$\gamma_x$ meets $\Sigma_2$ at a first point $Q(x)$, and we let $
r(x)=d(x,Q(x)).
$ For $x\in\Sigma_1$, we denote by $\theta_x$ the angle between the ray $\gamma'_x$ and the outer normal to $\Sigma_2$ at the point $Q(x)$, and we let $$m\doteq\min_{x\in\Sigma_1}{\cos\theta_x}.$$ We assume that $\Omega$ is strictly starlike, that is, $m>0$; in particular $Q(x)$ is unique. Recall also that: $$\label{annulus}
\beta=\min_{x\in\Sigma_1}r(x), \quad B=\max_{x\in\Sigma_1}r(x).$$ We construct a suitable smooth function $\psi$ and estimate the constant $K=K_{\Omega,\psi}$ with respect to the geometry of $\Omega$. The starlike assumption implies that each point in $\Omega$ belongs to a unique ray $\gamma_x$. Then we can define a function $\psi:\Omega\to [0,1]$ as follows: $$\psi=\threesystem
{0\quad\text{on}\quad\Sigma_1}
{1\quad\text{on}\quad\Sigma_2}
{\text{linear on each ray from $\Sigma_1$ to $\Sigma_2$}.}$$ Estimates and now follow from Theorem \[main3\] together with the following Proposition.

\[estimate doubly convex\] a At all points of $\Omega$ one has: $
\frac{1}{B}\leq{\lvert{\nabla\psi}\rvert}\leq\frac{1}{\beta m}.
$ Therefore: $$K_{\Omega,\psi}=\dfrac{\sup_{\Omega}{\lvert{\nabla\psi}\rvert}}{\inf_{\Omega}{\lvert{\nabla\psi}\rvert}}\leq\dfrac{B}{\beta m}.$$ b One has $$\sup_{r\in [0,1]}{\lvert{\psi^{-1}(r)}\rvert}=L={\lvert{\Sigma_2}\rvert}.$$ c If $\Sigma_2$ is also convex, then $m\geq \beta/B$ hence we can take $K=\beta^2/B^2$.

The proof of the Proposition \[estimate doubly convex\] depends on the following steps.

[**Step 1.**]{} [*On the ray $\gamma_x$ joining $x$ to $Q(x)$, consider the point $Q_t(x)$ at distance $t$ from $x$, and let 