Composition of differentiable with weakly differentiable function

Question

Let $\Omega\subset \mathbb{R}$ be open. If I define a weakly differentiable function $u$ as $L^1_{loc}(\Omega)$ function for which the typical partial integration formula holds for all test functions, I know that the following chain rule holds:

Let $f\in C^1$ with $f'$ bounded, then $f\circ u$ is again weakly differentiable with the usual derivative $f'(u)u'(x)$.

I can prove this via a density argument. However, I see no way to prove this without the boundedness of $f'$. Does this result hold without the condition or if not, can anybody state an easy "counter example" for where the weak differentiability of the composition with an arbitray $C^1$-map can go wrong?

Answer 1

Weak differentiability is true without assumptions on $f'$. Take $\Omega\subset \mathbb R$, $u\in L^1_{loc}(\Omega)$ with weak derivative $u'\in L^1_{loc}(\Omega)$. Let $f:\mathbb R\to \mathbb R$ be continuously differentiable.

Take $\phi \in C^\infty_c(\Omega)$. Then $supp(\phi) \subset [a,b] \subset \Omega$ is a compact set, the restriction of $u$ to $[a,b]$ is in $W^{1,1}(a,b)$, hence $u$ is continuous on $[a,b]$. Hence, $|u|\le M$ on $[a,b]$ for some $M>0$. And in the integral $\int_\Omega f(u(x)) \phi(x)dx$ only the values of $f$ on $[-M,M]$ matter, $f'$ is bounded there, so $f(u)$ is weakly differentiable.

To see this, define $f_M(s):=f(s)$ for all $|s|\le M$, $f_M(s):= f(M) + f'(M)(s-M)$ for $s>M$, and similarly for $s<-M$. Then $f_M$ is continuously differentiable, with bounded derivative, and $f_M(u(x)) = f(u(x))$ for all $x\in [a,b]$. Then we can perform the integration by parts.

And $f(u)$ is weakly differentiable with $f'(u)u'\in L^1_{loc}(\Omega)$.

Answer 2

As the other answer pointed out, the result is true for dimension 1. For an easy counterexample in $d=2$, let $\Omega=B_1(0)$ in dimension $2$, and consider the function $$ u(x)=\begin{cases} |x|^{-\alpha} & x\neq 0\\ 0 & x=0 \end{cases} $$ Convince yourself that this is weakly differentiable iff $\alpha <1$. Then simply take $f(x)=x^2$. Certainly this is $C^1$, but $$ f(u)=\begin{cases} |x|^{-2\alpha} & x\neq 0\\ 0 & x=0 \end{cases} $$ which is weakly differentiable iff $2\alpha <1$. So just take any $\alpha \in [\frac{1}{2},1)$ to get a counterexample.