# Sobolev Bound for solution to Dirichlet Problem from knowledge of boundary estimates

Take some bounded domain, $$\Omega \in \mathbb{R}^n$$, $$n \geq 3$$. Denote by $$\partial \Omega$$ the boundary of $$\Omega$$, which we take to be Lipschitz. Let $$L$$ be an elliptic operator that satisfies ellipticity and boundness conditions. Let $$u$$ be the solution to the following Dirichlet problem: $$$$\begin{cases} Lu = 0 \qquad & \mbox{in} \quad \Omega \\ u = -v &\mbox{on} \quad \partial \Omega \end{cases}$$$$ Assume that we know the following is true about $$v$$ $$$$\sup_{\partial \Omega} (|v| + |\nabla v|) \leq C_1$$$$ where $$C_1$$ is some positive constant. I am wondering, as a consequence of the inequality above, can we say $$$$|| u ||_{H^{1}(\Omega)} \leq C_2$$$$ where $$C_2$$ is a positive constant, possibly distinct from $$C_1$$?

There might be subtleties depending on the exact assumptions that you make, but the general reasoning is as follows.

When $$u \in H^1(\Omega)$$, its trace on $$\partial \Omega$$ belongs to $$H^{1/2}(\partial\Omega)$$. Conversely, if you take a $$v \in H^{1/2}(\partial\Omega)$$, you can find a $$\bar{v} \in H^1(\Omega)$$ such that $$\bar v = v$$ on $$\partial \Omega$$ and $$\| \bar{v} \|_{H^1(\Omega)} \leq C \| v \|_{H^{1/2}(\partial\Omega)}.$$ Then you look for $$u$$ under the form $$u = u' + \bar{v}$$ so that $$u'$$ is a solution to $$Lu' = f$$ with $$f = - L\bar{v}$$ and $$u' = 0$$ on $$\partial \Omega$$. For this problem, you might know that $$\| u' \|_{H^1(\Omega)} \leq C \| f \|_{H^{-1}(\Omega)}$$ and here you have $$\|f\|_{H^{-1}(\Omega)} \leq C \| \bar{v} \|_{H^1(\Omega)}.$$ Putting all of this together leads you to the estimate $$\| u \|_{H^1(\Omega)} \leq C \| v \|_{H^{1/2}(\Omega)}.$$ Now the $$H^{1/2}=W^{1/2,2}$$ fractional Sobolev norm is bounded above by the $$W^{1,\infty}$$ norm which you use in your assumption. So, essentially, what you are looking for is indeed valid.

As mentioned above, there are subtleties depending on the exact regularity of $$\partial\Omega$$ and on the coefficients of $$L$$.

• Thank you very much for your answer. There two small questions I have. 1. Do you have a reference for the first inequality relating the H^1 norm and the H^{1/2} norm? 2. I am not sure where the two inequalities under "For this problem, you might know that" come from? Is there a reference for this? @cs89 Commented May 17, 2023 at 11:34
• For trace lifting result, you can see e.g. the second part of Theorem 18.40 of Leoni's book A first course in Sobolev spaces. For the resolution of $L u' =f$ with $f \in H^{-1}$ you can look e.g. Theorem 4.22 of Hunter's lecture notes on PDEs. Note that, once you know $u'$ exists, the bound $\|u'\|_{H^1} \leq \|f\|_{H^{-1}}$ is a straightforward energy estimate (just multiply the PDE by $u$ and integrate).
– cs89
Commented May 17, 2023 at 12:08
• Thanks for the links to the references. Sorry for all the questions but I am looking at Theorem 18.40 of Leoni's book. It states that $||u||_{L^p(\Omega)} \leq C ||g||_{L^p(\partial \Omega)}$. Am I correct in saying that $||g||_{L^p(\partial \Omega)}$ is the $H^{1/2}$-norm? I haven't seen the $H^{1/2}(\partial \Omega)$- norm defined as the $L^2(\partial \Omega)$-norm before. I know that $g \in L^2(\partial \Omega)$ but I thought the $H^{1/2}(\partial \Omega)$-norm was defined using the $H^1(\Omega)$-norm. Commented May 18, 2023 at 8:48
• No. The $H^{1/2}$ norm is neither the $L^2$ nor the $H^1$ one, it is "in between". It is a "fractional Sobolev space". You can find many resources online or in textbooks or on this forum about them. Theorem 18.40 has two estimates, one that says $\|u\|_{L^2(\Omega)} \leq \|g\|_{L^2(\partial\Omega)}$ and one that says $\|u\|_{H^1(\Omega)} \leq \|g\|_{L^2(\partial\Omega)}+\|g\|_{B^{1-1/p,p}(\partial\Omega)}$, which is $H^{1/2}$ for $p = 2$. So heuristically, you can estimate a weak norm of $u$ by a weak norm of $g$ and a strong norm of $u$ by a strong one of $g$.
– cs89
Commented May 18, 2023 at 12:40
• The $H^1$ norm is precisely defined as the sum of the $L^2$ norm of $u$ and the $L^2$ norm of $\nabla u$.
– cs89
Commented May 22, 2023 at 21:44