Saturday, 14 March 2015 23:35

Carleson measure

Written by
Rate this item
(0 votes)

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.

Read 1883 times Last modified on Saturday, 28 March 2015 16:47