Motivation and Cotlar’s lemma
We always need to consider a transform on Hilbert space (this is a discrete model), or a finite dimensional space . If under a basis is given by a diagonal matrix this story is easy,
In fact, for is a transform of a finite dimensional space, is given by by duality we have , so we have,
If we have given and .
But in application of this idea, the orthogonal condition always seems to be too restricted and due too this we have the following lemma which is follow the idea but change the orthogonal condition by almost orthogonal.
Let be finitely many operators on some Hilbert space . Such that for some function one has,
for any . Let . then ,
tensor power trick + duality .
Singular integrals on
Define is a operator on measure space equipped positive product measure , via,
is a measurable kernel, then,
3). , .
1),2),4) merely due to Fubini theorem and Bath lemma.
3) proof by the interpolation and combine 1) and 2).
Let K be a Calderon-Zegmund operator, with the additional assumption
that . Then
Caldero ́n–Vaillancourt theorem
For any there is a constant with the prop-
for all .
[…] given by transform from version of orthogonal to almost orthogonal which was merely explained in the previous note. We could image the following corresponding almost orthogonal variate of “Bessel […]