# Ngô Quốc Anh

## June 16, 2010

### De Giorgi’s class and De Giorgi’s theorem

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 18:55

In this entry, we introduce that De Giorgi’s class $DG(\Omega)$ and prove that functions in $DG(\Omega)$ are Holder continuous. This has as a consequence the celebrated De Giorgi’s theorem saying that solutions of second-order elliptic equations with bounded measurable coefficients are Holder continuous.

Let us study $H^1$-solutions of

$\displaystyle\int_\Omega {{A^{\alpha \beta }}(x){\nabla _\alpha }u{\nabla _\beta }\varphi dx} = 0, \quad \varphi \in H_0^1(\Omega )$

where $A$ is assumed to be of class $L^\infty(\Omega)$. Our aim is to show that $u$ is in fact Holder continuous. The key idea is to use the Cacciopoli inequality on level sets of $u$.

Definition (De Giorgi’s class). We define the De Giorgi’s class ($DG(\Omega)$) to consist of all $u \in H^1(\Omega)$ which satisfy

$\displaystyle \int_{{B_\rho }} {{{\left| {\nabla {{(u - k)}^ + }} \right|}^2}dx} \leqslant \frac{c}{{{{(R - \rho )}^2}}}\int_{{B_R}} {{{\left| {{{(u - k)}^ + }} \right|}^2}dx} , \quad \forall k \in \mathbb{R}$.

The following remark is useful.

1. If $u$ is a sub-solution then $u \in DG(\Omega)$.
2. If $u$ is a super-solution then $-u \in DG(\Omega)$.
3. If $u \in DG(\Omega)$ then $u+{\rm const} \in DG(\Omega)$.
4. In the definition of the De Giorgi class, exponent $p=2$ can be replaced by any $p>1$.

If we denote by $A$ the level set of $u$, that is

$A(k,r)=\{x \in B_r : u(x)>k\}$

then if $u \in DG(\Omega)$ by choosing a cut-off function

$\displaystyle \eta \in C_0^\infty\left(B_\frac{R+\rho}{2}\right), \quad \eta \equiv 1, \quad {\rm on } B_\rho$

we get

$\displaystyle\int_{{B_{\frac{{R + \rho }}{2}}}} {{{\left| {\nabla {{(\eta (u - k))}^ + }} \right|}^2}dx} \leqslant \frac{c}{{{{(R - \rho )}^2}}}\int_{{B_R}} {{{\left| {{{(u - k)}^ + }} \right|}^2}dx} ,\quad \forall k \in \mathbb{R}$

or

$\displaystyle\int_{A(k,\rho )} {{{\left| {\nabla (u - k)} \right|}^2}dx} \leqslant \frac{c}{{{{(R - \rho )}^2}}}\int_{A(k,R)} {{{\left| {u - k} \right|}^2}dx} , \quad \forall k \in \mathbb{R}$.

By using the Sobolev inequality

$\displaystyle {\left( {\int_{{B_\rho }} {{{\left| {{{(u - k)}^ + }} \right|}^{{2^ \star }}}dx} } \right)^{\frac{2}{{{2^ \star }}}}} \leqslant \int_{{B_{\frac{{R + \rho }}{2}}}} {{{\left| {\nabla {{(\eta (u - k))}^ + }} \right|}^2}dx}$

and the Holder inequality

$\displaystyle \int_{{B_\rho }} {{{\left| {{{(u - k)}^ + }} \right|}^2}dx} \leqslant {\left| {x \in {B_\rho }:u(x) > k} \right|^{1 - \frac{2}{{{2^ \star }}}}}{\left( {\int_{{B_\rho }} {{{\left| {{{(u - k)}^ + }} \right|}^{{2^ \star }}}dx} } \right)^{\frac{2}{{{2^ \star }}}}}$

we conclude that

$\displaystyle\int_{A(k,\rho )} {{{\left| {u - k} \right|}^2}dx} \leqslant \frac{c}{{{{(R - \rho )}^2}}}{\left| {A(k,R)} \right|^{\frac{2}{n}}}\int_{A(k,R)} {{{\left| {u - k} \right|}^2}dx}$.

For $h>k$

$\displaystyle {\left| {h - k} \right|^2}\left| {A(h,\rho )} \right| = \int_{A(h,\rho )} {{{\left| {h - k} \right|}^2}dx} \leqslant \int_{A(k,\rho )} {{{\left| {u - k} \right|}^2}dx}$.

Set

$\displaystyle a(h,\rho ) = \left| {A(h,\rho )} \right|, \quad u(h,\rho ) = \int_{A(h,\rho )} {{{\left| {u - h} \right|}^2}dx}$

and write the above inequalities in the form

$\displaystyle\begin{gathered} u(h,\rho ) \leqslant \frac{c}{{{{(R - \rho )}^2}}}u(k,R)a{(k,R)^{\frac{2}{n}}}, \hfill \\ a(h,\rho ) \leqslant \frac{1}{{{{(h - k)}^2}}}u(k,R). \hfill \\ \end{gathered}$

For some positive numbers $\xi$ and $\eta$  we find

$\displaystyle u{(h,\rho )^\xi }a{(h,\rho )^\eta } \leqslant \frac{{{c^\xi }}}{{{{(R - \rho )}^{2\xi }}}}\frac{1}{{{{(h - k)}^{2\eta }}}}u{(k,R)^{\xi + \eta }}a{(k,R)^{\frac{{2\xi }}{n}}}$.

Now we choose $\xi$ and $\eta$ in such a way that for some $\theta>1$

$\displaystyle \xi + \eta = \theta \xi ,\frac{{2\xi }}{n} = \theta \eta$.

Solving these equations gives $\eta=1$ and $\xi=\frac{n}{2}\theta$.

Thus we arrive at

$\displaystyle\phi (h,\rho ) \leqslant \frac{{{c^\xi }}}{{{{(R - \rho )}^{2\xi }}{{(h - k)}^{2\eta }}}}\phi {(k,R)^\theta }, \quad \theta > 1$

where $\phi=u^\xi a^\eta$.

Proposition. For any $k_0$ and any $\sigma \in (0,1)$ we have

$\phi(k_0+d,R_0-\sigma R_0)=0$

for some $d$.

Consequently, if we choose $\sigma=\frac{1}{2}$ we get

Theorem. If $u \in DG(\Omega)$ then

$\displaystyle\mathop {\sup }\limits_{x \in {B_{\frac{R}{2}}}} |u(x)| \leqslant {k_0} + c{\left( {\frac{1}{{{R^n}}}\int_{A({k_0},R)} {{{\left| {u - {k_0}} \right|}^2}dx} } \right)^{\frac{1}{2}}}{\left( {\frac{{A({k_0},R)}}{{{R^n}}}} \right)^{\frac{{\theta - 1}}{2}}}$.

In particular when $k_0=0$ we have

Theorem. If $u \in DG(\Omega)$ then

$\displaystyle\mathop {\sup }\limits_{{B_{\frac{R}{2}}}} {u^ + } \leqslant c{\left( {\overline\int_{{B_R}} {{{\left| {{u^ + }} \right|}^2}dx} } \right)^{\frac{1}{2}}}$

and

Theorem. If $-u \in DG(\Omega)$ then

$\displaystyle\mathop {\sup }\limits_{{B_{\frac{R}{2}}}} {u^ - } \leqslant c{\left( {\overline\int_{{B_R}} {{{\left| {{u^-}} \right|}^2}dx} } \right)^{\frac{1}{2}}}$

From these two estimates we infer

Theorem. Solution $u$ of PDE

$\displaystyle\int_\Omega {{A^{\alpha \beta }}(x){\nabla _\alpha }u{\nabla _\beta }\varphi dx} = 0, \quad \varphi \in H_0^1(\Omega )$

is bounded and

$\displaystyle\mathop {\sup }\limits_{{B_{\frac{R}{2}}}} |u| \leqslant c{\left( {\overline\int_{{B_R}} {{{\left| u \right|}^2}dx} } \right)^{\frac{1}{2}}}$.

We now come to the main result of this entry

Theorem (De Giorgi). If $u$ and $-u$ belong to $DG(\Omega)$ then $u$ is $\alpha$-Holder continuous. Moreover, for $\rho we have

$\displaystyle\int_{{B_\rho }} {{{\left| {\nabla u} \right|}^2}dx} \leqslant c{\left( {\frac{\rho }{R}} \right)^{n + 2\alpha - 2}}\int_{{B_R}} {{{\left| {\nabla u} \right|}^2}dx}$.

The proof of the Holder continuity is based an estimate concerning the oscillation

$\displaystyle\mathop {\rm osc}\limits_{{B_R}} u = \mathop {\sup }\limits_{{B_R}} u - \mathop {\inf }\limits_{{B_R}} u$.

Precisely, there exists some constant $\alpha$ such that

$\displaystyle\mathop {\rm osc}\limits_{{B_\rho }} u \leqslant c{\left( {\frac{\rho }{R}} \right)^\alpha }\mathop {\rm osc}\limits_{{B_R}} u$.

Thus by scaling and iterating this estimate, one obtains the Holder continuity of $u$ since the oscillation of $u$ is easily seen to decay as a power of radius. To see this, let pick $x$ and $y$ arbitrary in $\Omega$. Denote $\rho=|x-y|$, then

$\displaystyle\left| {u(x) - u(y)} \right| \leqslant \mathop {\sup }\limits_{{B_\rho } \cap \Omega } u - \mathop {\inf }\limits_{{B_\rho } \cap \Omega } u \leqslant \mathop {\rm osc}\limits_{{B_\rho }} u \leqslant c\frac{{\mathop {\rm osc}\limits_{{B_R}} u}}{{{R^\alpha }}}{\rho ^\alpha } = c\frac{{\mathop {\rm osc}\limits_{{B_R}} u}}{{{R^\alpha }}}{\left| {x - y} \right|^\alpha }$.