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A basic problem in harmonic analysis (as well as in linear algebra, random matrix theory, and high-dimensional geometry) is to estimate the operator norm of a linear map between two Hilbert spaces, which we will take to be complex for sake of discussion. Even the finite-dimensional case is of interest, as this operator norm is the same as the largest singular value of the matrix associated to .
In general, this operator norm is hard to compute precisely, except in special cases. One such special case is that of a diagonal operator, such as that associated to an diagonal matrix . In this case, the operator norm is simply the supremum norm of the diagonal coefficients:
whenever and are sequences with ; but this easily follows from the arithmetic mean-geometric mean inequality
Schur’s test (4) (and its many generalisations to weighted situations, or to Lebesgue or Lorentz spaces) is particularly useful for controlling operators in which the role of oscillation (as reflected in the phase of the coefficients , as opposed to just their magnitudes ) is not decisive. However, it is of limited use in situations that involve a lot of cancellation. For this, a different test, known as the Cotlar-Stein lemma, is much more flexible and powerful. It can be viewed in a sense as a non-commutative variant of Schur’s test (4) (or of (1)), in which the scalar coefficients or are replaced by operators instead.
To illustrate the basic flavour of the result, let us return to the bound (1), and now consider instead a block-diagonal matrix
Indeed, the lower bound is trivial (as can be seen by testing on vectors which are supported on the block of coordinates), while to establish the upper bound, one can make use of the orthogonal decomposition
to decompose an arbitrary vector as
with , in which case we have
and the upper bound in (6) then follows from a simple computation.
When is large, this is a significant improvement over the triangle inequality, which merely gives
The reason for this gain can ultimately be traced back to the “orthogonality” of the ; that they “occupy different columns” and “different rows” of the range and domain of . This is obvious when viewed in the matrix formalism, but can also be described in the more abstract Hilbert space operator formalism via the identities
whenever . (The first identity asserts that the ranges of the are orthogonal to each other, and the second asserts that the coranges of the (the ranges of the adjoints ) are orthogonal to each other.) By replacing (7) with a more abstract orthogonal decomposition into these ranges and coranges, one can in fact deduce (8) directly from (9) and (10).
The Cotlar-Stein lemma is an extension of this observation to the case where the are merely almost orthogonal rather than orthogonal, in a manner somewhat analogous to how Schur’s test (partially) extends (1) to the non-diagonal case. Specifically, we have
on each component of , which by the triangle inequality gives the inferior bound
The Cotlar-Stein lemma was first established by Cotlar in the special case of commuting self-adjoint operators, and then independently by Cotlar and Stein in full generality, with the proof appearing in a subsequent paper of Knapp and Stein.
The Cotlar-Stein lemma is often useful in controlling operators such as singular integral operators or pseudo-differential operators which “do not mix scales together too much”, in that operators map functions “that oscillate at a given scale ” to functions that still mostly oscillate at the same scale . In that case, one can often split into components which essentically capture the scale behaviour, and understanding boundedness properties of then reduces to establishing the boundedness of the simpler operators (and of establishing a sufficient decay in products such as or when and are separated from each other). In some cases, one can use Fourier-analytic tools such as Littlewood-Paley projections to generate the , but the true power of the Cotlar-Stein lemma comes from situations in which the Fourier transform is not suitable, such as when one has a complicated domain (e.g. a manifold or a non-abelian Lie group), or very rough coefficients (which would then have badly behaved Fourier behaviour). One can then select the decomposition in a fashion that is tailored to the particular operator , and is not necessarily dictated by Fourier-analytic considerations.
Once one is in the almost orthogonal setting, as opposed to the genuinely orthogonal setting, the previous arguments based on orthogonal projection seem to fail completely. Instead, the proof of the Cotlar-Stein lemma proceeds via an elegant application of the tensor power trick (or perhaps more accurately, the power method), in which the operator norm of is understood through the operator norm of a large power of (or more precisely, of its self-adjoint square or ). Indeed, from an iteration of (14) we see that for any natural number , one has
Recall that when we applied the triangle inequality directly to , we lost a factor of in the final estimate; it will turn out that we will lose a similar factor here, but this factor will eventually be attenuated into nothingness by the tensor power trick.
On the other hand, we can group the product by pairs in another way, to obtain the bound of
for (16). Taking roots, we obtain
Sending , we obtain the claim.
Remark 1 As observed in a number of places (see e.g. page 318 of Stein’s book, or this paper of Comech, the Cotlar-Stein lemma can be extended to infinite sums (with the obvious changes to the hypotheses (11), (12)). Indeed, one can show that for any , the sum is unconditionally convergent in (and furthermore has bounded -variation), and the resulting operator is a bounded linear operator with an operator norm bound on .
Remark 2 If we specialise to the case where all the are equal, we see that the bound in the Cotlar-Stein lemma is sharp, at least in this case. Thus we see how the tensor power trick can convert an inefficient argument, such as that obtained using the triangle inequality or crude bounds such as (15), into an efficient one.
Remark 3 One can prove Schur’s test by a similar method. Indeed, starting from the inequality
(which follows easily from the singular value decomposition), we can bound by
Estimating the other two terms in the summand by , and then repeatedly summing the indices one at a time as before, we obtain
and the claim follows from the tensor power trick as before. On the other hand, in the converse direction, I do not know of any way to prove the Cotlar-Stein lemma that does not basically go through the tensor power argument.