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Let $\Sigma$ be a smooth open disk munished with a Riemannian metric $g$. We assume that $\Sigma$ has finite volume. Let $x$ be any point on $\Sigma$.

For $\epsilon>0$, let $m_\epsilon$ be the minimal length of a curve $\gamma$ that winds once around $x$ and stays at distance at most $\epsilon$ from the boundary (when $\epsilon$ is small, winding around $x$ is the same as "winding along the boundary").

question: Does there exists a constant $C_g$ such that $m_\epsilon\leq \frac{C_g}{\epsilon}$? If not, is there still a good bound on the divergence rate of $m_\epsilon$ as $\epsilon\to 0$?

remark: I know that $m_\epsilon$ is finite. Indeed, if $T$ is a tubular end of $\Sigma$, we double it across its boundary (including the boundary $\partial \Sigma$ "at infinity") to obtain a torus. We can then apply a refined version of the Loewner's torus inequality, to show that there is two curves that form a basis in homology and such that the product of there lengths is less than a constant times the volume. Back on $T$, it implies that there is a finite length curve between the two boundaries , and another one that winds along the boundary and stay at finite distance from the boundary. We can then go from "stay at finite distance" to "stay at small distance".

Isao
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    It seems we do have this inequality. For each $\epsilon$, let $\gamma_\epsilon = { x\in \Sigma : d(x, \partial \Sigma) = \epsilon}$. Assume that this is the minimal curve, then an $\epsilon$-thickening $U_\epsilon$ of $\gamma_\epsilon$ will be in $\Sigma$. It seems that $U_\epsilon$ has area around $m_\epsilon \cdot \epsilon$, so since $\Sigma$ has finite area we have $m_\epsilon \cdot \epsilon < A$. (Edited: not so sure now...) – Arctic Char Aug 24 '20 at 13:48
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    $\gamma_\epsilon$ is not necessarily a curve... For fixed $\epsilon$, the $\epsilon'$ thickening of $\gamma_\epsilon$ has area EXACTLY $2 m_\epsilon.\epsilon'$ if $\epsilon'$ is small enough (in general the volume of a thickening is polynomial for small enough parameters. The maximal $\epsilon'$ depends on $\epsilon$ though... – Isao Aug 24 '20 at 14:16
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    If $\Sigma$ is a domain on $\mathbb{R}^2$ which is $\alpha$ Holder only for $\alpha<\frac{1}{2}$, I think the area of $U_\epsilon$ is bounded by $C \epsilon^{2\alpha}$ but no more... – Isao Aug 24 '20 at 14:19

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