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 $\mathbb{P}[T_b < \infty] = 1$.

又到了夏天来临的日子

每年一调整夏令时，巴黎的白天就显得特别长，气温也非常配合地直接进入夏天的节奏，从四月开始，当真是一年中我最喜欢的季节了。

时间好快，现在是2018年了。

四年前的那个剧本如约走到了结局，但其中的变化好像也不曾是当时所想。之前只是想，还是要做个论文，后来觉得概率似乎比分析有意思，再后来迷上了随机几何，心心念念想做这个方向，也听了很多报告学了很多知识还去了一些会。再后来M2开学，发现之前两三年所学都好似花架子没一点真材实料的，心里很慌。

后来小半年时间每天上课记笔记，回来就复习琢磨证明，到现在基本上拿个思路都能够差不多补齐了。最后M2考试成绩也不错。

可是这个时候，老师和我说觉得有些方向人太多了，还是建议换个方向吧。

开始我是有点失落的，但可能就是在这样的过程中吧，对随机是什么，思考越来越多了，发现概率也不只是一个方向，也有很多很多不同的分支，应该都看看的。想到这里，似乎思路就打开了，也就不拘泥说非得做某一个课题了。

然后就到了现在的课题，似乎是之前所有所学的总和呢，也算无心插柳吧。

就是得把分析和方程都捡回来了，不怕，当年我就是从实变泛函助教做起的。

回到开头那个话题，为什么喜欢夏天呢？因为白天很长，有很多时间可以做自己喜欢的事情。恍惚又想到了2015的那个夏天，那个被压抑了特别久，特别想学数学的状态，每天在PC算题目，算到天黑11点，有一次甚至是两三点，然后醒了就继续算。把课后习题一道一道算过去……

写这段的时候，不为了别的什么，就是想让自己回想起那个勇敢的自己。有句歌词说“如果知道这些当时我到底去不去？”

当然去了，这几年挺开心的，也看着理想在实现，自己也在不断强大起来，我还想看看自己还能进化成什么样子呢，拭目以待。

夏天来了，抓紧时间再疯狂一次吧

（立个Flag：每周读至少一个和主线不是那么相关的证明，更新一个博客，还是要保持一点学习的劲头不能变成只会一个方向的傻瓜的。毕竟心里还有一个大问题啊！）

TCL theorem for one type Riemann integral of Brownian motion

This is one question in the exercise of "Local time and excursion", but I think it is very interesting.

We consider a measurable function $g$ and a Brownian motion $(B_t)_{t \geq 0}$ one integral defined as  

$$A_t = \int_0^t g(B_s)ds$$

means the integration along the path. We suppose that $g$ is intégrable then this formula makes sense. Well, if $g$ is continuous this is obvious : although there is random part, it's in fact a Riemann integral (or Lebesgue) of continuous function. In general case, we apply a very useful formule called time of occupation

$$\int_0^t g(B_s)ds = \int g(a) L^a_t(B)da$$

Then, since $\left(L^a_t(B)\right)_{a \geq 0}$ is continuous and zero at infinity, the one has a max so $A_t$ is well defined.
One more remark for this formule : One large advantage of Lebesgue integral is the introduction of measure, so when we compare two integral, we have not to compare it point-wisely, but cut them into blocks. However, the integral like Riemann isn't good, but local time transform it again with the style of Lebesgue one. The stochastic integral face the same problem, luckily we have Ito, Doob, BDG so we can make that one like "one deterministic term + one random error".

Our main theorem is to prove that

$$\frac{1}{\sqrt{t}} A_t \Rightarrow \int g(a)da |N|$$
where $N \sim \mathcal{N}(0,1)$. This convergence is in weak sense.

We remark why this formula should be correct. One important observation is one Levy's theorem that
$$\left( L^0_t, |B_t| \right)_{t \geq 0} = (\text{law})  \left( S_t, S_t - B_t \right)_{t \geq  0}$$
So we have obviously $\frac{1}{\sqrt{t}}L^0_t$ has the same law as $|N|$. For the local time at other level, once it is touched, it will behavior like $L^0_t$.

However, the problem is that : the convergence in law of sigle random variable doesn't mean the convergence in law of random process. I make the this phrase red to point out the danger. But we know that $a \rightarrow L^a_t(B)$ is also continuous, so what's the error term ? Could this error disappear after the normalization of $\frac{1}{\sqrt{t}}$ ?

We have to go back to the analysis of the regularity of the local time. Using the Tanaka formule
$$L^a_t(B) = 2(B_t - a)^ +  - 2(B_0 - a)^ + - 2\int_{0}^t \mathbb{1}_{\{B_s > a\}} dB_s$$
We  obtain that
$$L^a_t(B) - L^0_t(B) = \left[2(B_t - a)^ +  - 2(B_0 - a)^ + \right] - \left[2(B_t )^ +  - 2(B_0 )^ +\right] + 2\int_{0}^t \mathbb{1}_{\{0 < B_s \leq a\}} dB_s$$
We would like to say that $L^a_t(B) - L^0_t(B)$ is uniformly little. The difficulty is the last one stochastic integration. However, we see that when $s$ grows, it's very rare that the Brownian motion could stay in the interval so it contributes very little to the integral (even in a stochastic one !). The powerful tool like BDG inequality tells totally the moment of this random variable. We estimate the tail so with large probability $1 - \epsilon$, the process will converge uniform to a $|N|$ process. We write down the proof properly by the density argument etc and conclude the proof.

Finally, I have to say once I come to the part of analysis the size of a random variable, the training in the course of statistic helps really a lot. That may be why we say probability and statistics are always together. (Oui, et analyse est aussi son bon amie)

Est-ce que le processus Gaussien peut être différentiable ?

Comme le titre, c'est une très bonne question. Hier mon ami me demande cette question et au premier coup, j'avais envie de dire que c'est pas possible après plusieurs raisons.

1. Le modèle très simple est le mouvement brownien, qui n'est pas de classe $C^1$.

2. Un théorème de Dubins-Schwarts nous dit que une martingale locale continuée est presque un mouvement brownien après un changement du temps.

3. Oui, on parle de dérivée bruit blanc, mais c'est toujours de sens faible ou il veut dire intégration stochastique.

Si on dit la "dérivée" dans l'autre sens non standard, bon, c'est toujours possible. Mais finalement, on trouve un exo qui nous dit c'est possible lorsque la covariance $K(s,t)$ est de classe $C^2$ et en fait on a fait cet exo avant.

Argument est comme suivant :
1. $\left\{\frac{G_{t+ \delta - G_t}}{\delta}\right\}_{t \geq 0}$ est un processus Gaussien après la linéarité d'espace Gaussien.
2. On vérifie il est de suite de Cauchy dans $L^2$. Donc il admet une limite comme Gaussien.
3. La limite a une covariance  $\partial_s \partial_t K(s,t)$.

Donc, on voit dans le sens de limite $L^2$ il admet la limite. C'est tout de propriété de Gaussien, qui est fermé dans le sens limite.

Mais pourquoi on a tendance de mélanger le cas avec la martingale. Voilà, une martingale est un processus ssi il est de croissance indépendante. Mais ici, le processus a vraiment de mémoire et le cas PAIS ne peut pas avoir dérivée. Donc, même si le processus aléa est souvent fractal, il peut être régulier.