Does $f(u)\in H^1$ imply that also the "inner average" $f(u_h)\in H^1$?

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To add more details to the question: Let $u\in L^\infty(\Omega,[0,1])$ (bounded domain, space dimension $n$ larger than one), $f\in C^1([0,1])$ increasing but $f'(0)=f'(1)=0$ and such that $f(u)\in H^1(\Omega)$. Denote an average of $u$ along any direction by $u_h$. Can we deduce that $f(u_h)\in H^1(\Omega)$?

We could for example define $$u_h(x)=1/h \int_{x_j}^{x_j+h} u(x_1,...x_{j-1},\tau,x_{j+1}...x_n) d\tau$$ for any $j\in\{1,...n\}$.

It is quiet immediate to show that $f(u_h)\in L^2$. But for the existence of the weak derivative I see no way to involve the existence of $\partial_j f(u)$. In particular I'm not allowed to interchange the averaging process and $f$ since $f$ is not convex.

Alternatively we can choose an average about an $n$ dimensional set of measure $\approx h$