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Let $(w_k)$ be a sequence of positive continuous functions and $\Omega\subset\mathbb R^n$ be a bounded domain.

We know that for $x_0\in\Omega$ and $0<r<R\leq\operatorname{dist}(x_0,\partial\Omega)$ the inequality $$u(y)\leq u(z)\exp\left(\frac{|y-z|}{R-r}\right)$$ holds for all $y,z\in B_r(x_0)$. Additionally we have the estimate $$|Du(x)|\leq \frac{u(x)}{\operatorname{dist}(x,\partial\Omega)}$$ for almost every $x\in\Omega$.

Now I have a proof that says if $\left(\sup_\Omega w_k\right)_{k\in\mathbb N}$ is bounded , the estimates above yield that the sequence $(w_k)$ is locally equicontinuous and thus converges locally uniformly up to a subsequence.

I am not sure why this holds.

My thought was that if $\sup_\Omega w_k$ is bounded, we have $$|Dw_k(x)|\leq\frac{w_k(x)}{\operatorname{dist}(x,\partial\Omega)}\leq \frac{C}{\operatorname{dist}(x,\partial\Omega)}\quad\text{a.e.}$$ where $C$ is the bound of $\sup_\Omega w_k$. Now if $\operatorname{dist}(x,\partial\Omega)$ is bounded away from zero, $w_k$ would be locally lipschitz and thus locally equicontinuous. But I am not sure if dist is bounded away from zero.

Can anyone help me with the reasoning here or tell me another reason why the functions $w_k$ are locally equicontinuous? Thanks.

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  • $\begingroup$ So, is $u=w_k$ in the above? $\endgroup$ – Jose27 Mar 28 '15 at 20:24
  • $\begingroup$ Also, if $w_k$ are only continuous, what is the meaning of $Dw_k$? $\endgroup$ – Jose27 Mar 28 '15 at 20:26
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    $\begingroup$ As for the argument, just pick $x$ such that $d(x, \partial \Omega)>r>0$. For instance fix $x_0$ and set $2r<d(x_0, \partial \Omega)$, then for $x\in B_r(x_0)$ you have what you want. $\endgroup$ – Jose27 Mar 28 '15 at 20:30
  • $\begingroup$ You are right about the differentiability. I forgot to mention that this is indeed given. Thanks for your answer, I totally forgot that we only want to show this locally and thus can bound the distance function $\endgroup$ – dinosaur Mar 29 '15 at 9:48

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