Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 25 / 4

Put $Z=\lfloor N^{2/5}\rfloor$ and $S={\sum }_{N<n\le N+M}{n}^{-it}$. For any positive integer shift $h\le Z^2$, translating the interval changes its sum by at most $2h$. Averaging the shifts $h=xy$ gives the bilinear shift averaging for a logarithmic phase identitySince $xy/n\le N^{-1/5}$, the alternating Taylor expansion of the logarithm has remainder at most $(xy/n{)}^{r+1}/(r+1)$. ThusFor $r=\lfloor 5.01\log t/\log N\rfloor$, we have $r+1>5.01\log t/\log N$, hence the error is at most $t^{-1/500}$. The exponential function on an imaginary argument changes by at most the change in that argument. ThereforeUse $Z^2\asymp N^{4/5}$ and $N<n\le N+M\le 2N$. Taking absolute values provesThe boundary error comes from integer shifts, so this argument also covers intervals shorter than a shift. Here $x,y$ range over positive integers; no zero term is needed.

The Vinogradov mean value isBy orthogonality of integer Fourier modes, it counts the ordered integer solutions of ${\sum }_{i=1}^k{x}_i^j={\sum }_{i=1}^k{y}_i^j$ for $1\le j\le r$, with every coordinate between one and $\lfloor Z\rfloor$. The diagonal solutions $y_i=x_i$ give $J_{k,r}(Z)\ge \lfloor Z{\rfloor }^k$.

Put $D=r(r+1)/2$. The moment vector $({\sum }_i{x}_i^j{)}_{j=1}^r$ has at most ${\prod }_{j=1}^r(k\lfloor Z{\rfloor }^j+1){\ll }_{k,r}{Z}^D$ possible values. If $R(v)$ counts the tuples with vector $v$, then ${\sum }_{v}R(v)=\lfloor Z{\rfloor }^k$ and $J_{k,r}(Z)={\sum }_{v}R(v{)}^2$. The Cauchy-Schwarz inequality gives $J_{k,r}(Z){\gg }_{k,r}{Z}^{2k-D}$. Combining the two lower bounds, with one common positive constant for large $Z$, yields

Here is an explicit way that upper bounds enter the Vinogradov mean-value method for a bilinear exponential sum. Write $U={\sum }_{x,y\le Z}e({\sum }_j{\alpha }_j{x}^j{y}^j)$ and $L_j=k\lfloor Z{\rfloor }^j$. Two applications of the Holder inequality, followed by grouping equal differences of moment vectors, giveAt integer ${\alpha }_{j}v$ the minimum is defined as $2L_j+1$; $\Vert u\Vert$ means distance to the nearest integer. To explain the mean-value factor, let $r(v)$ count pairs of $k$-tuples with prescribed moment difference. It is an autocorrelation of $R$, so $r(v)\le {\sum }_{w}R(w{)}^2=J_{k,r}(Z)$ by Cauchy-Schwarz inequality. The first Holder inequality groups the $y$ tuples, and the second groups the $x$ tuples, providing the two factors $J$. The remaining sums over moment differences are bounded by the displayed finite geometric series estimates. Good Vinogradov mean value upper bounds, together with rational approximation or spacing bounds for ${\alpha }_j$, therefore give cancellation in $U$. The mean-value estimate alone does not force cancellation for arbitrary coefficients: when all ${\alpha }_j$ are integers, $U=\lfloor Z{\rfloor }^2$. For the paper take $Z=N^{2/5}$ and the specified ${\alpha }_j$.

Set $\epsilon =1-\sigma \in [0,1)$ and $L=\log (t+2)$. By the Hardy-Littlewood approximation to the Riemann zeta function at $x=t$, it is enough to bound ${\sum }_{n\le t}{n}^{-\sigma -it}$: the integral term has size $O(t^{-\sigma })$ because $|s-1|\ge t$.

On a dyadic interval $(N,N^{\prime }]\subset (N,2N]$ with $N^{\prime }\le t$, the assumed estimate holds for every initial subinterval. Abel summation with $n^{-\sigma }$ therefore givesIndeed the weighted endpoint and integral of the term proportional to the subinterval length are $O(N^{1-\sigma })$, and those of the constant $N^{4/5}$ term are $O(N^{4/5-\sigma })$. The constants can be uniform in $0<\sigma \le 1$.

For the first term, write $u=j\log 2$ and complete the square:The sum of a shifted Gaussian function on a fixed-spaced lattice is $O(\sqrt{L})$, uniformly in the shift. Thus these dyadic contributions are $O(\sqrt{L}{e}^{{\epsilon }^{2}L/(4c)})$. This is the Gaussian dyadic summation bound.

For the second term, if $\epsilon \le 1/10$ its dyadic sum is bounded. If $\epsilon >1/10$, a crude bound is $O(L\exp (\max (\epsilon -1/5,0)L))$. The positive exponent obeys $\epsilon -1/5\le (5/4){\epsilon }^2$, and $L$ can be absorbed into $\sqrt{L}{e}^{C{\epsilon }^{2}L}$ uniformly on $\epsilon >1/10$ by increasing the fixed constant $C$. The finitely many initial terms cause no problem. We conclude, with one fixed sufficiently large $C$,In particular the endpoint $\sigma =1$ gives $O(\sqrt{\log t})$ under the assumed exponential-sum hypothesis. This conditional conclusion uses that hypothesis, not an unconditional improvement of the stated Richert bound for the Riemann zeta function.