$\theta_x(t)$ be the angle between $\gamma'_x$ and $\nabla\psi(Q_t(x))$. Then the function $$h(t)=\cos(\theta_x(t))$$ is non-increasing in $t$. As $\theta_x(r(x))=\theta_x$ we have in particular: $$\cos(\theta_x(t))\geq \cos(\theta_x)\geq m$$ for all $t\in [0,r(x)]$ and $x\in\Sigma_1$.*]{}

[**Step 2.**]{} [*The function $r\to{\lvert{\psi^{-1}(r)}\rvert}$ is non-decreasing in $r$.*]{}

[**Step 3.**]{} [*If $\Sigma_2$ is also convex we have $m\geq \beta/B$.*]{}

We will prove Steps 1-3 below.

[**Proof of Proposition \[estimate doubly convex\]**]{}. a) At any point of $\Omega$, let $\nabla^R\psi$ denote the radial part of $\nabla\psi$, which is the gradient of the restriction of $\psi$ to the ray passing through the given point. As such restriction is a linear function, one sees that $$\dfrac{1}{B}\leq{\lvert{\nabla^R\psi}\rvert}\leq \dfrac{1}{\beta}.$$ Since ${\lvert{\nabla\psi}\rvert}\geq{\lvert{\nabla^R\psi}\rvert}$ one gets immediately $${\lvert{\nabla\psi}\rvert}\geq\dfrac{1}{B}.$$ Note that $\theta_x(t)$, as defined above, is precisely the angle between $\nabla\psi$ and $\nabla^R\psi$, so that, using Step 1,

$${\lvert{\nabla^R\psi}\rvert}={\lvert{\nabla\psi}\rvert}\cos\theta_x(t)\geq m{\lvert{\nabla\psi}\rvert}$$ hence: $${\lvert{\nabla\psi}\rvert}\leq \dfrac{1}{m}{\lvert{\nabla^R\psi}\rvert}\leq\dfrac{1}{\beta m}.$$ as asserted. It is clear that b) and c) are immediate consequences of Steps 2-3.

**Proof of Step 1.** We use a suitable parametrization of $\Omega$. Let $l$ be the length of $\Sigma_1$ and consider a parametrization $\gamma:[0,l]\to \Sigma_1$ by arc-length $s$ with origin at a given point in $\Sigma_1$. Let $N(s)$ be the outer normal vector to $\Sigma_1$ at the point $\gamma(s)$. 