Research Blog

De Giorgi

2021, January 22

Introduction

Let $\Omega \subset \Rd$ be a smooth bounded domain. Let $A: \Omega \to \mathcal M ^{d \times d} (\R)$ be a measurable matrix-valued function such that it is uniformly elliptic: there exists $0 < \lambda \le \Lambda$ such that $$\begin{align*} \lambda \ \Id \le A (x) \le \Lambda \ \Id, \qquad \forall x \in \Omega. \end{align*}$$

We say $u \in H ^1 (\Omega)$ is a subsolution in $\Omega$ of $$\begin{align*} - \div (A \grad u) = 0 \end{align*}$$ if for every $\vp \in H _0 ^1 (\Omega)$ such that $\vp \ge 0$, we have $$\begin{align*} \ang{\grad u, \grad \vp} _A := \int _{\Omega} (\grad u) ^\top (A \grad \vp) \dx \le 0. \end{align*}$$

De Giorgi’s First Lemma: Small Energy to Boundedness

Proof.

Define $$\begin{align*} r _k &= 1 + 2 ^{-k}, & B \bp k &= B _{r _k} (0), & \ind{B \bp {k + 1}} &\le \vp _k \le \ind{B \bp k}. \end{align*}$$ Denote $$\begin{align*} c _k &= 1 - 2 ^{-k}, & u _k &= (u - c _k) _+, & \Omega _k &= \set{\vp _k u _k > 0}, & \ind k &= \ind{\Omega _k}. \end{align*}$$

Denote energy $$\begin{align*} U _k = \nmL2{\vp _k u _k} ^2 = \int (\vp _k u _k) ^2 \dx. \end{align*}$$

• Hölder’s inequality. For $p = (1/2 - 1/d) \inv$, $$\begin{align*} \nmL2{\vp _k u _k} \le \nmL p{\vp _k u _k} \nmL d{\ind k}. \end{align*}$$

Conclusion: $$\begin{align*} U _k \lesssim C ^k U \bms k ^{1 + 2/d}. \end{align*}$$ If $U _0 < \delta$ sufficiently small, then $U _k \to 0$. That is, $\nor{(u - 1) _+} _{L ^2 (B _1)} = 0$, $u \le 1$ in $B _1$.

De Giorgi’s Second Lemma: Boundedness to Oscillation Decay

Proof.

Recall $$\begin{align*} c _k &= 1 - 2 ^{-k}, & u _k &= (u - c _k) _+, \end{align*}$$ we define $$\begin{align*} w _k &= 2 \cdot 2 ^k u _k = 2 (w \bms k - 1) _+ \le 2. \end{align*}$$

Claim that there exists a universal $K$ such that $w _{K + 2} \equiv 0$ in $B _{1/2}$.

Otherwise, for any $k \le K$, $w \bps[2] k \not\equiv 0$ in $B _{1/2}$, then $$\begin{align*} w \bps[2] k \not\equiv 0 \inn B _{1/2} & \Rightarrow \abs{\set{w \bps[2] k > 0} \cap B _{1/2}} > 0 \newline & \Rightarrow \abs{\set{w \bps k > 1} \cap B _{1/2}} > 0 \newline \text{(by the first lemma)} & \Rightarrow \nor{w \bps k} _{L ^2 (B _1)} > \delta \newline & \Rightarrow \nor{w \bps k} _{L ^\infty (B _1)} \abs{\set{w \bps k > 0} \cap B _1} ^\frac12 > \delta \newline & \Rightarrow \abs{\set{w \bps k > 0} \cap B _1} > \pthf{\delta}{2} ^2 \newline & \Rightarrow \abs{\set{w _k > 1} \cap B _1} > \pthf{\delta}{2} ^2. \end{align*}$$

Then

• $w _k$ has large top. $$\begin{align*} \abs{\set{w _k > 1} \cap B _1} > \pthf{\delta}{2} ^2. \end{align*}$$

• $w _k$ has large base. $$\begin{align*} \abs{\set{w _k < 0} \cap B _1} \ge \abs{\set{u < 0} \cap B _1} > \mu. \end{align*}$$

• $w _k$ has bounded energy. This is because $w _k$ is a subsolution in $B _2$, by energy inequality $$\begin{align*} \nor{\grad w _k} _{L ^2 (B _1)} \le C \nor{w _k} _{L ^2 (B _2)} \le C _0. \end{align*}$$

By isoperimetric inequality, $w _k$ has large sides, that is, $$\begin{align*} \abs{\set{0 < w _k < 1} \cap B _1} > \alpha. \end{align*}$$ Since $\set{0 < w _k < 1} = \set{c _k < u < c _{k + 1}}$ are pairwise disjoint, if we let $K$ be so large such that $\mu + K \alpha > |B _1|$, then we must have $w _{K + 2} \equiv 0$ in $B _\frac12$, so $u _{K + 2} = 0$, $u \le c _{K + 2}$. This finishes the proof by setting $\eta = 2 ^{-K - 2}$.

Tools (six inequalities, ethics).

Proof.

Since $\vp ^2 u \in H ^1 _0 (\Rn)$ is nonnegative, $$\begin{align*} 0 \ge \ang{\grad (\vp ^2 u), \grad u} _A &= \ang{\vp \grad u, \vp \grad u} _A + 2\ang{u \grad \vp, \vp \grad u} _A \newline &= \ang{\vp \grad u + u \grad \vp, \vp \grad u + u \grad \vp} _A - \ang{u \grad \vp, u \grad \vp} _A \newline &= \ang{\grad (\vp u), \grad (\vp u)} _A - \ang{u \grad \vp, u \grad \vp} _A. \end{align*}$$ By ellipticity of $A$, $$\begin{align*} \lambda \nmL2{\grad (\vp u)} ^2 \le \Lambda \nmL2{u \grad \vp} ^2. \end{align*}$$

Proof.

Choose $\vp$ to be a cutoff such that $\ind{B _1} \le \vp \le \ind{B _2}$, and use the Caccioppoli inequality.

Proof.

Without loss of generality assume $0 \le u \le 1$, so $A = \set{u = 0}$, $C = \set{u = 1}$. Fix $x \in C$.

If $\sigma$ is a direction such that the ray $\set{x + t \sigma} _{t \ge 0}$ pierces through $A$, then the variation on this ray is greater than 1, $$\begin{align*} 1 \le \int _0 ^{D _\sigma} \abs{\ddt u(x + r \sigma)} \d r \le \int _0 ^{D _\sigma} |\grad u (x + r \sigma)| \d r, \end{align*}$$ where $D _\sigma$ is the distance at which $x + D _\sigma \sigma \in \partial B _1$. If $\Sigma$ is the set of all such $\sigma$, then $$\begin{align*} \abs{\Sigma} \le \int _\Sigma \int _0 ^{D _\sigma} |\grad u (x + r \sigma)| \d r \d \sigma &= \int _\Sigma \int _0 ^{D _\sigma} \frac{|\grad u (x + r \sigma)|}{r ^{n - 1}} r ^{n - 1} \d r \d \sigma \newline &= \int _{B _1} \frac{|\grad u (y)|}{|y - x| ^{n - 1}} \dy. \end{align*}$$ Since a circular sector with center $x$, radius $2$ and angle $\Sigma$ will cover $A$, we have $|A| \lesssim |\Sigma|$. Now integrate $x$ over $C$, $$\begin{align*} \abs{A} \abs{C} &\lesssim \int _{C} \int _{B _1} \frac{|\grad u (y)|}{|y - x| ^{n - 1}} \dy \dx = \int _{B _1} |\grad u (y)| \int _{C} \frac{1}{|y - x| ^{n - 1}} \dx \dy. \end{align*}$$ For all set with the same measure as $C$, The integral $\int _{C} \frac{1}{|y - x| ^{n - 1}} \dx$ is minimized when $C$ is a ball centered at $y$, so $$\begin{align*} \int _{C} \frac{1}{|y - x| ^{n - 1}} \dx \lesssim |C| ^{\frac1n}. \end{align*}$$ Therefore, $$\begin{align*} \abs{A} \abs{C} &\lesssim |C ^\frac1n| \int _{B _1} |\grad u (y)| \dy. \end{align*}$$ Finally, using $\supp (\grad u) \subset D$, $$\begin{align*} \int _{B _1} |\grad u (y)| \dy \le \nor{\grad u} _{L ^2(B _1)} \nor{\ind D} _{L ^2 (B _1)} \le (C _0 |D|) ^\frac12. \end{align*}$$

Proof.

We show a $p < (1/2 - 1/d) \inv$ version. If $\grad u = f$ and $u$ is compactly supported, then $$\begin{align*} u = \La \inv \div \grad u = \div \La \inv f = f * \grad \Gamma \end{align*}$$ where $\Gamma$ is the fundamental solution to the Laplace equation, and $$\begin{align*} \grad \Gamma (x) = \frac{c _d}{|x| ^d}x \in L ^q _\loc \end{align*}$$ for any $q < \frac{d}{d - 1}$. By Young’s convolution inequality, $$\begin{align*} \nmL p u \le \nmL 2 f \nmL q{\grad \Gamma} \end{align*}$$ with $1 + p \inv = 2 \inv + q \inv$.

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