Given $\Omega\subset \mathbb R^N$ open bounded with nice boundary. Then for $u\in W^{1,p}(\Omega)$, $1\leq p\leq \infty$, we have $$\|u\|_{L^p(\Omega)}\leq C(\|T[u]\|_{L^p(\partial\Omega)}+\|\nabla u\|_{L^p(\Omega)}) $$ where $T$ denotes the trace operator and constant $C$ depends on $\Omega$, $\partial\Omega$, and $p$, $N$.

I proved this by using density argument, i.e., first show for $(u_n)\subset C^\infty (\bar{\Omega})$ and push to limit. Since $u_n\to u$ in $W^{1,p}(\Omega)$ implies that $T[u_n]\to T[u]$ in $L^p(\partial\Omega)$, so the density argument works.

I use this argument a lot in weak theorem for Robin's boundary problem, especially in non-linear case.

However, I never see this statement in any text book, at least in H. Brezis, Evans, Evans & Gariepy, Adams, and Leoni's, nor some exercises...

So I was wondering is my statement true? It intuitively makes sense, and if you want, I can write some details. Also, if you know this statement was stated in some books, please kindly direct me there. Thank you!


This is a known result. For example, Theorem 4.2.1 in Ziemer's Weakly Differentiable Functions, specialized to first-order Sobolev spaces, says that for any linear functional $\varphi:W^{1,p}\to\mathbb R$ that takes value $1$ on the constant function $1$, we have $$ \|u-\varphi(u)\|_{p;\Omega} \le C\|\nabla u\|_{p;\Omega} $$ Take $\varphi( u)=\frac{1}{|\partial \Omega|}\int_{\partial \Omega}T[u]$ to obtain your result.

A stronger statement (a form of the Poincaré-Sobolev inequality) can be found later in the same book:

4.4.7. Corollary If $\Omega$ is a bounded Lipschitz domain and $u\in W^{1,p}(\Omega)$, $p>1$, then $$\|u\|_{p^*;\Omega}\le C\left[\|\nabla u\|_{p;\Omega}+\left|\int_{\partial\Omega} u\right|\right]$$

(The statement in the book has a typo: the absolute value around the boundary integral is omitted.) Here subscripts mean Lebesgue norms, and $p^*=np/(n-p)$ is the Sobolev embedding exponent for $p$. The author then remarks that the inequality extends to $BV$, in particular to $W^{1,1}$.

  • $\begingroup$ I see. So this is even true in BV space. That is so nice. $\endgroup$ – spatially Dec 29 '14 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.