The Sobolev Imbedding Theorem says that for $\Omega\subset\mathbb{R}^n$ we have for $mp<n$ that $W^{m,p}(\Omega)\hookrightarrow L^q(\Omega)$, where $p\leq q \leq np/(n-mp)$.

The Sobolev Trace Theorem says that for $\Omega \subset \mathbb{R}^n$ with a $C^m$ class boundary and $mp<n$ that $W^{m,p}(\Omega)\hookrightarrow L^q(\partial \Omega)$, where $p\leq q \leq (n-1)p/(n-mp)$.

My question is for a very nice boundary (say $\Omega$ is a cube in $\mathbb{R}^3$) can the restriction on $q$ be relaxed such that $W^{m,p}(\Omega)\hookrightarrow L^q(\partial \Omega)$ for $p\leq q \leq np/(n-mp)$?

Perhaps I don't understand fully why we go from $np$ to $(n-1)p$ in the numerator of the upper bound on $q$ when going from the Embedding Theorem to the Trace Theorem. Is this not possible? Is a counterexample given in any texts of the trace theorem failing (for a nice enough boundary) when $q=np/(n-mp)$?


To get some intuition for the allowed exponents, you should play around with some model functions and see when the integrals converge. Power functions alone can tell you a lot.

Let's study the functions $f(x) = |x|^r, r<0$ on the unit cube $\Omega = [0,1]^n$. Some simple calculations should convince you that the iterated derivative $D^m f $ has norm proportional to $|x|^{r-m}.$ We know that $\int_0^1 x^a dx$ converges iff $a > -1$; so you should be able to convince yourself that $\int_\Omega |x|^a \, d^n x$ converges iff $a+n > 0$. (Use spherical coordinates.)

Since the highest-order term in the $W^{m,p}(\Omega)$ norm is $\int_\Omega |D^m f|^p,$ we see that $f \in W^{m,p}(\Omega)$ iff $r > m-n/p$, and thus $f \in L^q(\Omega)$ iff $r>-n/q$. Some simple algebra should now allow you to show the necessary condition $W^{m,p}(\Omega)\subset L^q(\Omega) \implies q\le np/(n-mp).$

When you want a trace theorem instead, the essential source of the difference in admissible exponents is that $\partial \Omega$ is of dimension $n-1$. In our example, the boundary of the $n$-cube is a collection of $n-1$-cube faces, with the problematic ones (those where $f$ and its derivatives are not bounded) being those that include the origin. On each of these faces, the function $f$ takes the form $f: [0,1]^{n-1} \to \mathbb R : f(x)=|x|^r;$ so by the same calculations we did before we know that $f \in L^q(\partial \Omega)$ if and only if $r>(1-n)/q,$ which is harder to achieve than $r > -n/q.$

As an explicit counterexample, let $\Omega = [0,1]^2$, $p=1$, $m=1,$ $q=np/(n-mp) = 2$ and consider the function $f(x) = |x|^{-1/2}.$ Then $|Df(x)|=\frac12 |x|^{-3/2}$, so we can verify $f \in W^{1,1}(\Omega)$ by checking that the integrals $\int_{[0,1]^2} |x|^{-1/2} dx_1\, dx_2, \int_{[0,1]^2} |x|^{-3/2} dx_1\, dx_2$ converge. Along the boundary component $[0,1] \times \{0 \}$, however, we find that $f$ is not $L^2$, since $\int_0^1 |f|^2 = \int_0^1 x^{-1} dx = \infty$. Thus $f \notin L^2(\partial \Omega)$, so the trace of a $W^{1,1} (\Omega)$ function is not necessarily in $L^2(\partial \Omega)$.

  • $\begingroup$ Thank you for the great answer! Makes a lot of sense! $\endgroup$ – Math244 Feb 20 '18 at 13:03

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