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.

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