Cotlar–Stein lemma

Remarks

The bound does not depend on $N$. So let $(T_i)$ be an infinite sequence with the same two hypotheses. Then the partial sums are bounded by $A$, and the series $\sum_i T_i$ converges in the strong operator topology to an operator of norm at most $A$.

Idea of the proof. Let $T = \sum_i T_i$ and let $m \ge 1$. Since $\norm{T}^{2m} = \norm{(T^* T)^m}$, expand $(T^*T)^m$ as a sum of $N^{2m}$ products $T_{i_1}^* T_{i_2} T_{i_3}^* \cdots T_{i_{2m}}$. Bound each product in two ways, by grouping the factors in pairs $T_{i_1}^* T_{i_2}, T_{i_3}^* T_{i_4}, \dots$ or in pairs $T_{i_2} T_{i_3}^*, \dots$, and take the geometric mean of the two bounds. Summing over the indices gives $\norm{T}^{2m} \le N A^{2m}$, and letting $m \to \infty$ gives $\norm{T} \le A$.

Example. If $T_i^* T_j = 0$ and $T_i T_j^* = 0$ for all $i \ne j$, the operators act on mutually orthogonal pieces. The two hypotheses then reduce to $\norm{T_i} \le A$ for each $i$, and the lemma gives $\norm{\sum_i T_i} \le \max_i \norm{T_i}$.

One hypothesis alone is not enough. Let $(e_k)$ be an orthonormal basis of a Hilbert space $H$ of dimension at least $N$, and let $T_i x = \ip{x, e_1} e_i$ for $i = 1, \dots, N$. Then $T_i^* T_j = 0$ for $i \ne j$ and $\norm{T_i^* T_i} = 1$, so the first sum equals $1$. But $\norm{T_i T_j^*} = 1$ for all $i, j$, so the second sum is $N$. Indeed $\sum_i T_i$ sends $e_1$ to $e_1 + \cdots + e_N$, so its norm is $\sqrt{N}$.

Going deeper

For a student

The lemma is the basic tool of almost orthogonality in harmonic analysis. A singular integral or pseudodifferential operator is cut into pieces that live at one scale or one frequency band, and the lemma gives $L^2$ boundedness of the sum. A classical use is the proof of the Calderón–Vaillancourt theorem on pseudodifferential operators. It only works on Hilbert spaces, since the proof uses the identity $\norm{T}^2 = \norm{T^*T}$.