# Why doesn't the trace operator preserve regularity?

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.

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 $$p=\infty$$.
• Interesting! My adapted question would then be: what's the intuition behind the term $-1/p$ ? Commented Jun 22, 2023 at 8:02
• @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. Commented Jun 22, 2023 at 8:19
• 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 $n/p$. Commented Jul 10, 2023 at 5:27
• 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. Commented Jul 10, 2023 at 5:34