If the first Dirichlet eigenfunction on a set $D$ is regular up to the boundary, is $D$ regular?

Question

Given any open set $D$ in $\mathbb R^n$, we can define the first Dirichlet eigenfunction $u$ of $-\Delta$ on $D$ as the minimizer of the Rayleigh quotient over $H_0^1(D)$. Interior regularity of $u$ follows easily, but in general it seems we cannot say anything about the regularity of $u$ up to the boundary.

But conversely, if we know $u$ is say globally Lipschitz continuous on the closure of $D$ (assume $D$ is bounded), does this tell us anything about the regularity of the boundary $\partial D$? For example, that it is locally a Lipschitz graph? What if $u$ has even more regularity?

This seems like a natural question, but I cannot find a reference or think of a way to construct counterexamples.

Answer 1

One cannot expect any fatness of the domain $D$: the thinner $D$ at some boundary point, the faster the decay of the eigenfunction near this point. In particular, the first eigenfunction will be Lipschitz continuous for any thorn-shaped domain $D = \{(x, y) \in \mathbf{R}^2 : |y| < f(x), \, x \in (0, 1)\}$, where $f$ is, say, positive, increasing and convex on $(0, 1]$.

However, regularity of the eigenfunction clearly implies some kind of fatness of the exterior of $D$. Since the first eigenfunction is superharmonic, its continuity on the boundary implies continuity of harmonic functions (indeed: the first eigenfunction is a barrier at any boundary point), and so every boundary point must be regular for the Dirichlet problem. If the first eigenfunction is in fact Lipschitz continuous, then harmonic functions which are equal to zero on some part of the boundary near $x_0$ have at most linear decay rate at $x_0$. This means, for example, that there is no interior cone at $x_0$ with aperture greater than $\pi$. However, I am not aware of any characterisation of domains for which the first eigenfunction is Lipschitz continuous.