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The trace operator maps $H^1(\Omega)$ into $H^{1/2}(\partial \Omega)$, I have used this fact several times, and know of references where to find the proof. Let us assume that the boundaries are arbitrarily smooth. The trace operator then maps $C^1(\Omega)$ into $C^1(\partial \Omega)$.

Do you have an intuitive reason why we lose regularity in the weak case?

I think this may be a good way to get valuable understanding of how weak and strong derivatives differ.

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The difference in regularity here is less about "weak derivatives vs strong derivatives", and more about $L^2$-based vs $L^\infty$-based spaces.

In general, the trace operator on suitable defined Sobolev spaces maps $$\mathcal{T}:W^{s,p}(\Omega)\to W^{s-1/p,p}(\partial \Omega).$$

From this we see that there is less loss the higher $p$ is, but the classical spaces $C^1$ are defined by a sup-norm estimate, which is a classical analogue to $p=\infty$.

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    $\begingroup$ Interesting! My adapted question would then be: what's the intuition behind the term $-1/p$ ? $\endgroup$
    – Lilla
    Commented Jun 22, 2023 at 8:02
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    $\begingroup$ @Lilla I am not sure I can give an illuminating (computation-free) intuition. As you have seen in the $L^2$ case, it boils down to what the concrete estimate for the $L^p$ norm on the boundary submanifold is. $\endgroup$
    – goonfiend
    Commented Jun 22, 2023 at 8:19
  • $\begingroup$ An other way to see it, is what happens when $\Omega=\mathbb{R}^n_+$ and $\partial\Omega = \mathbb{R}^{n-1}\times\{0\}$. In this case, taking the trace is the same as "fixing" a variable considering $u(\cdot,0)\in \mathrm{W}^{s-1/p,p}(\mathbb{R}^{n-1})$ (when $s>1/p$). Up to a translation, this is similar to prove that $x_n\mapsto u(\cdot,x_n) \in \mathrm{C}^0_b([0,+\infty), \mathrm{W}^{s-1/p,p}(\mathbb{R}^{n-1}))$. Now, we have the following insight : "fixing a variable cost $1/p$ derivative", and notice that for usual Sobolev embeddings one looks at $n$ var. hence the cost of $n/p$. $\endgroup$
    – ToGle
    Commented Jul 10, 2023 at 5:27
  • $\begingroup$ Getting back to what goonfiend said : taking the trace on the boundary is the restriction to a $n-1$-dimensional submanifold, which is morally (in fact, exactly, up to a localisation procedure) the same as fixing one variable. Thus the lost of $1/p$ derivative when you take a trace. $\endgroup$
    – ToGle
    Commented Jul 10, 2023 at 5:34

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