of Step 2.**]{} In the coordinates $(t,s)$ the curve $\psi^{-1}(r)$ is parametrized by $\alpha:[0,l]\to\tilde\Omega$ as follows: $$\alpha(u)=(r\rho(u),u)\quad u\in [0,l].$$ Then: $$\begin{aligned}
{\lvert{\psi^{-1}(r)}\rvert}&=\int_0^l\sqrt{g(\alpha'(u),\alpha'(u))}\,du\\
&=\int_0^l\sqrt{r^2\rho'(u)^2+(1+rk(u)\rho(u))^2}\,du
\end{aligned}$$ Convexity of $\Sigma_1$ implies that $k(u)\geq 0$ for all $u$; differentiating under the integral sign with respect to $r$ one sees that indeed $\frac{d}{dr}{\lvert{\psi^{-1}(r)}\rvert}\geq 0$ for all $r\in [0,1]$.

[**Proof of Step 3.**]{} Let $T_x$ be the tangent line to $\Sigma_2$ at $Q(x)$ and $H(x)$ the point of $T_x$ closest to $x$. As $\Sigma_2$ is convex, $H(x)$ is not an interior point of $\Omega$, hence $$d(x,H(x))\geq\beta.$$ The triangle formed by $x, Q(x)$ and $H(x)$ is rectangle in $H(x)$, then we have: $$r(x)\cos\theta_x=d(x,H(x)).$$ As $r(x)\leq B$ we conclude: $$B \cos\theta_x\geq \beta,$$ which gives the assertion.

Sharpness of the lower bound {#sharpness}
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An upper bound
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In this short paragraph, we give a simple way to get an upper bound when the potential $A$ is *closed*. Then, we will use this in different kinds of examples, in order to show that the assumptions of Theorem \[main2\] are sharp. The geometric idea is the following: if we have a region $D \subset \Omega$ such that the first absolute cohomology group $H^1(D)$ is $0$, then we can estimate from 