Intuition behind losing half a derivative via the trace operator

Question

This is an informal question, but here goes:

For a function $f \in H^s(\Omega)$ ($s > 1/2$), there is a well-defined operator (the trace) $T$ such that $Tf = f\vert_{\partial \Omega}$ if $f \in C^\infty \cap H^s(\Omega)$ (so that it agrees with 'restriction to the boundary' in the classical sense), and such that $Tf \in H^{s - 1/2}(\partial \Omega)$ (so that we 'lose half a derivative' when taking the trace).

My question is this: why half a derivative? If this is just a technical result of an interpolation theorem (or similar), then so be it, but is there a good intuition behind this?

Answer 1

Low-tech illustration (copied from comments):

Take one-dimensional situation: how good should $f(x)=\sum c_n e^{inx}$ be so that we can restrict it to a point, i.e., evaluate? We want the series $\sum |c_n|$ to converge. In terms of $f\in H^s$, this is guaranteed if $s>1/2$, because by Cauchy-Schwarz $$\left(\sum|c_n|\right)^2 \le \sum (1+n^2)^{-s} \sum (1+n^{2})^s|c_n|^2 =C \|f\|_{H^s} \tag1$$ In order to make Cauchy-Schwarz work, we need $|c_n|^2$ to be summable with the factor of $n$ (plus a tiny bit), which translates into half-derivative.

More general, medium-tech explanation. The reason we have to lose half-derivative when restricting is that we can gain half-derivative when extending. Take extension to half-spae for simplicity. Given $g$ defined on $\mathbb R^{n-1}$ (for simplicity, compactly supported), a natural way to extend it to half-space is via averaging: for $x\in\mathbb R^{n-1}$ and $t>0$, let $$f(x,t)=\frac{1}{t^{n-1}}\int \varphi((x-y)/t) g(y)\,dy\tag2$$ where $\varphi$ is some mollifier. Suppose we want $f\in H^1$. Differentiating $f$ and sweating through estimates we get something like $$\int_{\mathbb R^n_+}|\nabla f|^2\le C\sum_{k=1}^{n-1}\int_0^\infty \frac{dt}{t^2} \int_{\mathbb R^{n-1}}|g(y+te_k)-g(y)|^2\,dy \tag3$$ where on the right we have $B^{1/2,2}$-norm of $g$, which is same as $H^{1/2}$.