Posted by: Shuanglin Shao | May 1, 2009

## The Cotlar-Stein lemma

The Coltar-Stein lemma is a powerful tool to deal with $L^2$ boundedness of some translation-invariant operators such as convolution operators, which can be expressed as

$Tf(x)=\int K(x-y) f(y)dy.$

I recently began to understand how powerful it might be via the $TT^*$ method. I would like to reproduce its proof following Fefferman’s presentation.

$\mathbf{Statement.}$

Suppose $T:=\sum_{k=1}^M T_k$ is a sum of operators on Hilbert spaces (say the usual $L^2$). Assume that
$\|T^*_jT_k\|\le a(j-k)$ and $\|T_jT^*_k\|\le a(j-k)$. Then $\|T\|\le \sum_{|j|\le M }\sqrt{a(j)}$.
Here $a(\cdot)$ is a non-negative even function.

$\mathbf{Proof.}$ The argument follows an idea of iteration.

Step 1. $\|Tf\|^2=\langle Tf, Tf \rangle =\sum_{j=1}^M\sum_{k=1}^M \langle T_j f, T_k f\rangle$. We estimate $\langle T_j f, T_k f\rangle$ into two ways. Firstly it is easy to see that
$\langle T_j f, T_k f\rangle \le (\max \|T_j\|)^2 \|f\|^2$; secondly it is also trival that
$\langle T_j f, T_k f\rangle = \langle T^*_kT_j f, f\rangle \le a(j-k) \|f\|^2$. Hence by taking the geometric means of these two bounds, we have

$\langle T_j f, T_k f\rangle \le \max \|T_j\| \sqrt {a(j-k)} \|f\|^2$.

Hence

$\|T\|^2\le \sum_{j=1}^M\sum_{k=1}^M \max \|T_j\| \sqrt {a(j-k)} \le M \max \|T_j\| \sum_{|j|\le M} \sqrt{a(j)}$.

This gives that

$\|T\|\le (M\max\|T_j\|)^{1/2} (\sum_{|j|\le M} \sqrt{a(j)})^{1/2}.$

Step 2. We investigate one more iteration of $TT^*$. We write

$\langle TT^* f, TT^* f\rangle= \sum_{1\le k_1,k_2,j_1,j_2\le M} \langle T_{k_1}T^*_{j_1}f, T_{k_2}T^*_{j_2}f\rangle$
$\qquad = \sum_{1\le k_1,k_2,j_1,j_2\le M} \langle T_{j_2}T^*_{k_2}T_{k_1}T^*_{j_1}f, f\rangle.$

Also we estimate $\langle T_{j_2}T^*_{k_2}T_{k_1}T^*_{j_1}f, f\rangle$ by organizing the operators in two ways.
Firstly since $T_{j_2}T^*_{k_2}T_{k_1}T^*_{j_1}=\bigl(T_{j_2}T^*_{k_2}\bigr) \bigl(T_{k_1}T^*_{j_1}\bigr)$, we see that

$\langle T_{j_2}T^*_{k_2}T_{k_1}T^*_{j_1}f, f\rangle \le a(j_2-k_2)a(j_1-k_1) \|f\|^2;$

secondly since $T_{j_2}T^*_{k_2}T_{k_1}T^*_{j_1}=T_{j_2}\bigl(T^*_{k_2}T_{k_1}\bigr)T^*_{j_1}$,

$\langle T_{j_2}T^*_{k_2}T_{k_1}T^*_{j_1}f, f\rangle \le ( \max\|T_j\|)^2 a(k_2-k_1) \|f\|^2;$

By taking geometric means, we see that

$\|T\|^4 \le \sum_{1\le k_1,k_2,j_1,j_2\le M} \max\|T_j\| \sqrt{a(j_2-k_2)}\sqrt{a(k_2-k_1)}\sqrt{a(k_1-j_1)}$
$\qquad \le M\max(\sum_{|j|\le M}\sqrt{a(j)})^3 .$

Hence
$\|T\|\le (M\max\|T_j\|)^{1/4} (\sum_{|j|\le M}\sqrt{a(j)})^{3/4}$.

Step 3. by induction, we see that for any $n\ge 1$, we have

$\|T\|\le (M\max\|T_j\|)^{\frac 1{2n}} (\sum_{|j|\le M}\sqrt{a(j)})^{\frac {2n-1}{2n}}$.

Let $n\to\infty$, we see that

$\|T\|\le \sum_{|j|\le M} \sqrt{a(j)}$. The proof of this lemma is complete.

We remark that in the argument both bounds $\|T_jT^*_k\|$ and $\|T^*_jT_k\|$ are used.

$\mathbf{Example.}$

Let $P:=\sum_{k\in \mathcal{Z}}P_k$ where $P_k$ be the Littlerwood-Paley projection operator. It is easy to see that $P_k$ is a self-adjoint operator. Also $a(0)\le 1$ and $a(j)=0$ for $j\neq 0$.
Then Stein-Cotlar gives,

$\|P\|\le 1$,

which matches that $\|P\|=1$.