The most important background of analytic number theory is the new understanding of the multiplication function on the shared interval. This result is established by Kaisa Matomäki & Maksym Radziwill, two very young and intelligent superstars.
The main theorem in their article is:
Theorem (Matomaki, Radziwill):
As soon as $H \to \infty$ when $x \to \infty$, one has:
$$
\sum_{x\leq n\leq x+H}\lambda(n)= o(H)
$$
for almost all $x\sim X$.
In my understanding of the result, the main strategy is:
Step 1: Parseval identity, monotonically inequality
Parseval identity, monotonically inequality, this is something about the $L^2$ norms of the quality we wish to charge. It is just trying to understand:
$$
\frac{1}{X}\int_{X}^{2X}\left|\frac{1}{H}\sum_{x\leq n\leq x+H}\lambda(n)\right|^2dx
$$
as a fuzzy thing by a more chargeable quality:
$$
\frac{1}{X^2}\int_{0}^{\infty}\left|\sum_{n\leq X}\lambda(n)n^{it}\right|^2dt
$$
In fact, we do a cutoff, the quality we really consider is just:
$$
\frac{1}{X^2}\int_{|\log(X)|^{100}}^{\frac{X}{H}}\left|\sum_{n\leq X}\lambda(n)n^{it}\right|^2dt
$$
established the monotonically inequality:
$$
\frac{1}{X}\int_{X}^{2X}\left|\frac{1}{H}\sum_{x\leq n\leq x+H}\lambda(n)\right|^2dx << \frac{1}{X^2}\int_{|\log(X)|^{100}}^{\frac{X}{H}}\left|\sum_{n\leq X}\lambda(n)n^{it}\right|^2dt
$$
In my understanding, This is a perspective of the quality, due to the quality is a multiplicative function integral on a domain ($\mathbb{N}^*$) with additive structure, it could be looked at as a lot of waves with the periodic given by primes, so we could do an orthogonal decomposition in the fractional space, try to prove the cutoff is an error term and we get such a monotonically inequality.
But at once we get the monotonically inequality, we could look at it as a compactification process and this process still carries most of the information so lead to the inequality.
It seems something similar occurs in the attack of the moments estimate of the zeta function by the second author. And it is also could be looked at as something similar to the spectral decomposition with some basis come from multiplication unclear, i.e. primes.
Step 2: Involved by multiplication property, spectral decomposition
I called it “spectral decomposition”, but this is not very exact. Anyway, the thing I want to say is that for the multiplication function $\lambda(n)$, we have the Euler-product formula:
$$
\prod_{p,prime}\left(\frac{1}{1-\frac{\lambda(p)}{p^s}}\right)=\sum_{n=1}^{\infty}\frac{\lambda(n)}{n^s}
$$
But anyway, we do not use the whole power of multiplication just use it on primes, i.e. $\lambda(pn)=\lambda(p)\lambda(n)$ leads to the following result:
$$
\lambda(n)=\sum_{n=pm,p\in I}\frac{\lambda(p)\lambda(m)}{# {p|n, p\in I}+1}+\lambda(n)1_{p|n;p\notin I}
$$
This is an identity about the function $\lambda(n)$, the point is it is not just use the multiplication at a point, i.e. $\lambda(mn)=\lambda(m)\lambda(n)$, but take average at an area which is naturally generated and compatible with multiplication, this identity carries a lot of information of the multiplicative property. Which is crucial to get a good estimate for the quality we consider about.
Step 3: from linear to multilinear, Cauchy-Schwarz
Now, we do not use one set $I$, but use several sets $I_1,…,I_n$ which are carefully chosen. And we do not consider $[X,2X]$ with linear structure anymore, instead reconsider the decomposition:
$$
[X,2X]=\prod_{i=1}^n (I_i\times J_i) \prod U
$$
On every $I_i\times J_i$ it equipped with a bilinear structure. And $U$ is a very small set, $|U|=o(X)$ which is in fact have a much better estimate.
$$
\int_{|\log(X)|^{100}}^{\frac{X}{H}}\left|\sum_{n\leq X}\lambda(n)n^{it}\right|^2dt =\sum_{i=1}^n\int_{I_i\times J_i} \frac{1}{X^2}\int_{|\log(X)|^{100}}^{\frac{X}{H}}\left|\sum_{n\leq X}\lambda(n)n^{it}\right|^2dt +\int_N \left|\sum_{n\leq X}\lambda(n)n^{it}\right|^2dt
$$
Now we just use a Cauchy-Schwarz:
$$
\sum_{i=1}^n\int_{I_i\times J_i} \frac{1}{X^2}\int_{|\log(X)|^{100}}^{\frac{X}{H}}\left|\sum_{n\leq X}\lambda(n)n^{it}\right|^2dt +\int_N \left|\sum_{n\leq X}\lambda(n)n^{it}\right|^2dt
$$
Step 4: major term estimate
Step 5: minor term estimate
Step 6: estimate the contribution of the area which is not filled