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The Wikipedia article on Agomon's inequality states the following:

Let $u\in H^2(\Omega)\cap H^1_0(\Omega)$ where $\Omega\subset\mathbb{R}^2$. Then Agmon's inequality in 2D states that there exists a constant $C$ such that $$ \displaystyle \|u\|_{L^\infty(\Omega)}\leq C \|u\|_{L^2(\Omega)}^{1/2} \|u\|_{H^2(\Omega)}^{1/2}. $$

In Agmon's lecture notes, the general version is as follows:

Lemma 13.2. Let $m>n/2$ and let $u\in H_m(\Omega)$.Then there exists a constant $\gamma_s$, depending only on $\Omega$ and $m$, such that, after modification of $u$ on a set of measure zero, $$ |u(x)|\leq\gamma_s \|u\|_{m}^{n/2m}\|u\|_{0}^{1-(n/2m)},\quad x\in\overline{\Omega}. $$

In Agmon's notation, $n$ denotes the dimensional of the Euclidean space and $\|\cdot\|_k=\|\cdot\|_{H_k}$.

Question: Is the assumption $H_0^1$ redundant in the Wikipedia article?

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    $\begingroup$ There is probably a regularity assumption on $\Omega$ in the lecture notes, right? Zero traces are very convenient because then $\Omega$ may be very irregular and one may rely on results for the full space $\mathbb{R}^n$ by considering the zero extension in the proofs. If one wants the result thus obtained also for non-zero trace functions, one usually needs some boundary regularity for $\partial\Omega$ to have suitable extensions to the full space at hand, but then the proofs work very similarly. I would expect something like that here. $\endgroup$ – Hannes Nov 27 '18 at 16:15
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    $\begingroup$ @Hannes: Thanks for your comment! Indeed in Agmon's notes, he assumes that the domain has "nice" boundary. I overlooked this. I would accept your comment as an answer. $\endgroup$ – Jack Nov 27 '18 at 17:08
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There is probably a regularity assumption on $\Omega$ in the lecture notes, right?

Zero traces are very convenient in such proofs because then $\Omega$ may be very irregular and one may rely on results for the full space $\mathbb{R}^n$ by considering the zero extension of the respective functions. If one wants the result thus obtained also for non-zero trace functions, one usually needs some boundary regularity for $\partial\Omega$ to have suitable extensions to the full space at hand, but then the proofs work very similarly. I would expect something like that here.

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