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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?

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    $\begingroup$ I think it is false. Just consider the right cylinder $[0,h] \times \mathbb{S}^1(r)$ of radius $r$ and height $h$. The first eigenvalue is equal to a constant multiple of $1/r^2$. Even imposing that the area is equal to $1$, say, allows for unbounded values of $\lambda_1$. $\endgroup$
    – Eduardo Longa
    Commented Jan 3, 2023 at 2:20
  • $\begingroup$ What if the boundary of $\Sigma$ is connected? $\endgroup$
    – Eduardo Longa
    Commented Jan 3, 2023 at 2:21

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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.

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