Monthly Archives: August 2016

Easy proof of John-Nirenberg Inequality

One of fundamental theorem in the space of bounded mean oscilation is the John-Nirenberg inequality.

Theorem(John, Nirenberg). Let f\in\BMO\left(\mathbb{R}^{n}\right). Then

    \[ \frac{\left|\left\{ x\in Q:\left|f\left(x\right)-f_{Q}\right|>\lambda\right\} \right|}{\left|Q\right|}\le c_{0}e^{-\frac{c_{1}\lambda}{\norm{f}{\BMO}}} \]

for all Q where c_{0}=e, c_{1}=\frac{1}{2^{n}e}.

In particular,

    \[ \fint_{Q}\exp\left(\frac{\left|f-f_{Q}\right|}{c_{2}\norm{f}{\BMO}}\right)\le c_{2}, \]

where c_{2}=e^{2n-1}e\left(e+2\right).

There are several proofs. Stein(1994 Monograph) gives a proof by using H^1-BMO duality inequality. Here we give another proof based on dyadic decomposition given by Martell in the Lecture of MSRI.

Proof. We will use dyadic decomposition idea. By scaling, without loss of generality, we may assume \norm f_{\BMO}=1. Define

    \begin{align*} E\left(Q,\lambda\right) & =\left\{ x\in Q:\left|f\left(x\right)-f_{Q}\right|>\lambda\right\} ,\\ \varphi\left(\lambda\right) & =\sup_{Q}\frac{\left|E\left(Q,\lambda\right)\right|}{\left|Q\right|}. \end{align*}

We want to show that \varphi\left(\lambda\right)\apprle e^{-\frac{\lambda}{c}}. First take \lambda>e>1. Then

    \[ \fint_{Q}\left|f-f_{Q}\right|\le\norm{f}{\BMO}=1<\lambda \]

for any Q.

Subdivide Q dyadically and stop when \fint_{Q^{\prime}}\left|f-f_{Q}\right|>\lambda. Collect these cubes \mathcal{F}=\left\{ Q_{j}\right\} _{j}, where we have stopped (it could be \mathcal{F}=\varnothing). Note that \mathcal{F}\subset\mathbb{D}_{Q}\setminus\left\{ Q\right\}, where \mathbb{D}_{Q} denotes the family of all dyadic cubes of Q.

Now we introduce the following type of dyadic maximal function

    \[ M_{Q}^{d}g\left(x\right)=\sup_{Q^{\prime}\in\mathbb{D}_{Q},x\in Q^{\prime}}\fint_{Q}g, \]

where g=\left|f-f_{Q}\right|. Then by definition of Q_{j}, we have

    \[ \left\{ x\in Q:M_{Q}^{d}g\left(x\right)>\lambda\right\} =\bigcup_{Q_{j}\in\mathcal{F}}Q_{j}. \]

For almost every x\in E\left(Q,\lambda\right), we have

    \[ \lambda<\left|f-f_{Q}\right|=g\left(x\right)\le M_{Q}^{d}g\left(x\right). \]

So

(1)   \begin{equation*} E\left(Q,\lambda\right)\subset\bigcup_{Q_{j}\in\mathcal{F}}Q_{j}\quad\text{for almost every }x. \end{equation*}

Let \hat{Q}_{j} be a parent cube of Q_{j} Then by definition of cube

(2)   \begin{equation*} \lambda<\fint_{Q_{j}}\left|f-f_{Q}\right|\le\frac{\left|\hat{Q}_{j}\right|}{\left|Q_{j}\right|}\fint_{\hat{Q}_{j}}\left|f-f_{Q}\right|\le2^{n}\lambda. \end{equation*}

Now

    \begin{align*} \left|f\left(x\right)-f_{Q}\right| & \le\left|f\left(x\right)-f_{Q_{j}}\right|+\left|f_{Q_{j}}-f_{Q}\right|\\  & \le\left|f\left(x\right)-f_{Q_{j}}\right|+\fint_{Q_{j}}\left|f-f_{Q}\right|\\  & \le\left|f\left(x\right)-f_{Q_{j}}\right|+2^{n}\lambda. \end{align*}

For x\in E\left(Q,t\right), t>2^{n}\lambda, we have

    \[ t<\left|f\left(x\right)-f_{Q}\right|\le\left|f\left(x\right)-f_{Q_{j}}\right|+2^{n}\lambda. \]

Hence \left|f\left(x\right)-f_{Q_{j}}\right|>t-2^{n}\lambda.

Since t>\lambda, by (1) we have

    \begin{align*} \left|E\left(Q,t\right)\right| & =\left|E\left(Q,t\right)\cap E\left(Q,\lambda\right)\right|\\  & \le\sum_{j}\left|E\left(Q,t\right)\cap Q_{j}\right|\\  & \le\sum_{j}\frac{\left|\left\{ x\in Q_{j}:\left|f\left(x\right)-f_{Q_{j}}\right|>t-2^{n}\lambda\right\} \right|}{\left|Q_{j}\right|}\left|Q_{j}\right|\\  & \le\varphi\left(t-2^{n}\lambda\right)\sum_{j}\left|Q_{j}\right|. \end{align*}

Then by (2), we have

    \begin{align*} \left|E\left(Q,t\right)\right| & \le\varphi\left(t-2^{n}\lambda\right)\frac{1}{\lambda}\sum_{j}\int_{Q_{j}}\left|f-f_{Q}\right|\\  & \le\frac{1}{\lambda}\varphi\left(t-2^{n}\lambda\right)\left|Q\right|. \end{align*}

Hence for t>2^{n}\lambda, we obtain

    \[ \frac{\left|E\left(Q,t\right)\right|}{\left|Q\right|}\le\frac{1}{\lambda}\varphi\left(t-2^{n}\lambda\right). \]

By taking supremum over Q, we have

    \[ \varphi\left(t\right)\le\frac{\varphi\left(t-2^{n}\lambda\right)}{\lambda}. \]

Put \lambda=e. Note that \varphi\left(t\right)\le1 for all t>0. So for 0<t\le e\cdot2^{n}, we see that

    \[ \varphi\left(t\right)\le e\cdot e^{-\frac{t}{2^{n}e}}. \]

Note that

    \[ \left(0,\infty\right)=(0,e\cdot2^{n}]\cup\left[\bigcup_{k=1}^{\infty}\left(e\cdot2^{n+k-1},e\cdot2^{n+k}\right]\right]. \]

So for e\cdot2^{n}<t<e\cdot2^{n+1}, we obtain for t>0, \varphi\left(t\right)\le e\cdot e^{-\frac{t}{2^{n}e}}. Since we have

    \[ \varphi\left(t\right)\le\frac{1}{e}\varphi\left(t-e\cdot2^{n}\right),\quad t>e\cdot2^{n}, \]

we see that \varphi\left(t-e\cdot2^{n}\right)\le e\cdot e^{-\frac{\left(t-2^{n}e\right)}{2^{n}e}} for t>e\cdot2^{n}. Hence, for e\cdot2^{n}<t<e\cdot2^{n+1}, we have

    \[ \varphi\left(t\right)\le e\cdot e^{-\frac{t}{2^{n}e}}. \]

Iterate this procedure. So we obtain the desired claim. This leads to

    \[ \frac{\left|E\left(Q,\lambda\right)\right|}{\left|Q\right|}\le e\cdot e^{-\frac{t}{2^{n}e}}\quad\text{for every cube }Q. \]

This proves the first part.

For the second part, use

    \[ \int\left|f\right|^{p}dx=\int p\lambda^{p-1}\left|\left\{ x:\left|f\left(x\right)\right|>\lambda\right\} \right|d\lambda.\qedhere \]