Almost orthogonality


Motivation and Cotlar’s lemma

We always need to consider a transform $latex T$ on Hilbert space $latex l^2(\mathbb Z)$ (this is a discrete model), or a finite dimensional space $latex V$. If under a basis $latex T$ is given by a diagonal matrix this story is easy,

$latex \displaystyle A = \begin{pmatrix} \Lambda_1 & 0 & \ldots & 0 \\ 0 & \Lambda_2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & \Lambda_n \end{pmatrix} \ \ \ \ \ (5)$

Then $latex ||T||=\max_{i}\lambda_i$.

In fact, for $latex T$ is a transform of a finite dimensional space, $latex T$ is given by $latex (a_{ij})_{n\times n}$ by duality we have $latex ||T||=||TT^*||$, so we have,

$latex ||T||=||TT^*||=|(\sum a_{ij}x_j)y_i|\leq |\sum_{i,j}\frac{1}{2}(|a_{ij}(|x_i|^2+|y_j|^2)|\leq M$

If we have given $latex \sum_{i}|a_{ij}|\leq M$ and $latex \sum_{j}|a_{ij}|\leq M$ $latex \forall i,j\in \{1,2,…,n\}$.

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 $latex \{T_j\}_{j=1}^N$ be finitely many operators on some Hilbert space $latex H$. Such that for some function $latex \gamma : \mathbb Z\to R^+$ one has,

$latex ||T_j^*T_k||\leq \gamma^2(j-k),||T_jT_k^*||\leq \gamma^2(j-k)$

for any $latex 1\leq j,k\leq N$. Let $latex \sum_{l=-\infty}^{\infty}\gamma(l)=A<\infty$. then ,

$latex ||\sum_{j=1}^NT_j||\leq A$


tensor power trick + duality $latex ||T||=||TT^*||^{\frac{1}{2}}$.

Singular integrals on $latex L^2$



Define $latex T$ is a operator on measure space $latex X\times Y$ equipped positive product measure $latex \mu\wedge \nu$, via,

$latex (Tf)(x)=\int_YK(x,y)f(y)\nu(dy)$

$latex K$ is a measurable kernel, then,

1). $latex ||T||_{1\to 1}\leq \sup_{y\in Y}\int_{X}|K(x,y)|\mu(dx)=:A$.

2). $latex ||T||_{\infty\to \infty}\leq \sup_{x\in X}\int_{Y}|K(x,y)|\nu(dy)=:B$.

3). $latex ||T||_{p\to p}\leq A^{\frac{1}{p}}B^{\frac{1}{p’}}$, $latex  \forall 1\leq p\leq \infty$.

4). $latex ||T||_{1\to \infty}\leq ||K||_{L^{\infty}(X\times Y)}$.


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 $latex |\nabla K(x)|\leq B|x|^{-d-1}$. Then

$latex ||T||_{2\to 2} \leq CB$

with $latex C = C(d)$.

Caldero ́n–Vaillancourt theorem


Hardy’s inequality

Theorem(Hardy inequality)

For any $latex 0 \leq s < \frac{d}{2}$ there is a constant $latex C(s, d)$ with the prop-

arty that,

$latex |||x|^{-s} f||_2 \leq C(s,d)||f||_{H^s(R^d)}$

for all $latex f \in H^s(R^d)$.