Are Neumann Laplacian eigenfunctions in $C(\overline{\Omega})$?

Consider that $$u\in H^1(\Omega)$$ with $$\Delta u\in L^2(\Omega)$$ (in the distributional sense) such that for some $$\lambda>0$$ we have that:

$$\begin{cases} \Delta u(x)=\lambda u(x), & x\in\Omega\\ \dfrac{\partial u}{\partial\nu} (x)=0, & x\in\partial\Omega\end{cases}$$

We assume $$\Omega\subseteq\mathbb{R}^2$$ to be an open, connected, bounded and has a uniform Lipschitz boundary.

1) Is it true that $$u\in H^2(\Omega)$$? If this is not true:

2) Is it true that $$u\in C(\overline{\Omega})$$? If this is not true:

3) How can we prove that $$u\in L^{\infty}(\Omega)$$?

I know that (3) is valid from the inequality posted here: Contractivity of Neumann Laplacean

But I do not know how to prove that inequality. Maybe it can be done in an easier way...

I wonder if there is an estimate of the form:

$$\Vert u\Vert_{\infty}\leq c\lambda^{\alpha}$$, where $$c$$ is a constant depending on $$\Omega$$?

I know that such estimates hold for for the Dirichlet laplacian. I found some references about the problem here:

https://math.stackexchange.com/questions/2309436/regularity-of-laplacian-eigenfunctions-in-convex-polygon

but they do not represent counter-examples for any of my questions.

P.S. I found that (2) might also be true from that post: Is the linear span of the Neumann eigenfunctions dense in $C(\overline{D})$ but I did not understand the argument. Is there any clear way of getting more information about the regularity of the eigenfunction $$u$$ of $$\lambda$$?

• What is the regularity of $\Omega$? Elliptic regularity strongly depends on the regularity of the boundary of $\Omega$... Commented Jan 25, 2023 at 17:15
• $\Omega$ is a bounded and connected Lipschitz domain. Commented Jan 25, 2023 at 17:40
• Domains with corners are well studied (there is a monograph by Grisvard), and they show that in general we do not get $H^2$. Commented Jan 25, 2023 at 19:52
• If $\Omega$ has a $C^2$ boundary, you can conclude that $u \in H^2(\Omega)$. Commented Jan 25, 2023 at 20:18
• @RomainGicquaud I know from the Grisvard book about that. Do you know if the inequality of the $L^{\infty}$ norm of $u$ satisfies the inequality written in the bold zone of the statement? Commented Jan 26, 2023 at 9:08

For (3), this follows from De-Giorgi-Nash-Moser so long as you are OK with dependence of the constant on the $$C^{0, 1}$$ character of the domain, not just e.g. the size.
There are some tricks that will give a power dependence on $$\lambda$$ like stated in the question for some large, explicit $$\alpha$$. First treat $$\lambda u$$ as a zero-order term and run a finite number of Moser iterations (can be skipped in 2D to get $$\alpha = 1$$) to show that $$\|u\|_{L^p} \leq C \lambda^\alpha \|u\|_{L^2}$$ for a $$p > \frac{n}{2}$$, then treat $$\lambda u = f$$ in $$- \Delta u = f$$ and show that $$\|u\|_{L^\infty} \leq C \|f\|_{L^p}$$. This will generally not give the optimal value $$\alpha$$.
This also answers (2); De Giorgi-Nash-Moser gives that $$u \in C^{0, \beta}$$ for some $$\beta$$ which will depend on the Lipschitz constant of the domain.
(1) is false, as can be checked on non-convex polygons in the plane by blowing up at the corners. The blow-up limit is essentially explicit (like $$z^\alpha$$ for some $$\alpha > \frac{1}{2}$$, but tending to $$\frac{1}{2}$$ as the corner becomes more oblique), so locally this has $$|D^2 u|^2 \approx |z|^{2\alpha - 4}$$, which fails to be integrable. While obviously not a proof, this is the correct heuristic here.
It's unclear to me whether or not (3) remains true with $$c=c(|\Omega|)$$ only, at least specifically in 2D.