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Carleson measure

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This note is taken in the PDE discussion group in 2012, on the topic of important spaces in fluid dynamics.

Lecture 2: Carleson measure

• Non-tangential maximal functions
• Unit cone at point $$x$$: $$\Gamma(x):=\{(y,t)\in\mathbb{R}^n\times\mathbb{R}^+~:~|y-x|<t\}$$.
• Non-tangential MF: $$\displaystyle F^*(x)=\sup_{(y,t)\in\Gamma(x)}|F(y,t)|$$.
• Carleson measure
• Tent mapping on balls: $$T:\text{Ball}(\mathbb{R}^n)\to\text{Cylinder}(\mathbb{R}^n\times\mathbb{R}^+)$$,
$T(B(x,r))=\{(y,t)~:~y\in B(x,r), t\in[0,2r]\}.$
• Tent mapping on cubes: $$T:\text{Cube}(\mathbb{R}^n)\to\text{Cube}(\mathbb{R}^n\times\mathbb{R}^+)$$, $$Q(x,l)$$ is a cube centered at $$x$$, side length $$l$$.
$T(Q(x,l))=\{(y,t)~:~y\in B(x,r), t\in[0,l]\}.$

Definition (Carleson measure): A positive measure $$\mu$$ is called a Carleson measure if

$\|\mu\|_C:=\sup_Q\frac{1}{|Q|}\mu(T(Q))<\infty.$

• $$\displaystyle C(\mu)(x):=\sup_{\{Q:x\in Q\}}\frac{1}{|Q|}\mu(T(Q))$$. Note that $$\|\mu\|_C=\sup_x C(\mu)(x)$$.
• Whitney decomposition

Let $$\Omega$$ be an open, nonempty, proper subset of $$\mathbb{R}^n$$. There exists a family of closed cubes $$\{Q_j\}_j$$ such that

1. $$\cup_j Q_j=\Omega$$, $$Q_j$$'s have disjoint interiors.
2. $$\sqrt{n}~l(Q_j)\leq dist(Q_j,\Omega^c)\leq4\sqrt{n}~l(Q_j)$$.
3. If the boundaries of $$Q_j$$ and $$Q_k$$ are touched, then $$\displaystyle\frac{1}{4}\leq\frac{l(Q_j)}{l(Q_k)}\leq4$$.
4. Given $$Q_j$$, there exists at most $$12^n$$ $$Q^k$$'s that touch it.
• Carleson measure in $$\mathbb{R}^{n+1}$$ can be controlled by $$C(\mu)$$ in $$\mathbb{R}^n$$.

Theorem 1: There exists a constant $$C_n$$ such that for any $$\alpha>0$$, any positive measure $$\mu$$ on $$\mathbb{R}_+^{n+1}$$ and any $$\mu$$-measurable function $$F$$ on $$\mathbb{R}_+^{n+1}$$, we have

$\mu\left(\{(x,t)\in\mathbb{R}_+^{n+1}:|F(x,t)|\geq\alpha\}\right)\leq C_n\int_{\{F^*\geq\alpha\}}C(\mu)(x)dx.$

Proof: Let $$\Omega=\{F^*>\alpha\}$$ be an open set in $$\mathbb{R}^n$$.

Step 1: Prove $$\{|F|>\alpha\}\subset\cup_{z\in\Omega}T(B(z,\delta(z))$$, where $$\delta(z)=dist(z,\Omega^c)$$.

Suppose $$(x,t)\in\{|F|>\alpha\}$$. As $$(x,t)$$ lies inside the cone of any point inside $$B(x,t)$$, we have $$B(x,t)\subset\Omega$$. Then, $$t\leq\delta(z)$$, which implies $$(x,t)\in T(B(x,\delta(x))$$.

Step 2: Prove $$B(z,\delta(z))\subset 12B_k$$, where $$B_k$$ is the smallest ball containing $$Q_k$$, and $$Q_k$$ is one of the cube from Whitney decomposition which contains $$z$$.

$$\delta(z)\leq\sqrt{n}~l(Q_k)+dist(Q_k,\Omega^c)\leq5\sqrt{n}~l(Q_k)$$.

For any point $$y\in B(z,\delta(z))$$, the distance between $$y$$ and the center of $$Q_k$$ is bounded by $$\delta(z)+\sqrt{n}~l(Q_k)\leq6\sqrt{n}~l(Q_k)$$. As the radius of $$B_k$$ is $$(\sqrt{n}/2)l(Q_k)$$, $$y\in 12B_k$$.

Step 3: From the first two steps, we get: LHS$$\leq\sum_k\mu(T(12B_k))\leq\sum_k\mu(T(12\sqrt{n}Q_k))$$.

By definition of $$C(\mu)$$, $$\mu(T(12\sqrt{n}Q_k))\leq |12\sqrt{n}Q_k|C(\mu)(x)$$, for all $$x\in Q_k$$. Finally,

$\text{LHS}\leq\sum_k|12\sqrt{n}Q_k|\inf_{x\in Q_k}C(\mu)(x)\leq\sum_k\frac{|12\sqrt{n}Q_k|}{|Q_k|}\int_{Q_k}C(\mu)(x)dx=(12\sqrt{n})^n\int_\Omega C(\mu)(x)dx.$

Corollary 1 ($$L^p$$ Control): $$\displaystyle \int_{\mathbb{R}_+^{n+1}}|F(x,t)|^pd\mu(x,t)\leq C_n\|\mu\|_C\int_{\mathbb{R}^n}(F^*(x))^pdx$$.

Proof:

\begin{align*}LHS=&\int_0^\infty\alpha^p\mu(|F(x,t)|\geq\alpha|) d\alpha\leq C_n\int_0^\infty\alpha^p\int_{F^*\geq\alpha}C(\mu)(x)dxd\alpha\\ \leq&C_n\|\mu\|_C\int_0^\infty\alpha^p|\{F^*\geq\alpha\}|d\alpha=RHS.\end{align*}

•  Carleson measure generated by a BMO function

Theorem 2: Let $$b$$ be a BMO function on $$\mathbb{R}^n$$ and let $$\psi$$ be an integrable function with mean value zero on $$\mathbb{R}^n$$ that satisfies

$|\psi(x)|\leq A(1+|x|)^{-n-1},\qquad\sup_{\xi\in\mathbb{R}^n}\sum_j|\hat{\psi}(2^{-j}\xi)|^2\leq A^2<\infty.$

Then, there exists a constant $$C_n$$ such that

$d\mu(x,t)=\sum_{j\in\mathbb{Z}}|\Delta_j(b)|^2dx\cdot\delta_{2^{-j}}(t)$

is a Carleson measure on $$\mathbb{R}_+^{n+1}$$ with norm at most $$C_n A^2\|b\|_{BMO}^2$$. Here, $$\Delta_j$$ is the Littlewood-Paley operator $$\Delta_j(b)=b\star\psi_{2^{-j}}$$, and $$\delta$$ is the Dirac mass.

Proof: We are going to show that, $$\mu(T(Q))\leq|Q|C_nA^2\|b\|_{BMO}^2$$, for all cubes $$Q\subset\mathbb{R}^n$$.

By definition $$\mu(T(Q))=\sum_{2^{-j}\leq l(Q)}\int_Q|\Delta_j(b)(x)|^2dx$$.

Let $$Q^*$$ be a larger cube with the same center as $$Q$$ and side length $$3\sqrt{n}~l(Q)$$. Decompose $$b$$ into 3 parts.

$b=(b-b_Q)\cdot\textbf{1}_{Q^*}+(b-b_Q)\cdot\textbf{1}_{(Q^*)^c}+b_Q.$

Then, we have

$|\Delta_j(b)|^2\leq3\big[|\Delta_j((b-b_Q)\cdot\textbf{1}_{Q^*})|^2+\Delta_j((b-b_Q)\cdot\textbf{1}_{(Q^*)^c})|^2+\Delta_j(b_Q)|^2\big].$

The third part $$\Delta_j(b_Q)=0$$ as $$\psi$$ has mean zero.

Now, we estimate the first term $$\Sigma_1=\sum_{j\in\mathbb{Z}}\int_{\mathbb{R}^n}|\Delta_j((b-b_Q)\cdot\textbf{1}_{Q^*})|^2dx$$.

\begin{align*}\Sigma_1\leq\sum_{j\in\mathbb{Z}}\int_{\mathbb{R}^n}|\hat{\psi}(2^{-j}\xi)|^2|\widehat{(b-b_Q)\cdot\textbf{1}_{Q^*}}|^2d\xi\leq A^2\int_{Q^*}|b-b_Q|^2dx.\end{align*}

Here, $$\int_{Q^*}|b-b_Q|^2\leq\int_{Q^*}|b-b_{Q^*}|^2+|Q^*||b_{Q}-b_{Q^*}|^2\lesssim_n |Q^*||b|_{BMO}^2 \lesssim_n |Q|\|b\|_{BMO}^2$$.

We are left with the estimate of $$\Sigma_2=\sum_{2^{-j}\leq l(Q)}\int_Q|\Delta_j((b-b_Q)\cdot\textbf{1}_{(Q^*)^c})|^2dx$$. As $$|\psi(x)|\leq A(1+|x|)^{-n-1}$$, we get $$|\psi_{2^{-j}}(x)|\leq2^{jn}A(1+|2^jx|)^{n+1}$$.

\begin{align*}\Sigma_2\leq&\sum_{2^{-j}\leq l(Q)}\int_Q\left|\int_{(Q^*)^c}\frac{A2^{jn}|b(y)-b_Q|}{(1+|2^j(x-y)|)^{n+1}}dy\right|^2dx\\ \leq&\sum_{2^{-j}\leq l(Q)}2^{-2j}\int_Q\left|\int_{(Q^*)^c}\frac{A|b(y)-b_Q|}{(2^{-j}+|(x-y)|)^{n+1}}dy\right|^2dx\\ \leq& \frac{4}{3}l(Q)^2\int_Q\left|\int_{(Q^*)^c}\frac{A|b(y)-b_Q|}{(l(Q)+|(x-y)|)^{n+1}}dy\right|^2dx.\end{align*}

To get a bound of the inside integral, we decompose $$(Q^*)^c$$ to ring in order to get a better bound for the denominator. $$\{2Q^*\backslash Q^*,\cdots,2^kQ^*\backslash 2^{k-1}Q^*,\cdots\}$$.

$\int_{2^kQ^*\backslash2^{k-1}Q^*}\frac{A|b(y)-b_Q|}{(l(Q)+|(x-y)|)^{n+1}}dy\leq\left(\frac{l(Q)}{2}+2^{k-1}l(Q^*)\right)^{-n-1}\int_{2^kQ^*}|b(y)-b(Q)|dy.$

Here, $$\int_{2^kQ^*}|b-b_Q|\leq A\int_{2^kQ^*}|b-b_{2^kQ^*}|+|2^kQ^*||b_{Q}-b_{2^kQ^*}|\lesssim_n k2^{kn}|Q|\|b\|_{BMO}$$. Therefore, we have

$\int\star~dy\lesssim_n A2^{-(k-1)(n+1)}l(Q)^{-n-1}\cdot k2^{kn}|Q|\|b\|_{BMO}=Ak2^{n+1-k}l(Q)^{-1}\|b\|_{BMO}.$

Sum on $$k$$ yields a finite number. Plug into $$\Sigma_2$$, the $$l(Q)$$ term is cancelled. It gives the desired estimate.