# Ngô Quốc Anh

## April 18, 2013

### A lower bound for solutions involving distance functions

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 21:14

In this note, we discuss an useful lemma and its beautiful proof given by Brezis and Cabré in a paper published in Boll. Unione Mat. Ital. Sez. B Artic. Ric. Mat. in 1998. The full article can be freely downloaded from here. Before saying anything further, let us state the lemma.

Lemma 3.2. Suppose that $h \geq 0$ belongs to $L^\infty (\Omega)$. Let $v$ be the solution of

$\left\{\begin{array}{rcl} - \Delta v &=& h \quad \text{ in }\Omega , \hfill \\ v &=& 0 \quad \text{ on }\partial \Omega . \hfill \\ \end{array}\right.$

Then

$\displaystyle\frac{{v(x)}}{{\text{dist}(x,\partial \Omega )}} \geqslant c\int_\Omega {h\text{dist}(x,\partial \Omega )} ,\qquad\forall x \in \Omega,$

where $c>0$ is a constant depending only on $\Omega$.

This type of estimate frequently uses in the literature. We now show the proof of the lemma.

The proof basically consists of two steps:

Step 1. For any compact subset $K \subset\Omega$, we show

$\displaystyle v(x) \geqslant c \int_\Omega h \delta$

for any $x \in K$ where the constant $c>0$ depends only on $K$ and $\Omega$. This and the fact that $K$ stays away from the boundary of $\Omega$ allow us to have the desired estimate. To do so, let

$\displaystyle \varrho = \frac{1}{2}\text{dist}(K,\partial\Omega),$

and take $m$ balls of radius $\varrho$ such that

$\displaystyle K \subset B_\varrho(x_1) \cup \cdots \cup B_\varrho (x_m) \subset \Omega.$

This can be done since $K$ is compact. Now, instead of proving our estimate in the whole $K$, we shall prove our estimate for each ball $B_\varrho(x_i)$. To do that, let $\xi_1,...,\xi_m$ be the solutions of

$\left\{\begin{array}{rcl} - \Delta \xi_i &=& \chi_{B_\varrho (x_i)} \quad \text{ in }\Omega ,\\ \xi_i &=& 0 \quad \text{ on }\partial \Omega . \end{array}\right.,$

where $\chi_A$ denotes the characteristic function of $A$. The Hopf boundary lemma implies that there is a constant $c>0$ such that

$\displaystyle \xi_i (x) \geqslant c \text{dist}(x,\partial\Omega)$

for all $x \in \Omega$ and all $i=\overline{1,m}$. Let now $x \in K$ and take a ball $B_\varrho (x_i)$ containing $x$. Then

$\displaystyle B_\varrho (x_i) \subset B_{2\varrho} (x) \subset \Omega,$

and since $-\Delta v \geq 0$ in $\Omega$, we conclude

$\begin{array}{rcl} v(x) &\geqslant & \displaystyle\frac{1}{{\text{vol}({B_{2\varrho }}(x))}}\int_{{B_{2\varrho }}(x)} v \hfill \\ &\geqslant & \displaystyle c\int_{{B_{2\varrho }}(x)} v \geqslant c\int_{{B_\varrho }({x_i})} v \hfill \\ &=& \displaystyle c\int_\Omega {v( - \Delta {\xi _i})} = \displaystyle c\int_\Omega {( - \Delta v){\xi _i}} \hfill \\ &=& \displaystyle c\int_\Omega {h{\xi _i}} \hfill \\ &\geqslant & \displaystyle c\int_\Omega {h \text{dist}(x,\partial\Omega )} .\end{array}$

Step 2. Fix a smooth compact subset $K \subset\Omega$, we need to prove

$\displaystyle\frac{{v(x)}}{{\text{dist}(x,\partial \Omega )}} \geqslant c\int_\Omega {h\text{dist}(x,\partial \Omega )}$

for any $x\in\Omega \backslash K$. Let $w$ be the solution of

$\left\{\begin{array}{rcl} - \Delta w &=& 0 \quad \text{ in }\Omega , \hfill \\ w &=& 0 \quad \text{ on }\partial \Omega , \hfill \\ w &=& 1 \quad \text{ on }\partial K. \hfill \\ \end{array}\right.$

The Hopf boundary lemma now gives

$\displaystyle w(x) \geqslant c \text{dist}(x,\partial M)$

for any $x\in\Omega \backslash K$. Since $v$ is superharmonic and $v \geqslant c\int_\Omega k \text{dist}(x,\partial \Omega$ on $\partial K$, the maximum principle implies

$\displaystyle v(x) \geqslant c \left(\int_\Omega h\text{dist}(x,\partial \Omega) \right)w(x) \geqslant c\left(\int_\Omega h\text{dist}(x,\partial \Omega) \right)\text{dist}(x,\partial \Omega)$

for any $x\in\Omega \backslash K$. The proof is now complete.

1. Very nice. It’s amazing the power of the maximum principle.

Comment by Fran — April 19, 2013 @ 7:59

2. Hi, I have a question. $\xi$ is not a $C^2$ function, so Hopf lemma can not be used directly. Is there a smoothing argument?

Comment by fanicolas — April 28, 2013 @ 2:13

• Hi, thanks for your question. It looks like we have a typo here since $\chi$ does not belong to any Sobolev space. I think we can change the definition of $\xi_i$ a little bit as follows

$\displaystyle -\Delta \xi_i = \eta \in [0,1]$

where $\eta$ is a cut-off function which equals $1$ in ${B_\frac{\varrho}{2} (x_i)}$ and $0$ outside the ball $B_\varrho (x_i)$. Then the following estimate remains valid as shown below

$\begin{array}{rcl} v(x) &\geqslant & \displaystyle\frac{1}{{\text{vol}({B_{2\varrho }}(x))}}\int_{{B_{2\varrho }}(x)} v \hfill \\ &\geqslant & \displaystyle c\int_{{B_{2\varrho }}(x)} v \geqslant c\int_{{B_\varrho }({x_i})} v \hfill \\ &\geqslant & \displaystyle c\int_{{B_\varrho }({x_i})} {v( - \Delta {\xi _i})} \\ &=& \displaystyle c\int_\Omega {v( - \Delta {\xi _i})} = \displaystyle c\int_\Omega {( - \Delta v){\xi _i}} \hfill \\ &=& \displaystyle c\int_\Omega {h{\xi _i}} \hfill \\ &\geqslant & \displaystyle c\int_\Omega {h \text{dist}(x,\partial\Omega )} .\end{array}$

Comment by Ngô Quốc Anh — April 28, 2013 @ 20:48

3. HI I have a query. How does the hopf lemma imply $w(x)\geq C dist(x,\partial\Omega)$? Thanks.
One more thing while constructing $w$ it should be $-\Delta w=0$ in $\Omega - K$

Comment by Santu — July 24, 2016 @ 12:18

• Hello Santu,

Thanks for asking. The use of Hopf’s lemma turns out to be standard for PDErs :).

You may argue in a “geometric way” as follows: Suppose there does not exist such a constant $C$; hence the function $w$ decays faster than the distance function to the boundary. By decay I mean there exists a sequence of points converging to the boundary. Therefore, if you compare the distance to the boundary and the amplitude of $w$, you will get a curve which is convex. Therefore, the directional derivative with respect to the outer normal at the limit point (on the boundary) tends to be non-negative, which contradicts to what Hopf’s lemma says.

For the second part of you question, this is standard too. Let me know if you still cannot figure out how.

Comment by Ngô Quốc Anh — July 25, 2016 @ 3:32

4. Yeah I got it. Thanks a lot.

Comment by Santu — July 25, 2016 @ 20:30