# (Explicit) Tauberian theorems: removing $(\log x/n)$

Say that $$\{a_n\}_{n\geq 1}$$, $$|a_n|\leq 1$$, are such that $$\left|\sum_{n\leq x} a_n \log \frac{x}{n}\right|\leq \epsilon x\quad\text{for all x\geq x_0.}$$ What sort of bound can we deduce on $$S(x)=\sum_{n\leq x} a_n$$?

Naïve answer. It is easy to give a simple bound: since, for $$y>1$$, \begin{aligned} \left|\sum_{n\leq x y} a_n \log \frac{x y}{n} - \sum_{n\leq x/y} a_n \log \frac{x/y}{n} \right.&\left. - 2 (\log y) \sum_{n\leq x} a_n\right| \\ &= \left|\sum_{x/y < n\leq x y} a_n \log \frac{x y}{n}\right|\\ &= \left|\sum_{x/y < n\leq x y} \log \frac{x y}{n}\right|\\ &\leq \int_{x/y}^{x y} \log \frac{x y}{t} dt + \log y^2 \\ &= \left(y-\frac{1}{y}\right) x - \left(\frac{x}{y}-1\right) \log y^2, \end{aligned} we know that, for $$x\geq x_0/y$$, assuming $$y\leq x/e$$, $$\frac{1}{x}\left|\sum_{n\leq x} a_n\right|\leq \frac{\epsilon}{2 \log y} \left(y + \frac{1}{y}\right) + \frac{y - \frac{1}{y}}{\log y} - \frac{2}{ y} +\frac{2}{x}.$$ Writing $$y = \exp(\delta)$$, we easily see that the leading term of the right side is $$\epsilon/\delta + 2 \delta$$. Hence, setting $$\delta = \sqrt{\epsilon/2}$$ and thus obtain that $$|S(x)|$$ is at most roughly $$\sqrt{8 \epsilon}\cdot x$$.

Now, that is not just a simple bound, but a downright simple-minded one. Can one do better?

$$\newcommand\ep\epsilon$$You cannot improve the upper bound $$c\sqrt\ep\,x$$ on $$|S(x)|$$ (where $$c>0$$ is a universal real constant factor) by more than a universal positive real constant factor.

Indeed, take any $$\ep\in(0,1/4)$$ and let $$a_n:=\sum_{j\ge1}(-1)^j1(c^{j-1} Let $$$$T(x):=\sum_{n\le x} a_n\ln\frac{x}n.$$$$ Let $$h(u):=1-u+u\ln u\sim(1-u)^2/2$$ as $$u\to1$$.

Then for natural $$k\to\infty$$ \begin{aligned} T(c^k) &= \sum_{j\le k}(-1)^j\sum_{c^{j-1} Note that, for $$x$$ between $$c^{k-1}$$ and $$c^k$$, we have $$T(x)$$ between $$T(c^{k-1})$$ and $$T(c^k)$$. It follows that for $$x\to\infty$$ $$$$\sum_{n\le x} a_n\ln\frac{x}n=T(x)=O(x\ep),$$$$ as required.

On the other hand, for natural $$j$$ and $$x=c^{j-1}$$, $$|S(cx)-S(x)|=\sum_{c^{j-1} so that the upper bound $$\asymp\sqrt\ep\,x$$ on $$|S(x)|$$ cannot be improved, as was claimed.

• I'm confused about the order of quantifiers here. I thought were supposed to fix the sequence first and then let $x\to\infty$, whereas in this example for each $x$ we get a different sequence. For the given $a_n$ (with $k$ fixed) we would have $xh(k/x) \to 1$ since $u=k/x\to 0$. So here both the smoothed sum and the usual sum are approximately $x$ and there is no $\epsilon$. Commented Aug 31, 2022 at 5:33
• @LiorSilberman : This should now be fixed. Thank you for your comment. Commented Aug 31, 2022 at 15:21
• This seems to work nicely. Thanks! Commented Aug 31, 2022 at 15:55
• I wonder whether one can do better by adding a condition, in true Tauberian fashion? I suppose your counterexample hints that we should be looking at $\sum_{n\leq x} a_n n^{i t} \log \frac{x}{n}$. Then it all becomes a matter of which interval $[-T,T]$ it is enough to check. (Of course the Dirichlet series $\sum_n a_n n^{-s}$ is lurking in the background.) Commented Aug 31, 2022 at 16:08
• @HAHelfgott : Introducing the factor $n^{it}$ may help, I guess, and I also guess that $T$ may play a role. However, I have done no work on Tauberian theorems or on their use, and at this point do not have any good ideas on how to proceed with $n^{it}$. Do you want to post the amended question separately, so that experts have a better chance to see it? Commented Aug 31, 2022 at 17:23