# Almost orthogonality

Motivation and Cotlar’s lemma

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

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

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

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

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

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

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.

Lemma(Catlar-Stein)

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

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

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

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

Pf:

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

Singular integrals on $L^2$

Lemma(Schur)

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

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

$K$ is a measurable kernel, then,

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

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

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

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

Pf:

1),2),4) merely due to Fubini theorem and Bath lemma.

3) proof by the interpolation and combine 1) and 2).

Theorem

Let K be a Calderon-Zegmund operator, with the additional assumption

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

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

with $C = C(d)$.

Caldero ́n–Vaillancourt theorem

Hardy’s inequality

Theorem(Hardy inequality)

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

arty that,

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

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

# The correlation of Mobius function and nil-sequences in short interval

I wish to establish the following estimate:

Conjecture :(correlation of Mobius function and nil-sequences in short interval)

$\lambda(n)$ is the liouville function we wish the following estimate is true.

$\int_{0\leq x\leq X}|\sup_{f\in \Omega^m}\sum_{x\leq n\leq x+H}\lambda(n)e^{2\pi if(x)}|dx =o(XH)$.

Where we have $H\to \infty$ as $x\to \infty$, $\Omega^m=\{a_mx^m+a_{m-1}x^{m-1}+...+a_1x+a_0 | a_m,...,a_1,a_0\in [0,1]\}$ is a compact space.

I do not know how to prove this but this is result is valuable to consider, because by a Fourier identity we could transform the difficulty of (log average) Chowla conjecture to this type of result.

There is some clue to show this type of result could be true, the first one is the result established by Matomaki and Raziwill in 2015:

Theorem (multiplication function in short interval)

$f(n): \mathbb N\to \mathbb C$ is a multiplicative function, i.e. $f(mn)=f(n)f(m), \forall m,n\in \mathbb N$. $H\to \infty$ as $x\to infty$, then we have the following result,

$\int_{1\leq x\leq X}|\sum_{x\leq n\leq x+H}f(n)|=o(XH)$.

And there also exists the result which could be established by Vinagrodov estimate and B-S-Z critation :

Theorem(correlation of multiplication function and nil-sequences in long interval)

$f(n): \mathbb N\to \mathbb C$ is a multiplicative function, i.e. $f(mn)=f(n)f(m), \forall m,n\in \mathbb N$. $g(n)=a_n^m+...+a_1n+a_0$ is a polynomial function then we have the following result,

$\int_{1\leq n \leq X}|f(n)e^{2\pi i g(n)}|=o(X)$.