Laplace on $\epsilon$-net (or the convergence of discrete harmoic function)

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Consider a domain $\Omega$ as the left in the below picture. Consider a PDE on it, for example $$ \Delta u=0~~~~x\in \Omega \\ u|_{\partial \Omega}= f(x)~~~~~~ $$ denote the solution as $U$.

Then consider a net on $\Omega$, for convience, we denote it as $\epsilon$-net, as the the right in below picture. the $\epsilon$ means the width of the net is $\epsilon$. We only need the intersection in $\Omega$. Denote them as $\Omega_\epsilon$. Besides, if a intersection has eight points around it, we call it as interior point. If not, we call it boundary point. We denote all boundary point of $\Omega_\epsilon$ as $\partial\Omega_\epsilon$. We define a function $f_\epsilon$ on $\partial\Omega_\epsilon$,
$$ f_\epsilon(a_\epsilon)=f(a) $$ where $a_\epsilon\in \partial \Omega_\epsilon$, $a\in\partial \Omega$ is the nearest point of $a_\epsilon$. Maybe, there are many nearest points, if so , take their average.

Now we define a function $U_\epsilon$ on $\Omega_\epsilon $:

  1. If $(x,y)$ is interior point of $\Omega_\epsilon$, $U_{\epsilon}(x,y)=\frac{1}{4}[U_{\epsilon}(x,y+1)+U_{\epsilon}(x,y-1)+U_{\epsilon}(x+1,y)+U_{\epsilon}(x-1,y)]$.

  2. If $(x,y)$ is the boundary point of $\Omega_\epsilon$, $U_{\epsilon}(x,y)=f_{\epsilon}(x,y)$.

Then, define a metric from $U_\epsilon$ to $U$ as $$ d(U_\epsilon,U)= \max_{(x,y)\in\Omega_\epsilon}|U_\epsilon(x,y)-U(x,y)| $$ Then how to show $$ \lim_{\epsilon\rightarrow 0} U_\epsilon= U $$ up to the above metric ?

PS: I think this question from the discrete Laplace. I think the above guess should be right, but don't know how to begin to prove it. Thanks for any help.

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