Let $\Sigma$ be a compact smooth surface with boundary. Is it true that the supremum


$$\sup \{ \lambda\_1(\Sigma,g) \operatorname{Area}(\Sigma,g) : g \text{ smooth Riemannian metric on $\Sigma$} \}$$


is finite? Here, $\lambda\_1(\Sigma,g)$ denotes the first eigenvalue of the Laplacian associated to the metric $g$ with Dirichlet ($=0$) boundary condition. This is true if $\Sigma$ has no boundary, as shown by Yang-Yau, for instance.


According to the answer by Gabe (see below), this is false for a cylinder. Is it false for **every** compact surface with boundary?



Eigenvalues of the Laplacian on surfaces with boundary
This is not true without making some type of stronger assumption on the geometry. For instance, if $\Sigma,g$ is a rectangle with sides $\epsilon$ and $1/\epsilon$, the area is 1 whereas the first Dirichlet eigenvalue is $\frac{\pi^2}{\epsilon^2}+\pi^2 \epsilon^2$. This isn’t a smooth domain, but we can round the corners to smoothen it while keeping the product of the area and eigenvalue large. And by varying the metric so that $\epsilon$ gets arbitrarily small, we can make the product as big was we like.


