Deduction of the classical Halasz's inequality from the new form

In the new proof of Halasz's Theorem, the authors give a statement in terms of a quantity $L(x)$ defined by $$L(x)^2=\sum_{|N|\leq\log^2x+1}\frac1{N^2+1}\max_{|t-N|\leq1/2}|F_x(1+it)|^2,$$

where $F_x(s)$ denotes the truncated Euler's product for the arithmetic function $f(n)$.

This quantity seems rather mysterious to me. As I understand it, instead of taking maximum over the whole interval as in some classical statements, which might be wasteful because large values of $F(s)$ are typically clustered into small intervals (as I read somewhere), here we divide the large interval into subintervals of length 1 and take maximums on each of them, which might be better.

I am however not able to use this quantity to obtain the classical form of the Theorem $$\left|\frac1x\sum_{n\leq x}f(n)\right|\ll (1+M(x,T))e^{-M(x,T)}+\frac1T+\frac{\log\log x}{\log x},$$ where $$M(x,T)=\min_{|t|\leq T}\sum_{p\leq x}\frac{1-\text{Re}(f(p)p^{-it})}{p}$$ and it holds that $$\log x \cdot e^{-M(x,T)}\asymp\max_{|t|\leq T}|F_x(1+it)|\asymp\max_{|t|\leq T}\left|F\left(1+\frac1{\log x}+it\right)\right|,$$ where $F(s)$ denotes the Dirichlet series of $f$.

The classical form follows if it holds that $$L(x)\asymp \max_{|t|\leq \log^2 x}|F_x(1+it)|,$$ but I do not know how to obtain this asymptotic (in particular the lower bound) or whether it is even true.

In the other paper, the authors prove a different form of Halász's Theorem, from which obtaining the classical statement is (at least for me) not straightforward.

• The authors say that the two versions are ''essentially'' the same. Therefore, I think that the trivial upper bound $L(x)\ll \max_{|t|\leq \log^{2}x}|F_{x}(1+it)|\ll e^{-M(x,\log^{2}x)}\log x$, is enough to derive the classical version of Halasz's Theorem from the new one. – The Number Theorist Apr 30 '18 at 20:32