Consider $K \subset \mathbb R^n$ a compact set . Let $R > 0 $ such that $B(0,R) \supset K$ and $\partial B(0,R) \cap \partial K = \emptyset .$
Let $u : \overline{B(0,R)} \rightarrow \mathbb R$ a function satifying
$u$ is harmonic in $B(0,R) \setminus K$
$u = 1 $ in $K$
$u = 0$ in $\partial B(0,R)$
Exists a neighbourhood $V$ of $K$ and a constant $\alpha >0$ ($\alpha$ depends only of $u$) such that $|\nabla u(x)| \geq \alpha$ for all $x \in V \setminus K.$
Let $\epsilon >0$ small such that $\{ u > 1- \epsilon \} \subset V$ (I know this epsilon exists)
Define $u_{\epsilon} (z) = \displaystyle\frac{u(z) - (1- \epsilon)}{\epsilon} , z \in \{ u > 1 - \epsilon\}$.
I am studying an article, and appears that the author is using this
$$ |u_{\epsilon} (x) - u_{\epsilon} (y)| \geq \displaystyle\frac{\alpha}{\epsilon} |x - y| $$
for $x \in \partial \{ u > 1 - \epsilon \}$ and $y \in \{u > 1 - \epsilon \}$.
I am trying to show this for days, but I am not seeing how ... If I understand then I understand Lemma 3.2 of this article.
Someone can give me a help to prove or disprove the inequality ?
No, the claimed inequality is not that strong. It says that for each $x$ with $u_\epsilon(x)=0$ there exists $y$ with $u_\epsilon(y)=l$ such that $|x-y|\le cl$. Here I write $c=\frac{\epsilon }{\alpha}$ for brevity.
To see why, note that $|\nabla u_\epsilon|\ge c^{-1}$ in the relevant region. Starting from $x$, follow the curve of steepest ascent, i.e., the direction of $\nabla u_\epsilon$. Traveling along this curve with unit speed, we see the values of $u_\epsilon$ increasing at the rate of at least $c^{-1}$. Therefore, we will reach the level $l$ by traveling no further than $cl$.
More details: The curve of steepest ascent is the solution of the ODE $$x'(t) = \frac{\nabla u_\epsilon(x(t))}{ |\nabla u_\epsilon(x(t))|}$$ starting at a point $x$ with $u_\epsilon(x)=0$. By the chain rule, $$ \frac{d}{dt} u_\epsilon(x(t)) = |\nabla u_\epsilon(x(t)| \ge c^{-1}$$ Therefore, $u(x(t))$ will reach $l$ on the interval $ 0\le t\le cl $.