Can you explain to me this property! True or false?

Question

If $z_n\to z$ strongly in $C_0([0,T]; W^{1,\infty}(\Omega))$ then for any continuous function $f\in C_0(\mathbb{R})$ then $f(z_n)\to f(z)$ strongly in $L^\infty(\Omega\times (0,T))$. Here $\Omega$ beinng a bounded smoth domain of $\mathbb{R}^N$.

Answer 1

We have $$\sup_{t\in [0,T]}\|z_n(t)-z(t)\|_{1,\infty}\to 0\tag{1}$$

$(1)$ is i equal to $$\sup_{t\in [0,T]}\left(\|z_n(t)-z(t)\|_\infty+\|\nabla z_n(t)-\ \nabla z(t)\|_\infty\right)\to 0\tag{2}$$

Before answering your question, I would like to point out that your comment about the convergencee of $z_n$ is (a priori) not right, because the norm is in $L^\infty$, hence the correct limit is $(2)$. If $z_n$ and $\nabla z_n$ are continuous, then $(2)$ is the same as your comment.

Now, let's get to the answer. First note that $W^{1,\infty}(\Omega)=C^{0,1}(\overline{\Omega})$ (see Evans page 294), hence, in our case we have convergence in the sup norm: $$\tag{3}\sup_{t\in [0,T]}\sup_{x\in \Omega}|z_n(t)(x)-z(t)(x)|\to 0$$

The last convergence is true because of $(2)$. From $(3)$ and from the fact that $|z_n(t,x)-z(t,x)|$ is bounded independetly of $n,t,x$ we conclude that $$f(z_n)\to f(z)\ \mbox{in}\ L^\infty((0,t)\times\Omega)$$