Gradient estimates for a boundary value problem $\newcommand{\avint}{⨍}$
Let $B_r$ be a call of radius $r$ and centre origin and $k<1$.$w$ satisfy the following PDE:
$$
\begin{cases}
-\Delta w = 0 \qquad \mbox{in $B_r\setminus B_{kr}$}\\
w=0 \qquad \mbox{on $\partial B_{kr}$}\\
w=\varphi\qquad \mbox{on $\partial B_r$}
\end{cases}
$$
Show that,  for all $x\in \partial B_{kr}$,
$$
|\nabla w(x)| \,\le\, \frac{C}{r} \avint_{\partial B_r} \varphi\,d\sigma
$$
$d\sigma$ is the surface measure.
(one can take $\varphi\geq 0$ is it helps in the proof. The issue has occured in the Theorem 3.1 of [1], page 436, the estimate mentioned between equation 3.2 and 3.3)
[1] Alt, Hans Wilhelm; Caffarelli, Luis A.; Friedman, Avner, Variational problems with two phases and their free boundary, Trans. Am. Math. Soc. 282, No. 2, 431-461 (1984). ZBL0844.35137.
 A: OK, here are some additional details to what I wrote in my comments.
Due to scaling, we can choose $r = 1$. Let $D = B_1 \setminus B_k$, and let $P_D(x, y)$ be the Poisson kernel of $D$. Thus,
$$ w(x) = \int_{\partial B_1} P_D(x, y) \varphi(y) \sigma(dy) $$
for $x \in D$. Let $n$ denote the unit normal vector at a boundary point of $D$. Note that $\nabla w(x) = (\partial_n w(x)) n$ for $x \in \partial B_k$. It follows that
$$ |\nabla w(x)| = \biggl| \int_{\partial B_1} \partial_n P_D(x, y) \varphi(y) \sigma(dy) \biggr| $$
for $x \in \partial B_k$; here and below the derivative acts on the $x$ variable. Finally, by the boundary Harnack inequality,
$$ \partial_n P_D(x, y) $$
is a bounded function of $x \in \partial B_k$ and $y \in \partial B_1$ (I can give more details on this if needed). We conclude that
$$ |\nabla w(x)| \leqslant C \int_{\partial B_1} |\varphi(y)| \sigma(dy) ,$$
as desired.
A: Below is a maximum principle-based alternative to the proof of Mateusz.
We may assume that $r = 1$ by scaling. Let $w$ be the harmonic function on $B_1 \subset \mathbb{R}^n$ with boundary data $|\varphi|$. By the usual representation formula we have
$$w|_{\partial B_{\frac{k+1}{2}}} \leq \frac{C(n,k)}{|\partial B_1|}\int_{\partial B_1}|\varphi| := A.$$
By the maximum principle, $|u| \leq w$ in $B_1 \backslash B_k$ (in particular, on $\partial B_{\frac{k+1}{2}}$). We conclude using the maximum principle again that
$$|u| \leq A\frac{k^{2-n}-|x|^{2-n}}{k^{2-n}-\left(\frac{k+1}{2}\right)^{2-n}} := v$$
on $B_{\frac{k+1}{2}} \backslash B_k$. (Here we assumed $n \geq 3$; when $n = 2$, the function $v$ is obtained in similar way using $\log$). Since $u = v = 0$ on $\partial B_k$ it follows that
$$|\nabla u| \leq |\nabla v| = \frac{(n-2)k^{1-n}}{k^{2-n}-\left(\frac{k+1}{2}\right)^{2-n}}A$$
on $\partial B_k$, which is an estimate of the desired form.
