Category Archives: Harmonic Analysis

Compensated compactness

Suppose T is a vector field and \nabla\cdot T = 0. E= \nabla \psi and \psi is a scalar function. We have following theorem(Coifman-Lions-Meyers-Semmes)

Theorem: If T\in L^2(\mathbb{R}^n) and T\in L^2(\mathbb{R}^n), then E\cdot T\in \mathcal{H}^1(\mathbb{R}^n), which is the hardy space.
Given f(x)\in L^1(\mathbb{R}^n), it has harmonic extension \mathbb{R}^{n+1}_+=\{(x,t)|x\in\mathbb{R}^n, t>0\}

\tilde{f}(x,t)=c_n\int_{\mathbb{R}^n}\frac{ f(x-y)t}{(t^2+|x|^2)^{\frac{n+1}{2}}}dy

Definition: the non-tangential maximal function

N(f)=\sup_{(\xi,t)\in \Gamma(x)}|\tilde f(\xi, t)|

It is easy to prove that N(f)\leq c_n f^*(x) the Hardy-Littlewood maximal function. From this we can Hardy norm as

||f||_{\mathcal{H}^1}=||f||_{L^1}+||N(f)||_{L^1}

Hardy space consists of all f having finite hardy norm. There is well know fact that the dual space of \mathcal{H}^1 is BMO, which is defined as the following.

Define f\in L^1_{loc}(\mathbb{R}^n), if for any cube Q,

\sup_Q\frac{1}{|Q|}\int_Q|f-f_Q|<\infty,\quad \text{where }f_Q=\frac{1}{|Q|}\int_Qf

then f\in BMO. L^\infty \subset BMO and \log|x|\in BMO but not in L^\infty.

Let us see how do we use the main theorem. Suppose on \mathbb{R}^2, u is the solution of the following elliptic equation

\displaystyle\frac{\partial}{\partial x_i}\left(a_{ij}(x)\frac{\partial u}{\partial x_j}\right)=\frac{\partial f}{\partial x_1}\frac{\partial g}{\partial x_2}-\frac{\partial f}{\partial x_2}\frac{\partial f}{\partial x_2}

where ||\nabla f||_{L^2}<\infty, ||\nabla g||_{L^2}<\infty and (a_{ij}) is uniform elliptic. YanYan Li and Sagun Chanillo proved that the green function of this elliptic operator belongs to BMO. The right hand side of this equation can be rewritten as T\cdot E, where

T=\left(\frac{\partial f}{\partial x_2}, -\frac{\partial f}{\partial x_1}\right),\quad E=\left(\frac{\partial g}{\partial x_1},\frac{\partial g}{\partial x_a}\right)

therefore the right hand side belong to \mathcal{H}^1. Since

u(x)=\int G_x(y)T\cdot E(y)dy

therefore from the theorem we stated at the beginning, we get

||u||_\infty\leq C||\nabla f||_{L^2}||\nabla g||_{L^2}

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A rigorous way to define the singular integral

Suppose K(x) satisfies the following conditions

(1)|K(x)|\leq \frac{C}{|x|^n}

(2)|\nabla K(x)|\leq frac{C}{|x|^{n+1}}

(3)\displaystyle \int_{r_1<|x|<r_2}K(x)dx=0

Then we can define the singular integral operator

\displaystyle Tf=\int_{\mathbb{R}^n}K(x-y)f(y)dy

On the face of it, this integrand may not be Lebesgue integrable. There is a way to interpret this integral

\textbf{Thm: } Suppose (1),(2),(3) are satisfied, define

\displaystyle T_{\epsilon,R}f(x)=\int_{\epsilon<|x-y|<R}K(x-y)f(y)dy

Then ||Tf||_2\leq C||f||_2 with C independent of \epsilon,R,f. Moreover, one can prove T_{\epsilon,R}f is a cauchy sequence, then there exist g\in L^2(\mathbb{R}^n), such that

\displaystyle T_{\epsilon,R}f\to g in L^2

From this, we define Tf=g.

Partial sum operator of fourier series and Hilbert transform

\mathbf{Problem:} Let D_N=\sum\limits_{-N}^{N}e^{inx}, the partial sum operator of fourier series can be represented by

\displaystyle S_Nf(x)=\int_{-\pi}^\pi f(t)D_N(x-t)dt=f\ast D_N

actually \displaystyle D_N=\frac{\sin(N+\frac{1}{2})x}{\sin\frac{x}{2}}

\displaystyle S_Nf(x)=\int_{-\pi}^\pi f(t)\frac{\sin(N+\frac{1}{2})(x-t)}{\sin\frac{x-t}{2}}

\displaystyle =\int_{-\pi}^\pi f(t)\frac{\cos(N+\frac{1}{2})x\sin(N+\frac{1}{2})t-\sin(N+\frac{1}{2})x\cos(N+\frac{1}{2})t}{\sin\frac{x-t}{2}}

\displaystyle =P.V.\cos(N+\frac{1}{2})x\int_{-\pi}^\pi\frac{f(t)\sin(N+\frac{1}{2})t}{\sin\frac{x-t}{2}}+P.V.\sin(N+\frac{1}{2})x\int_{-\pi}^\pi\frac{f(t)\cos(N+\frac{1}{2})t}{\sin\frac{x-t}{2}}

\displaystyle =P.V.\frac{1}{\sin\frac{t}{2}}\ast f_N^1+P.V.\frac{1}{\sin\frac{t}{2}}\ast f_N^2

Since \displaystyle |\frac{1}{\sin\frac{t}{2}}-\frac{1}{\frac{t}{2}}|\leq ct, for t\in (-\pi,\pi), then S_Nf behavior much like Hilbert transform

\displaystyle Hf(x)=P.V.\int_{-\pi}^\pi\frac{f(x)}{x-t}

\mathbf{Thm:}

(a) ||Hf||_p\leq c_p||f||_p, 1<p<\infty

(b) \{x\in(-\pi,\pi)||Hf(x)|>\lambda\}\leq \frac{C||f||_1}{\lambda}

By the fact in this post, S_Nf\to f in L^p(-\pi,\pi)

\text{Q.E.D}\hfill \square

\mathbf{Remark:}

Partial sum operator of fourier series and L^p boundedness

Suppose f\in L^1(-\pi,\pi) , consider the fourier series of f

\displaystyle f\sim \sum \limits_{n\in\mathbb{Z}}c_ne^{inx}

 and its partial sum \displaystyle S_N(f)=\sum \limits_{-N}^Nc_ne^{inx}

\mathbf{Problem:} The following statements are equivalent

(a)   ||S_Nf||_p\leq c_p||f||_p, 1<p<\infty

(b)  ||S_Nf-f||_p\to \infty as N\to \infty

(c)   ||S_Nf||_{p'}\leq ||f||_{p'}, \frac{1}{p}+\frac{1}{p'}=1

\mathbf{Proof:}

(a)\Rightarrow(b). Since trigonometric polynomial are dense in L^p, then we can choose g to be a trigonometric polynomial are sufficiently close to f in L^p.

\displaystyle ||S_Nf-f||_p\leq ||S_Nf-S_Ng||_p+||S_Ng-g||_p+||g-f||_p

\displaystyle = ||S_N(f-g)||_p+||S_Ng-g||_p+||g-f||_p

\displaystyle \leq c_p||f-g||_p+||S_Ng-g||_p+||g-f||_p\to \infty

(b)\Rightarrow(a). Uniform boundedness principle.

(a)\Rightarrow(c) and (c)\Rightarrow(a) are quite easy.

\text{Q.E.D}\hfill \square

\mathbf{Remark:}

Poincare inequality from different approach

\mathbf{Problem:} Let \Omega be convex and u\in C^1(\Omega). Suppose S\subset\Omega with \displaystyle \frac{|S|}{|\Omega|}>0,

\displaystyle |u(x)-u_S|\leq \frac{d^n}{n|S|}\int_{\Omega}\frac{|\nabla u(y)|}{|x-y|^{n-1}}dy

where d is the diameter of \Omega and \displaystyle u_S=\frac{1}{|S|}\int_S u(y)dy.

\mathbf{Proof\,1: }

\displaystyle u(y)-u(x)=\int_0^1\frac{d}{dt}\left(u(ty+(1-t)x)\right)dt=\int_0^1\nabla u(ty+(1-t)x)\cdot(x-y)dt

\displaystyle |u(x)-u_S|\leq\frac{1}{|S|}\int_S|u(x)-u(y)|dy\leq \frac{1}{|S|}\int_S\int_0^1|\nabla u(ty+(1-t)x)||x-y|dtdy

Set z=ty+(1-t)x, then t|x-y|=|z-x|. Since |x-y|\leq d, \displaystyle t\geq \frac{|z-x|}{d}. We also have t^ndy=dz

Then

\displaystyle |u(x)-u_S|\leq \frac{d}{|S|}\int_\Omega\int_\frac{|z-x|}{d}^\infty \frac{|\nabla u(z)|}{t^n}\,dtdz=\frac{d^n}{(n-1)|S|}\int_\Omega\frac{|\nabla u(z)|}{|z-x|^{n-1}}dz.

\mathbf{Proof\,2: } Using

\displaystyle u(x)-u(y)=\int_{0}^{|x-y|}\frac{d}{dt}u(x+tw)dt=\int_0^{|x-y|}\nabla u(x+tw)\cdot wdt

where \displaystyle w=\frac{x-y}{|x-y|}. Then

\displaystyle |u(x)-u_S|\leq\frac{1}{|S|}\int_S |u(x)-u(y)|dy\leq \frac{1}{|S|}\int_S\int_0^{|x-y|}|\nabla u(x+tw)|dtdy

\displaystyle \leq \frac{1}{|S|}\int_\Omega\int_0^{\infty}|\nabla u(x+tw)|\chi_{\{t<|x-y|\}}dtdy

\displaystyle \leq \frac{1}{|S|} \int_0^\infty\int_{|w|=1}\int_0^d|\nabla u(x+tw)|\chi_{\{t<|x-rw|\}}r^{n-1}dwdrdt

\displaystyle \leq \frac{d^n}{n|S|}\int_0^\infty\int_{|w|=1}|\nabla u(x+tw)|\chi_{\{t<|x-\xi(w)|\}}dwdt

\displaystyle =\frac{d^n}{n|S|}\int_\Omega \frac{|\nabla u(z)|}{|z-x|^{n-1}}dz

where \xi(w) is the intersection of ray x+tw with \partial \Omega (it is unique because \Omega is convex).

\text{Q.E.D}\hfill \square

\mathbf{Remark:}

Weak (1,1) boundedness of Hardy Littlewood maximal function

f is a locally integrable function in \mathbb{R}^d, define the maximal function as

\displaystyle M(f)=\sup\limits_{x\in B}\frac{1}{|B|}\int_B|f(y)|dy

here B is an  arbitrary ball.

\textbf{Thm: } Maximal operator is weakly (1,1) bounded. That is \exists C=C(d)>0 such that for \forall\lambda>0,

\displaystyle |\{x:M(f)(x)>\lambda\}|\leq\frac{C||f||_1}{\lambda}

\textbf{Proof: } Let E_\lambda=\{x:M(f)(x)>\lambda\}, K\subset E_\lambda is a compact set.

Then K\subset\bigcup\limits_{i=1}^N B_i, where B_i=B_i(x_i) with \displaystyle \frac{1}{|B_i|}\int_{B_i}|f(y)|dy>\lambda. Apply the Vitali covering lemma

\textbf{Vitali covering lemma: } Given \{B_1,\cdots,B_n\} a finite collection of balls in \mathbb{R}^n, then there exists a subcollection B_{ij}, j=1,\cdots,s disjoint from each other and

\bigcup\limits_{i=1}^N B_i\subset \bigcup\limits_{j=1}^s 3B_{i_j}

We can conclude that  K\subset \bigcup\limits_{j=1}^s 3B_{i_j}. So

\displaystyle |K|\leq \sum\limits_{j=1}^s 3^d|B_{i_j}|\leq \frac{3^d}{\lambda}\sum\limits_{j=1}^s\int_{B_{i_j}}|f(y)|dy\leq \frac{3^d||f||_1}{\lambda}

Since K is arbitrary, we know |E_\lambda| is bounded by the right hand side.\hfill\square

Actually the decay of maximal function is a little faster than $\frac{1}{\lambda}$

\textbf{Thm: } There exists a constant C=C(d)>0 such that

\displaystyle |\{x:M(f)(x)>\lambda\}|\leq \frac{C}{\lambda}\int_{|f|>\frac{\lambda}{2}}|f(y)|dy

\textbf{Proof: } Define

\displaystyle g=\begin{cases}\frac{\lambda}{2},\quad |f(x)|>\frac{\lambda}{2}\\ 0, \quad |f(x)|\leq\frac{\lambda}{2}\end{cases}

Then |f(x)|\leq |g(x)|+\frac{\lambda}{2}. This means

\displaystyle \{x:M(f)(x)>\lambda\}\subset \{x:M(g)(x)>\frac{\lambda}{2}\}

So \displaystyle |\{x:M(f)(x)>\lambda\}|\leq |\{x:M(g)(x)>\frac{\lambda}{2}\}|\leq \frac{C}{\lambda}\int_{|f|>\frac{\lambda}{2}}|f(y)|dy

\hfill\square

Boundedness of fractional integration operator

Suppose f is locally integrable, define the fractional integration operator

\displaystyle I_\alpha(f)(x)=\int_{\mathbb{R}^n}\frac{f(y)}{|x-y|^{n-\alpha}}dy, 0<\alpha<n

\mathbf{Thm:} Suppose f\in L^p(\mathbb{R}^n) and 0<\alpha<n, then

\displaystyle ||I_\alpha(f)||_q\leq C(p,n,\alpha)||f||_p

where \displaystyle \frac{1}{q}=\frac{1}{p}-\frac{\alpha}{n} and p>1. If p=1, then I_\alpha(f) is a weak type (1,\frac{n}{n-\alpha}).

\bf{Lemma:} Suppose \psi:\mathbb{R}^n\to\mathbb{R}^n is radially decreasing function, then

\displaystyle \left|\int_{\mathbb{R}^n}f(x)\psi(x-y)dy\right|\leq M(f)(x)||\psi||_1

when ||\psi||_1<\infty.

\textbf{Proof of Theorem:} Write

\displaystyle I_\alpha(f)(x)=\int_{\mathbb{R}^n}\frac{f(x-y)}{|y|^{n-\alpha}}dy=\int_{|y|\leq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy+\int_{|y|\geq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy

By the lemma,

\displaystyle \int_{|y|\leq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy\leq M(f)(x)\int_{|y|\leq A}\frac{1}{|y|^{n-\alpha}}dy\leq \frac{nw_n}{\alpha}A^\alpha M(f)(x)

By holder inequality, if p>1

 \displaystyle \int_{|y|\geq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy\leq ||f||_p\left(\int_{|y|\geq A}\frac{1}{|y|^{(n-\alpha)p'}}\right)^\frac{1}{p'}\leq \left(\frac{qw_n}{p'}\right)^\frac{1}{p'}A^{-\frac{n}{q}}||f||_p

If p=1

\displaystyle \int_{|y|\geq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy\leq A^{-\frac{n}{q}}||f||_1 with \displaystyle q=\frac{n}{n-\alpha}.

Combing the preceeding result,

\displaystyle |I_\alpha(f)(x)|\leq C(n,p,\alpha)(A^{-\frac{n}{q}}||f||_p+A^\alpha M(f)(x)) for \forall\,A>0

Choose A=Mf(x)^{-\frac{p}{n}}||f||_p^{\frac{p}{n}},

\displaystyle |I_\alpha(f)(x)|\leq C(n,\alpha,p)M(f)(x)^{\frac{p}{q}}||f||_p^{1-\frac{p}{q}}

When p>1,  maximal function is bounded from L^p to L^p,

\displaystyle ||I_\alpha(f)||_q\leq C(n,\alpha,p)||f||_p^\frac{p}{q}||f||_p^{1-\frac{p}{q}}=C(n,\alpha,p)||f||_p

When p=1, maximal function is weak (1,1),

\displaystyle |\{x||I_\alpha(f)(x)|>\lambda|\}|\leq \{x|C(n,\alpha)M(f)(x)^{\frac{n-\alpha}{n}}||f||_1^{\frac{\alpha}{n}}>\lambda\}

\displaystyle =|\{x|M(f)(x)>\left(\frac{\lambda}{C(n,\alpha)||f||_1^{\frac{\alpha}{n}}}\right)^{\frac{n}{n-\alpha}}\}|

\displaystyle \leq 3^n||f||_1\left(\frac {C(n,\alpha)||f||_1^{\frac{\alpha}{n}}}{\lambda} \right)^\frac{n}{n-\alpha}=C'(n,\alpha)\left(\frac{||f||_1}{\lambda}\right)^\frac{n}{n-\alpha}

  \square

When the domain is not \mathbb{R}^n but a bounded one, we can obtain the boundedness of \displaystyle \frac{1}{q}<\frac{1}{p}-\frac{\alpha}{n}.

\textbf{Thm: } Suppose f\in L^p(\Omega), p>1. We have I_\alpha(f) is bounded from L^p to L^q  with

\displaystyle 0\leq \delta=\frac{1}{p}-\frac{1}{q}<\frac{\alpha}{n}, 1\leq q\leq \infty

\displaystyle ||I_\alpha f||_q\leq Cw_n^{1-\alpha/n}|\Omega|^{\alpha/n-\delta}||f||_p

\textbf{Remark:} Loukas Grafakos: Modern fourier analysis.

Mixed norm and basic property

In the analysis of evolution equation, we will constantly treat this kind of function f(x,t) where x\in \mathbb{R}^d, t\in \mathbb{R}. Considering the difference of time and space variable, we will introduce this kind of norm

\displaystyle ||f||_{L^q_t L^r_x}=\left(\int_\mathbb{R}\left(\int_{\mathbb{R}^d}|f|^rdx\right)^{\frac{q}{r}}dt\right)^{\frac 1q}  for q,r\geq 1

This kind of norm also has similar properties of L^p norm.

\mathbf{Proposition 1:} \displaystyle \iint |fg|dxdt \leq ||f||_{L^q_t L^r_x}||g||_{L^{q'}_t L^{r'}_x} here q and q' are conjugate  and so are r,r'.

\mathbf{Proof:} \displaystyle \iint |fg|dxdt =\int\left(\int |fg|dx\right)dt\leq \int\left(||f||_{L^r_x}||g||_{L^{r'}_x}\right)dt\leq||f||_{L^q_t L^r_x}||g||_{L^{q'}_t L^{r'}_x}

\mathbf{Corollary 1:} \displaystyle ||f||_{L^q_t L^r_x}\leq ||f||_{L^{q_1}_t L^{r_1}_x}||f||_{L^{q_2}_t L^{r_2}_x}, where \displaystyle \frac 1q=\frac {1}{q_1}+\frac{1}{q_2} and \displaystyle \frac 1r=\frac{1}{r_1}+\frac{1}{r_2}. All the powers are bigger than or equal 1.

\mathbf{Proposition 2:} \displaystyle ||f||_{L^q_t L^r_x}=\sup_{||g||_{L^{q'}_t L^{r'}_x}}\iint fgdxdt

\mathbf{Proof:} From the proposition 1, LHS\leq ||f||_{L^q_t L^r_x}.

Define \displaystyle g=sgn(f)|f|^{r-1}\left(\int |f|^r\right)^{\frac{q}{r}-1}||f||^{\frac{1}{q}-1}_{L^q_t L^r_x} when ||f||_{L^q_t L^r_x}\neq 0, otherwise the proposition holds trivially.
Then ||g||_{L^{q'}_t L^{r'}_x}=1 and \displaystyle \iint fgdxdt=||f||_{L^q_t L^r_x}.

\text{Q.E.D}\hfill \square

\mathbf{Remark:}

Various proofs of Hilbert transform: Hardy’s inequality, Minkowski integral inequality

\mathbf{Problem:} For x,y>0, define

\displaystyle Tf(x)=\int_0^\infty \frac{f(y)}{x+y}dy

prove that T:L^p\to L^p is a bounded operator.

Proof by Hardy’s inequality

Hardy’s inequality: Let 1\leq p<\infty and f\geq 0

1. \forall\, \alpha<-1 

 \displaystyle \int_0^\infty\left(\int_0^x f(t)dt\right)^px^\alpha dx\leq \left(\frac{p}{1+|\alpha|}\right)^p\int_0^\infty f^p(t)t^{\alpha+p}dt

2. \forall\, \alpha>-1

\displaystyle \int_0^\infty\left(\int_x^\infty f(t)dt\right)^px^\alpha dx\leq \left(\frac{p}{1+\alpha}\right)^p\int_0^\infty f^p(t)t^{\alpha+p}dt

\mathbf{Proof:} Split Tf(x) as \displaystyle Tf(x)=\int_0^x \frac{f(y)}{x+y}dy+\int_{x}^\infty\frac{f(y)}{x+y}dy. Then

\displaystyle \left\|\int_0^x \frac{f(y)}{x+y}dy\right\|_p^p=\int_0^\infty \left|\int_0^x \frac{f(y)}{x+y}dy\right|^pdx\leq \int_0^\infty \left(\frac{1}{x}\int_0^x |f(y)|dy\right)^pdx\leq C_p\int_0^\infty|f(y)|^pdy

Here we used Hardy’s inequality case (1) with \alpha=-p<-1.

\displaystyle \left\|\int_x^\infty \frac{f(y)}{x+y}dy\right\|_p^p=\int_0^\infty \left|\int_x^\infty \frac{f(y)}{x+y}dy\right|^pdx\leq \int_0^\infty \left(\int_x^\infty\frac{|f(y)|}{y}dy\right)^pdx\leq C_p\int_0^\infty|f(y)|^pdy

Here we used Hardy’s inequality case (2) with \alpha=0.

So \displaystyle \|Tf\|_p\leq \left\|\int_0^x\frac{f(y)}{x+y}dy\right\|_p+\left\|\int_x^\infty\frac{f(y)}{x+y}dy\right\|_p\leq C\|f\|_p.

\text{Q.E.D}\hfill \square

Proof by Minkowski integral inequality

\mathbf{Proof:} Changing variable by y=tx

\displaystyle \int_0^\infty \frac{f(y)}{x+y}dy=\int_0^\infty\frac{f(tx)}{1+t}dt

So by Minkowski Integral Inequality

\displaystyle \|Tf\|_p=\left( \int_0^\infty \left(\int_0^\infty \frac{f(tx)}{1+t}dt\right)^pdx\right)^\frac{1}{p}\leq \int_0^\infty\left(\int_0^\infty \left|\frac{f(tx)}{1+t}\right|^pdx\right)^\frac{1}{p}dt\\=\int_0^\infty \frac{1}{(1+t)t^\frac{1}{p}}\left(\int_0^\infty|f(y)|^p\right)^\frac{1}{p}dt=\|f\|_p\int_0^\infty \frac{1}{(1+t)t^\frac{1}{p}}dt

Since p>1, \displaystyle \int_0^\infty \frac{1}{(1+t)t^\frac{1}{p}}dt<\infty. We are done.

\text{Q.E.D}\hfill \square

\mathbf{Remark:}

Pay attention to the tricks using here. The proof by Hardy’s inequality is hinted from Prof. Chanillo’s notes. For Minkowski method, see stein’s book.