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

Question

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 $S(x)=\sum_{n\leq x} a_n$?

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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 $y\leq x/e$, $$\frac{1}{x}\left|\sum_{n\leq x} a_n\right|\leq \frac{\epsilon}{2 \log y} \left(y + \frac{1}{y}\right) + \frac{y - \frac{1}{y}}{2 \log y} - \frac{1}{ y} +\frac{1}{x}.$$ Writing $y = \exp(\delta)$, we easily see that the leading term of the right side is $\epsilon/\delta + \delta$. Hence, setting $\delta = \sqrt{\epsilon}$ and thus obtain that $|S(x)|$ is at most roughly $2 \sqrt{\epsilon}\cdot x$.

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

Answer 1

$\newcommand\ep\epsilon$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 $$a_n:=\sum_{j\ge1}(-1)^j1(c^{j-1}<n<c^j),\quad c:=e^{\sqrt\ep}.$$ Let \begin{equation} T(x):=\sum_{n\le x} a_n\ln\frac{x}n. \end{equation} Let $h(u):=1-u+u\ln u\sim(1-u)^2/2$ as $u\to1$.

Then for natural $k\to\infty$ \begin{equation} \begin{aligned} T(c^k) &= \sum_{j\le k}(-1)^j\sum_{c^{j-1}<n<c^j}\ln\frac{c^k}n \\ &= \sum_{j\le k}(-1)^j\Big(o(c^k/k)+\int_{c^{j-1}<n<c^j}\ln\frac{c^k}n\,dn\Big) \\ &=o(c^k)+ \sum_{j\le k}(-1)^j[c^j h(c)+(c^j-c^{j-1})(k-j)\ln c] \\ &=o(c^k)+ \sum_{j\le k}(-1)^j c^j\ep[1/2+(k-j)+O(\ep)] \\ &=O(c^k\ep). \end{aligned} \end{equation} 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 $x\to\infty$ \begin{equation} \sum_{n\le x} a_n\ln\frac{x}n=T(x)=O(x\ep), \end{equation} as required.

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