Analysis and PDE : Harmonic function

$\Delta u = 0$ may be one of the most important function in the PDE since it appears many times in different context and has nice properties : well, we have to say that its beautiful property implies the interesting result in physics and our natural. Here, we recall some basic property and proof strategy of this topic.

Classical solution

Although the existence should be considered as the first property necessary for study the object, maybe it's nicer to give some interesting property at first. We suppose that the solution is of class $C^2$, then one important formula is Stokes formula that 

$$\begin{eqnarray*}\int_{\partial \Omega} F \cdot \nu  d\sigma = \int_{\Omega} \nabla \cdot F(x) dx \end{eqnarray*}$$

$$\int_{\partial \Omega} u \nu  d\sigma = \int_{\Omega} \nabla u(x) dx$$

we apply this and obtain that in a domain of harmonic function we have 

$$\int_{\partial \Omega} \nabla u(x) \cdot \nu d\sigma = \int_{\Omega} \Delta u(x) dx = 0$$

Furthermore, we obtain some useful formula as Green formula, we obtain the mean value principle that 

$$u(x) = \frac{1}{|\partial B_R(x)|}\int_{\partial B_R(x) }u(y) dy$$

One remarkable result is that this property improve the regularity as $C^{\infty}$ and it also implies the function is harmonic. (The proof is similar by the convolution below). Since if we do derivative, we get 

$$0 = \frac{d}{dr} \int_{\partial B_1(x) }u(x + ry) d\sigma = \int_{\partial B_1(x) }\nabla u(x + ry) \cdot \nu d\sigma = \frac{d}{dr} \int_{ B_r(x) } \Delta u(y) dy$$

A second important property is the Liouville proeprty. It says that a bounded harmonic function is trivial and is constant. The proof uses the fact that all the derivative are also harmonic 

$$|\partial_i u| \leq \frac{1}{\omega_d R^d} \int_{\partial B_R} |u| d\sigma \leq \frac{N}{R}\sup|u| \rightarrow 0$$

and then we use the mean value principle to analysis its size.

A third important property is the maximum principle. Idea is simple : the maximum and minimum of the function can only be attended at its boundary. Use the maximum principle, we prove that the uniqueness of the Dirichlet problem.

Weak solution is also classical solution

One very famous theorem Weyl states that all the weak solution that 

$$\int_{\Omega} u(x) \Delta \phi(x) dx = 0$$

for the test function $\phi \in C_c^{\infty}(\Omega)$ is also a strong solution of class $C^{\infty}$. The proof is very classical : we do convolution $u_{\epsilon} = u \ast \psi_{\epsilon}$. Then we prove that this function satisfies the harmonic by weak relation. Finally,  we pass this limit to the mean value principle. 

$$u_{\epsilon}(x) = \frac{1}{| B_R(x)|}\int_{ B_R(x) }u_{\epsilon}(y) dy \rightarrow u(x) = \frac{1}{| B_R(x)|}\int_{ B_R(x) }u(y) dy$$

Since this property doesn't require the regularity, we reprove that is is strong solution.

Existence

Finally, we come back to the problem of existence. There are two ways to prove the existence and it works in some more general framework. 

First one is the Lax-Milgram theorem. It treat the problem as find the inverse of operator in some function space 

$$a(u, v) = L(v)$$

The second one is the variational formation that we treat the solution  as a minimum of 

$$J(u) = \frac{1}{2}\int_{\Omega} |\nabla u|^2 dx - \int_{\Omega} f u dx$$

One can also deduce the characterization from one to another.

Analysis and PDE : Covering and decomposition lemmas - Vitali, Calderon-Zygmund, Whitney

Some estimations a priori are important in PDE, while they come from some functional inequality, and some of them come from harmonic analysis and require some combinatoric covering lemma to prove it. This is a subject that I have learned several years ago from Prof. Hongquan LI when I was in Fudan. Today I spent a whole day to review them.

Vitali covering

When we study the Hardy-Littlewood maximal function 

$$\mathcal{M} f(x) = \sup_{r > 0} \frac{1}{|B_r(x)|} \int_{B_r(x)} f(y) dy$$

We would like to prove that this operator is of type strong $(p,p)$. The strategy is to prove that it is weak $(1,1)$ and strong $(\infty, \infty)$ and then uses the interpolation inequality. However, the weak $(1,1)$ isn't very clear. One tool used in the proof is so called Vitali covering. It says that for a covering $\{B_i\}_{i \geq 0}$ we could abstract a disjoint sub-covering $\{B_{kj}\}_{j \geq 0}$such that 

$$\bigcup_j^{\infty} B_{kj} \subset \bigcup_i^{\infty} B_i \subset \bigcup_j^{\infty} 3B_{kj}$$

This nice structure makes some sub-additive into additive and is very  useful in the proof.

Calderon-Zygmund decomposition

Calderon-Zygmund decomposition could be seen as a dyadic version of maximal inequality. We start by the system of dyadic cubes 

$$\mathcal{D} = \bigcup_{k \in \mathbb{Z}} \mathcal{D_k}$$  

where $\mathcal{D_k}$ is the collection of cubes of  length $2^{-k}$. Then for any point $x$, it belongs to only one cube in the system $\mathcal{D_k}$. And any two cubes could be disjoint or one included in another - they could not have non-trivial intersection and difference at the same time. This decomposition is used to study singular integral, but at first we could use it to build a generalized cubic version of maximal function defined as 

$$\mathcal{M}_Q f(x) = \sup_{x \in Q \in \mathcal{D}} \frac{1}{|Q|} \int_{Q} f(y) dy$$

The first fact is that any open set in $\mathbb{R}^d$ could be built by the disjoint union of cubes. Idea is simple, we give one cube for each point $x$ such that $x \in Q \subset \Omega$. Then we mesh them if one belongs to another.

We apply this idea to the open set 

$$A_{\lambda} = \{x | \mathcal{M}_Q f(x)  \geq \lambda \}$$

and for each point, we choose the largest $Q$ admit. This works once we suppose $f \in L^1(\mathbb{R}^d)$ where an infinite cubes make it null. Then we know that this gives a perfect partition of the open set.

$$A_{\lambda} = \bigsqcup_{i=1}^{\infty}Q_i$$

Moreover, we know that the average at each cube is less than $2^d \lambda$ if $f$ is positive by the stopping time property, and $\mathcal{M}_Q f(x) < \lambda \text{ a.e }$ for the point outside $A_{\lambda}$ by the Lebesgue differential theorem.  

Whitney decomposition

Finally, we give a stronger version of Whitney decomposition. It is also a dyadic decomposition but has two more properties

1, The distance of one cube from $\Omega^c$ is comparable to its length.

2, If two cubes are neighbors, their lengths are comparable.

So this decomposition looks more balanced.

Its construction is a little tricky : it applies a "level by level peeling" to cover the level by the distance to the boundary. Then we do mesh. This make the decomposition always comparable by some distance. 

Analysis and PDE : Caccioppoli, Morrey, Holder and ergodic of heat equation

Since I will study my PhD in SPDE, I have to find back my once very solid capacity in analysis and PDE. Maybe, the best way is to record some nice estimation that I have met in exercises and articles.

Functional inequality Morrey

We start from some functional inequality, the most classical but powerful tools of all analysis. We state the Morrey inequality. Generally speaking, if in domain $\Omega$ and a $L^2(\Omega)$ function, all the oscillation satisfies $\forall x \in \Omega, B_r(x) \subset \Omega,$
$$\frac{1}{|B_r(x)|} \int_{B_r(x)} |f(y) - (f)_{ B_r(x) } |^2 dy \leq M^2 r^{2 \alpha}$$
then we could say that this function has a a.e modification of class $C^{0, \alpha}$ Holder. A very easy corollary is to replace the oscillation by a gradient function and Poincare. The idea of proof comes from the Lebesgue differential theorem :
$$a.e \lim_{r \rightarrow 0} (f)_{ B_r(x) } = f(x)$$
so it suffices to pass all the estimation to the function $(f)_{ B_r(x) }$ and use "three difference trick" to the theorem.

One Holder interpolation

As the Soblev injection tells us, roughly speaking, the Holder space is a little better than all the $L^{p}$ space. Inspired by the interpolation theorem, we would like to obtain the Holder interpolation. One is
$$\|f\|_{L^{\infty}(B_r)} \leq C \|f\|^{\frac{2\alpha}{d+2\alpha}}_{\underline{L}^{2}(B_r)}\left(r^{\alpha}[f]_{C^{0, \alpha}(B_r)}\right)^{\frac{d}{d+2\alpha}}$$
This tells us that if we have the function Holder + $L^2$ implies also $L^{\infty}$. This may be seen as one part of the Soblev injection, but we recall a little its proof. Idea isn't difficult but wise : we find a small ball such that the value in it is at least 1/2 maximum and then we compare its $L^2$ norm. Thanks to the regularity, the radius of ball should not be so small and we get the result.

Caccioppoli inequality

Then we come to elliptic equation :
$$- \nabla \cdot (a(x) \nabla u(x)) = h$$
The regularity is one heart question in the research and one estimation used many many times in it is the Caccioppoli inequality
$$\|\nabla u\|_{\underline{L}^{2}(B_{r/2})} \leq C \left( \frac{1}{r} \|u -  (u)_{ B_r(x) }\|_{\underline{L}^{2}(B_r)} + \|h\|_{\underline{H}^{-1}(B_r)}\right)$$
The interpretation is very natural : this bound doesn't require the regularity of the coefficients but enlarge a little the domain. In fact, it tells us the interior of the solution is more regular while the outside may be a little pike.

Functional inequality Nash

Finally, we come to study the behavior of heat equation. We know that it will decrease generally, and one useful functional inequality is Nash that $\forall f \in L^1(\mathbb{R}^d) \bigcap H^1(\mathbb{R}^d)$ we have
$$\|f\|^{1+2/d}_{L^2} \leq \|f\|^{2/d}_{L^1} \|\nabla f\|_{L^2}$$ For the solution of heat equation, we can apply easily the bound of $L^1 \rightarrow L^\infty, L^1 \rightarrow L^1$. By the decreasing of heat flow
$$\frac{d}{dt}\|u(t, \cdot)\|^2_{L^2} = - \|\nabla u(t, \cdot)\|^2_{L^2}$$
and Nash inequality we obtain
$$\|u(t, \cdot)\|_{L^2} \leq C t^{-d/4}\|u(0, \cdot)\|_{L^1}$$
The interpolation works and we obtain
$$\|u(t, \cdot)\|_{L^2} \leq C t^{-\frac{d}{2}(1-1/p)}\|u(0, \cdot)\|_{L^1}$$

Ergodic of heat equation

Finally, we would like to study the ergodic property. We suppose that the initial data $u(0, \cdot)$ is Z-periodic. In this case, we have $\forall t > 1$
$$\|u(t, \cdot) - (u)_{\Box}\|_{L^{\infty}(\Box)} \leq \exp(-ct)$$
Idea comes from the $L^2$ estimates and Holder estimates. The former is the result of decreasing of heat flow and the latter is the result of convolution. Then the two combine. We remark that when the time is small, we could not expect a better one since the average is of range $1/\sqrt{t}$ and could not regularize better.

One not so simple question of probability

The sum of i.i.d random variable is one of most classical topic in probability. However, there are always questions not so easy even in the first year course.

Question : Let $X_n$ be a series of independent random variables and $S_n = \sum_{i=1}^{n}X_i$. We would like to study the behavior of $\limsup S_n, \liminf S_n$ in the following two situations : 

(1) $$\begin{eqnarray}\mathbb{P}[X_n = n] &=& \frac{1}{n+1} \\ \mathbb{P}[X_n = -1] &=& \frac{n}{n+1}\end{eqnarray}$$  

(2) $$\begin{eqnarray}\mathbb{P}[X_n = n^2] &=& \frac{1}{n^2+1} \\ \mathbb{P}[X_n = -1] &=& \frac{n^2}{n^2+1}\end{eqnarray}$$  

In another word, we would like to study the behavior of a random walk.

Even we have some advanced tools, we may make mistakes in this questions. Here, I give two false proofs that I have thought.

False proof (1) : OK, the sum $S_n$ is a martingale, so we apply the representation of martingale and we know that we can embed this martingale to the Brownian motion by Dubins-Schwartz theorem. (It's false since it's discrete and we may choose some specific moment in the random process)

False proof (2) : We use the theorem of optional stopping theorem to the stopping time $T = T_a \wedge T_{-b}$ and study it as the ruin of gambler. (This time the errors in the proof is harder to find. In fact, since the positive jump becomes bigger and bigger, even the martingale $S_{n \wedge T}$ isn't U.I. This isn't the same case as simple random walk or continuous martingale) 

Let's give a correct proof :

The question (2) isn't so difficult. We apply Borel-Cantelli lemma that

$$\sum_{n=1}^{\infty} \mathbb{P}[X_n = n] = \sum_{n=1}^{\infty} \frac{1}{n^2+1} < \infty$$

this means that the positive only appears finite times almost surely. So, we have of course $\lim_{n \rightarrow \infty} S_n = - \infty$.

The question (1) has an easy part. We apply once again the Borel-Cantelli lemma that 

$$\sum_{n=1}^{\infty} \mathbb{P}[X_n = n] = \sum_{n=1}^{\infty} \frac{1}{n^2+1} = \infty$$

so it will happen infinite times. Moreover, we know once the positive jump occurs, it will cover all the negative jump before this step. So as the positive jump accumulates, we conclude that

$$\limsup_{n \rightarrow \infty} S_n = \infty$$

The hard part, we apply another sub-martingale $\exp(- \lambda S_{n \wedge T})$. This time, a very big positive jump doesn't matter as we know this guy is smaller than $\exp(\lambda b)$. So by the optional stopping theorem, we get

$$(1 - \mathbb{P}[T_b < T_a])e^{-\lambda a} + \mathbb{P}[T_b < T_a] \geq 1 \Rightarrow \mathbb{P}[T_b < T_a] \geq \frac{1- e^{-\lambda a}}{e^{\lambda b} - e^{- \lambda a}}$$

We take $a \rightarrow \infty$ and get $\mathbb{P}[T_b < \infty]  \geq e^{-\lambda b}$ then we take $\lambda \rightarrow 0$ and get $\mathbb{P}[T_b < \infty] = 1$.