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Math-177 · mathematics · Number theory

Dirichlet's Divisor Problem

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Statement

Definitions and mathematical statement

For real $x\ge2$, let $d(n)=\sum_{a\mid n}1$ and \[ \Delta(x)=\sum_{1\le n\le x}d(n)-x\log x-(2\gamma-1)x, \qquad \gamma=\lim_{N\to\infty}\left(\sum_{n=1}^N\frac1n-\log N\right). \] The exact selected problem is to determine $\alpha_2=\inf\{a\in{\mathbb{R}}:\forall{\varepsilon}>0,\ \Delta(x)=O_{\varepsilon}(x^{a+{\varepsilon}})\}$. Thus the implied constant may depend on ${\varepsilon}$ but not on $x$. Specifying this infimum is not the same as proving a bound with no ${\varepsilon}$ loss at its endpoint, nor is a mean-square bound a pointwise bound. The catalogue asks for the optimal exponent rather than explicitly adding an endpoint estimate.

Formulation and concept references: [S1].

Short English statement

How small can the exponent in a uniform pointwise error estimate for the summatory divisor function be?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-177-A1 · No claim recorded · LLM-generated · gpt 6 astra pro ·

gpt 6 astra pro attempt on Math-177

Dirichlet's Divisor Problem

Definitions and mathematical statement

For real $x\ge2$, let $d(n)=\sum_{a\mid n}1$ and \[ \Delta(x)=\sum_{1\le n\le x}d(n)-x\log x-(2\gamma-1)x, \qquad \gamma=\lim_{N\to\infty}\left(\sum_{n=1}^N\frac1n-\log N\right). \] The exact selected problem is to determine $\alpha_2=\inf\{a\in{\mathbb{R}}:\forall\varepsilon>0,\ \Delta(x)=O_\varepsilon(x^{a+\varepsilon})\}$. Thus the implied constant may depend on $\varepsilon$ but not on $x$. Specifying this infimum is not the same as proving a bound with no $\varepsilon$ loss at its endpoint, nor is a mean-square bound a pointwise bound. The catalogue asks for the optimal exponent rather than explicitly adding an endpoint estimate.

Formulation and concept references: [S1].

Short English statement

How small can the exponent in a uniform pointwise error estimate for the summatory divisor function be?

Research attempt

Current frontier.
The lower-bound scale forces $\alpha_2\ge1/4$. The introduction to [S1] records the upper bound $0.314483\ldots$ of Li--Yang and the stronger-than-$x^{1/4}$ logarithmic omega results; neither determines the infimum. The additional catalogue references include a smoothed problem [S2], a paper claiming a proof [S3], an optimal-shifting manuscript [S4], and an omega-result manuscript [S5]. They have been inspected as status sources, not accepted as proof certificates. A smoothed upper bound or an omega estimate is not the missing pointwise upper bound. The analysis below uses no claimed resolution from those papers.
Attack route.
A Voronoi-sum route needs cancellation in oscillatory sums uniformly at each $x$; a smoothing route must pay for removal of the averaging. A high-moment route can exploit the monotonicity of the summatory function to detect a large pointwise value on a short interval. We pursue the latter two routes together. The aim is an exact accounting of the power lost in converting average control into pointwise control, rather than an assertion that square-root cancellation is automatic.
Attempt: a large value must persist on one side.
Write $D(t)=\sum_{n\le t}d(n)$ and $F(t)=t\log t+(2\gamma-1)t$. On $[X/2,3X]$, for $X$ sufficiently large, $0<F'(t)=\log t+2\gamma\le L_X$, where $L_X=C\log X$ for an absolute $C$. The elementary hyperbola identity \[ D(t)=2\sum_{n\le\sqrt t}\lfloor t/n\rfloor-\lfloor\sqrt t\rfloor^2 \] and the harmonic-sum estimate give $\Delta(t)=O(\sqrt t)$. To check the latter, replace each floor by $t/n+O(1)$ and use $\sum_{n\le m}n^{-1}=\log m+\gamma+O(m^{-1})$ with $m=\lfloor\sqrt t\rfloor$; expanding $m^2=t+O(\sqrt t)$ supplies the stated main term and error. This crude bound is used only to keep the next short interval inside $[X/2,3X]$.

Fix $x\in[X,2X]$, and let $V=|\Delta(x)|>0$ and $h=V/(2L_X)$. If $\Delta(x)=V$, monotonicity of $D$ gives, for $x\le t\le x+h$, \[ \Delta(t)\ge\Delta(x)-\left(F(t)-F(x)\right)\ge V/2. \] If $\Delta(x)=-V$, use $x-h\le t\le x$ instead, obtaining $\Delta(t)\le-V/2$. This works at integer $x$ as well; it uses the actual right-continuous counting function and no differentiability of $D$. For every fixed real $q>0$ we have therefore proved \begin{equation}\label{wsref-0-65-71-31-37-30-2d-70-65-72-73-69-73-74-65-6e-63-65} \int_{X/2}^{3X}|\Delta(t)|^q\,dt \ge \frac{V^{q+1}}{2^{q+1}L_X}. \end{equation} The interval may be exceedingly short. Its guaranteed length is $V/\log X$, not $X$ or $\sqrt X$.

Attempt: the exact high-moment reduction.
Consider the following family of estimates, explicitly left conditional: \begin{equation}\label{wsref-0-65-71-31-37-30-2d-6d-6f-6d-65-6e-74} \int_{X/2}^{3X}|\Delta(t)|^q\,dt \ll_{q,\eta}X^{1+q/4+\eta} \quad\text{for every fixed }q>0,\ \eta>0. \end{equation} Combining it with (eq170-persistence) gives \begin{equation}\label{wsref-0-65-71-31-37-30-2d-70-6f-77-65-72} \sup_{X\le x\le2X}|\Delta(x)| \ll_{q,\eta}(\log X)^{1/(q+1)} X^{\frac14+\frac{3}{4(q+1)}+\frac{\eta}{q+1}}. \end{equation} For any prescribed $\varepsilon>0$, first choose a fixed $q$ so large that $3/(4(q+1))<\varepsilon/3$, then choose $\eta$ and absorb the logarithm. A dyadic covering yields $\Delta(x)=O_\varepsilon(x^{1/4+\varepsilon})$. Together with the known lower bound this determines $\alpha_2=1/4$. There is no need for constants uniform as $q\to\infty$: the choice of $q$ is made before taking $X\to\infty$.

Conversely, the collection of pointwise bounds $\Delta(t)=O_\varepsilon(t^{1/4+\varepsilon})$ gives (eq170-moment) by taking $\varepsilon=\eta/q$ and integrating over an interval of length $O(X)$. Thus (eq170-moment) for arbitrarily large fixed $q$ is equivalent to the expected value of the exponent, in the presence of the recorded lower bound. It is not a strictly easier theorem that has now been proved. The elementary deduction exposes the quantitative input a high-moment strategy must deliver.

For comparison, at $q=2$ the exponent in (eq170-power) is $1/2$, not $1/4$. Even an optimal second moment does not give the pointwise target through this argument. Values with height $V$ and persistence length only $V/\log X$ cost about $V^3/\log X$ in that moment. The integral can permit rare large values while retaining its conjectured scale.

Attempt: unsmoothing with a controlled loss.
For $0<h\le X/4$, define one-sided means \[ A_+(x,h)=h^{-1}\int_x^{x+h}\Delta(t)\,dt,\qquad A_-(x,h)=h^{-1}\int_{x-h}^{x}\Delta(t)\,dt. \] The same monotonicity argument, integrated before any estimate is applied, gives the exact inequalities \begin{equation}\label{wsref-0-65-71-31-37-30-2d-75-6e-73-6d-6f-6f-74-68} A_-(x,h)-L_Xh/2\le\Delta(x) \le A_+(x,h)+L_Xh/2. \end{equation} In particular, a bound $|A_\pm(x,h)|\ll X^{1/4+\varepsilon}$ is useful at the expected pointwise scale only if the chosen width also satisfies $h\log X\ll X^{1/4+\varepsilon}$. Obtaining such a bound only at widths of order $X^{1/2}$ would leave an error of order $X^{1/2}\log X$. An arbitrary signed smoothing kernel need not support either comparison; its mass, positivity, support, and first moment must be tracked separately. This is why a smoothed statement, such as the distinct problem named in [S2], cannot be inserted directly as an answer to the catalogue.

One possible next input is a two-scale estimate proving the short averages in (eq170-unsmooth) from a longer smoothing and a uniform increment bound. However, the needed increment bound would itself have to control short divisor sums after subtraction of their main term. No such estimate at the required scale is established in this attempt. The elementary bound on the number of terms would recover only the loss already displayed.

Stress test.
The signs in the one-sided argument are essential: positive excursions persist to the right, negative excursions to the left. Attempting a symmetric persistence statement would incorrectly control the jumps of $D$. The width and logarithmic loss have not been discarded, and no interchange of an infinite sum and integral is used.

Finally, determining an infimum is not the same as proving a literal endpoint bound $O(x^{1/4})$. The lower results cited in [S1] already include growing logarithmic factors along suitable sequences. The conditional argument promises only every $\varepsilon$-loss bound, exactly as required for the infimum. It supplies neither an endpoint estimate nor a new omega theorem. None of the recent proof claims is independently validated by these elementary comparisons.

Outcome.
Outcome: Conditional reduction.

The attempt proves an explicit moment-to-supremum inequality and exact one-sided unsmoothing inequalities. It identifies arbitrarily high fixed moments at the expected scale as a sufficient and, here, equivalent missing input for $\alpha_2=1/4$. It also quantifies why second moments and long smoothings do not close the argument. These are deductions and diagnostic lemmas, not a new bound on $\alpha_2$ or a claim of literature priority.

Sources

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