Cotlar–Stein lemma

Let E,\,F be two Hilbert spaces. Consider a family of operators T_j, j\geq 1, with each T_j a bounded linear operator from E to F. Denote : a_{jk}=\Vert T_j T_k^\ast\Vert, \qquad b_{jk}=\Vert T_j^\ast T_k\Vert. The family of operators T_j:\;E\to F, j\ge 1, is almost orthogonal if :A=\sup_{j}\sum_{k}\sqrt{a_{jk}} The Cotlar–Stein lemma states that if T_j are almost orthogonal, then the series \sum_{j}T_j converges in the strong operator topology, and :\Vert \sum_{j}T_j\Vert \le\sqrt{AB}. ==Proof==

If T_1,\ldots,T_n is a finite collection of bounded operators, then (as will be proven below) :\displaystyle{\sum_{i,j} |(T_i v,T_jv)| \le \left(\max_i \sum_j \|T_i^*T_j\|^{1\over 2}\right)\left(\max_i \sum_j \|T_iT_j^*\|^{1\over 2}\right)\|v\|^2.} So under the hypotheses of the lemma, :\displaystyle{\sum_{i,j} |(T_i v,T_jv)| \le AB\|v\|^2.} It follows that :\displaystyle{\|\sum_{i=1}^n T_iv\|^2 \le AB \|v\|^2,} and that :\displaystyle{\|\sum_{i=m}^n T_iv\|^2 \le \sum_{i,j\ge m} |(T_iv,T_jv)|.} Hence, the partial sums :\displaystyle{s_n=\sum_{i=1}^n T_iv} form a Cauchy sequence. The sum is therefore absolutely convergent with the limit satisfying the stated inequality. To prove the inequality above set :\displaystyle{R=\sum a_{ij}T_i^*T_j} with |aij| ≤ 1 chosen so that :\displaystyle{(Rv,v)=|(Rv,v)|=\sum |(T_iv,T_jv)|.} Then :\displaystyle{\|R\|^{2m} =\|(R^*R)^m\|\le \sum \|T_{i_1}^* T_{i_2} T_{i_3}^* T_{i_4} \cdots T_{i_{4m}}\| \le \sum \left(\|T_{i_1}^*\|\|T_{i_1}^*T_{i_2}\|\|T_{i_2}T_{i_3}^*\|\cdots \|T_{i_{4m-1}}^* T_{i_{4m}}\|\|T_{i_{4m}}\|\right)^{1\over 2}.} Hence :\displaystyle{\|R\|^{2m} \le n \cdot \max \|T_i\| \left(\max_i \sum_j \|T_i^*T_j\|^{1\over 2}\right)^{2m}\left(\max_i \sum_j \|T_iT_j^*\|^{1\over 2}\right)^{2m-1}.} Taking 2mth roots and letting m tend to ∞, :\displaystyle{\|R\|\le \left(\max_i \sum_j \|T_i^*T_j\|^{1\over 2}\right)\left(\max_i \sum_j \|T_iT_j^*\|^{1\over 2}\right),} which immediately implies the inequality. ==Generalization==

The Cotlar-Stein lemma has been generalized, with sums being replaced by integrals. Let X be a locally compact space and μ a Borel measure on X. Let T(x) be a map from X into bounded operators from E to F which is uniformly bounded and continuous in the strong operator topology. If :\displaystyle{A= \sup_x \int_X \|T(x)^*T(y)\|^{1\over 2} \, d\mu(y),\,\,\, B= \sup_x \int_X \|T(y)T(x)^*\|^{1\over 2}\, d\mu(y),} are finite, then the function T(x)v is integrable for each v in E with :\displaystyle{\|\int_X T(x)v\, d\mu(x)\| \le \sqrt{AB} \cdot \|v\|.} The result can be proven by replacing sums with integrals in the previous proof, or by utilizing Riemann sums to approximate the integrals. ==Example==

Here is an example of an orthogonal family of operators. Consider the infinite-dimensional matrices. : T=\left[ \begin{array}{cccc} 1&0&0&\vdots\\0&1&0&\vdots\\0&0&1&\vdots\\\cdots&\cdots&\cdots&\ddots\end{array} \right] and also : \qquad T_1=\left[ \begin{array}{cccc} 1&0&0&\vdots\\0&0&0&\vdots\\0&0&0&\vdots\\\cdots&\cdots&\cdots&\ddots\end{array} \right], \qquad T_2=\left[ \begin{array}{cccc} 0&0&0&\vdots\\0&1&0&\vdots\\0&0&0&\vdots\\\cdots&\cdots&\cdots&\ddots\end{array} \right], \qquad T_3=\left[ \begin{array}{cccc} 0&0&0&\vdots\\0&0&0&\vdots\\0&0&1&\vdots\\\cdots&\cdots&\cdots&\ddots\end{array} \right], \qquad \dots. Then \Vert T_j\Vert=1 for each j, hence the series \sum_{j\in\mathbb{N}}T_j does not converge in the uniform operator topology. Yet, since \Vert T_j T_k^\ast\Vert=0 and \Vert T_j^\ast T_k\Vert=0 for j\ne k, the Cotlar–Stein almost orthogonality lemma tells us that :T=\sum_{j\in\mathbb{N}}T_j converges in the strong operator topology and is bounded by 1. ==Notes==