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In my textbook, it said that we have the trace operator on Sobolev space like this: (Suppose $\Omega$ is a nice domain in $R^d$) \begin{equation*} H^{s}(\Omega) \hookrightarrow H^{s-\frac{1}{2}}(\partial \Omega), \forall s > \frac{1}{2} \end{equation*}

I'm wondering if it is still valid in the case $s = \frac{1}{2}$, where we consider $H^0(\Omega)$ as $L^2(\Omega)$. I think it will have counterexample for this one, while I can not find any.

And I'm also looking for a proof of the following problem:

\begin{equation*} \text{ There exists $C>0$ s.t. for any $f \in H^1(\Omega)$,we have } \|f\|_{L^2(\partial \Omega)}^2 \leq C \|f\|_{L^2(\Omega)} \|f\|_{H^1(\Omega)} \end{equation*}

This problem is related to the previous problem in the following sense: suppose the trace theorem holds for $s = \frac{1}{2}$, then the left hand side is controlled by the $H^{1/2}$ norm, and the results can be proved by interpolation inequality on Sobolev Space.

Openminded
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  • Can you give me a reference for that trace theorem as you stated it? I was looking for it recently. – Overflowian Apr 20 '22 at 09:44
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    @Overflowian Quarteroni Valli Numerical Approzimation of Partial Differential Equations theorem 1.3.1 – Luigi May 02 '24 at 21:43

2 Answers2

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The borderline case $s=\frac12$ is false, even for the one-dimensional case. One way to see this is to consider the function $$ u(x) = \log \log \frac1{|x|} $$ on $B^+_{1/e}(0) = \{ x \in \Bbb R^2 : x_2 > 0, \lvert x \rvert< 1/e \}.$ This lies in $W^{1,2},$ so by the trace theory we have $$ x_1 \mapsto \log \log \frac1{|x_1|} $$ restricts to a function on $H^{\frac12}((-1/e,1/e)).$ However the existence of a bounded trace operator $H^{\frac12}((a,b)) \to L^2(\{a\})$ would imply this function is bounded in this region, which is evidently not the case. There should also be counterexamples in higher dimensions, but I can't recall the references for these at the moment.

What we have instead is the existence of a surjective trace operator $$ \mathrm{Tr} \colon B^{\frac12}_{2,1}(\Omega) \longrightarrow L^p(\partial\Omega).$$ Here $B^{s}_{p,q}$ denotes the Besov spaces, and the aformentioned space is slightly smaller than $H^{\frac12}(\Omega) = B^{\frac12}_{2,2}(\Omega).$ This is discussed for instance in Chapter 7 of

Adams, Robert A.; Fournier, John J. F., Sobolev spaces, Pure and Applied Mathematics 140. New York, NY: Academic Press (ISBN 0-12-044143-8/hbk). xiii, 305 p. (2003). ZBL1098.46001.

Note the inequality you seek can be proven this way via interpolation, since $B^{\frac12}_{2,1}$ is also an interpolation space between $L^2$ and $H^1$ (see the above text for details).

ktoi
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There is no trace theorem for $H^{1/2}(\Omega)$ since $\mathcal{C}^\infty_c(\Omega)$ is dense in $H^{1/2}(\Omega)$. Thus you can find, as close as needed (wrt the $H^{1/2}$ norm) to any given function of that Sobolev space, a smooth function with compact support, thus vanishing on $\partial\Omega$. A good reference for that is Lions-Magenes book "Nonhomogenous boundary value problems and applications" where you could find that half-integer Sobolev spaces have special properties.

Emmanuel
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