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The trace operator $T$ is defined for bounded domain $U$ with $C^1$ boundaries as the linear, continuous operator $T: W^{1,p}(U) \rightarrow L^p(\partial U)$ such that $$ Tu=u\;\text{ on }\partial U $$ whenever $u$ has a version that can be continuously extended on the closure of $U$.
Intuitively, this is supposed to describe $u$ on its boundary, so I would expect that if, for a suitable real valued function $g$, we have that $g(u)$ is in $W^{1,p}$ we would have furthermore $$ Tg(u)=g(T(u))\;\text{ for all }u\text{ on }W^{1,p}.\label{1}\tag{1} $$ This happens for instance in the case when $u$ has a continuous extendable version.

So the question is: does \eqref{1} actually hold?

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    $\begingroup$ It is true if $g \in C^1$. In fact the identity is true for smooth functions and extends to $W^{1,p}$ by continuity since both $T\circ g$ and $g \circ T$ are continuous. $\endgroup$
    – Giorgio Metafune
    Commented Dec 13, 2023 at 21:30
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    $\begingroup$ Extending what Giorgio said: All you really need is that the function $g$ has the property that $u\mapsto g(u)$ is a continuous mapping of $W^{1,p}(U)$ and that $v\mapsto g(v)$ is a continuous mapping of $L^p(\partial U)$. This should also be true when $g$ is Lipschitz continuous. $\endgroup$
    – Willie Wong
    Commented Dec 14, 2023 at 4:18

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