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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{equation} \begin{cases} Lu = 0 \qquad & \mbox{in} \quad \Omega \\ u = -v &\mbox{on} \quad \partial \Omega \end{cases} \end{equation} Assume that we know the following is true about $v$ \begin{equation} \sup_{\partial \Omega} (|v| + |\nabla v|) \leq C_1 \end{equation} where $C_1$ is some positive constant. I am wondering, as a consequence of the inequality above, can we say \begin{equation} || u ||_{H^{1}(\Omega)} \leq C_2 \end{equation} where $C_2$ is a positive constant, possibly distinct from $C_1$?

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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 $L$.

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  • $\begingroup$ 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 $\endgroup$ Commented May 17, 2023 at 11:34
  • $\begingroup$ 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). $\endgroup$
    – cs89
    Commented May 17, 2023 at 12:08
  • $\begingroup$ 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. $\endgroup$ Commented May 18, 2023 at 8:48
  • $\begingroup$ 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 $g$. $\endgroup$
    – cs89
    Commented May 18, 2023 at 12:40
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    $\begingroup$ The $H^1$ norm is precisely defined as the sum of the $L^2$ norm of $u$ and the $L^2$ norm of $\nabla u$. $\endgroup$
    – cs89
    Commented May 22, 2023 at 21:44

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