1,1}((0,1))$

Question

I'm searching for an example for $u\in W^{1,1}((0,1))$ and $f\in C^1(\mathbb R)$ such that the composition $f\circ u$ does not belong to $W^{1,1}((0,1))$.

The chain rule I know states that there is no such example if $f'$ is bounded. The answer to this question gives an example on a non-connected subset of $(0,1)$.

Is there an example on $(0,1)$ or holds the chain rule in this simple case also for $f$ with non-bounded derivative?

Answer 1

Such an example is not possible.

Since $(0,1)$ is one dimensional, your function $u$ is continuous, thus bounded. Hence, $-M \le u(x) \le M$ for all $x \in [0,1]$. Hence, only the values $f(t)$ for $t \in [-M,M]$ are of interest. Thus, you could change $f$ by extending it linearly outside of this interval. This will yield a function $\hat f \in C^1(\mathbb R)$ with bounded derivative and $f(u) = \hat f(u) \in W^{1,1}(0,1)$.