# Smoothed exponential sums: bounds and sources?

Let $f:\mathbb{R}\to\mathbb{C}$ be differentiable $k$ times, with $f, f',\dotsc,f^{(k)}\in L^1$. Let $\alpha\in \mathbb{R}/\mathbb{Z}$, $\alpha\ne 0$. In "Every odd number..." (Math. Comp. 83, 2014), Lemma 3.1, Tao shows that

$$\left|\sum_{n\in \mathbb{Z}} f(n) e(\alpha n)\right|\leq \frac{1}{|2 \sin(\pi \alpha)|^k} |f^{(k)}|_1,$$

where $e(t) = e^{2\pi i t}$. The proof goes essentially by summation by parts.

(a) Are there older sources for this? Somewhat confusingly, Tao credits Gallagher ("The large sieve") and Lemma 1.1 in Montgomery's Topics in Multiplicative Number Theory, but they give only equation (3.1) in Tao's papers, not the inequality above.

(b) For $k=2$, this is not in general optimal: one can show

$$\left|\sum_{n\in \mathbb{Z}} f(n) e(\alpha n)\right|\leq \frac{1}{|2 \sin(\pi \alpha)|^2} |\widehat{f''}|_\infty,$$

which is no weaker and often strictly stronger, since $|\widehat{f''}|_\infty\leq |f''|_1$. This is Lemma 2.1 in my three-prime book draft on the arxiv; the proof I give goes by the Poisson summation formula, plus Euler's formula for the cotangent.

Are similar bounds true for general $k$? (Is $\left|\sum_{n\in \mathbb{Z}} f(n) e(\alpha n)\right| \leq |\widehat{f'}|_\infty/|2 \sin \pi \alpha|$, for instance?) Again, can such results be found in older sources?

• A partial answer to my own question: the bound I give for $k=2$ implies similar bounds for $k>2$. This goes as follows: by summation by parts, $$\sum_n (f(n+1)-f(n)) e(\alpha n) = \sum_n f(n) (e(\alpha n)- e(\alpha (n+1))) = (1-e(-\alpha)) \sum_n f(n) e(\alpha n);$$ hence, $\sum_n f(n) e(\alpha n) = \frac{1}{1-e(-\alpha)} \int_0^1 \left(\sum_n f'(n+t) e(\alpha n)\right) dt$. Now use the inequality for $k=2$, with $f'(x+t)$ instead of $f(x)$; we obtain that $|\sum_n f(n) e(\alpha n)|\leq |\widehat{f^{(3)}}|_\infty/|2 \sin(\pi \alpha)|^3$. Iterate to obtain the result for all $k>2$. – H A Helfgott Sep 14 '15 at 15:12
• In summary (given the answer below and other comments): we can do better than the above for $k\geq 4$ (and better than Tao for $k\geq 2$, and the proposed inequality for $k=1$ isn't true. But what about Tao's bound for $k=1$? Was this known before? (Note that it is precisely what one gets by applying bounds for $\sum e(\alpha n)$ (non-smoothed) over intervals - not that this is genuinely different from Tao's proof by summation by parts.) – H A Helfgott Sep 14 '15 at 23:17

By Poisson: $$\sum_{n \in \mathbb Z} f(n) e(\alpha n)= \sum_{m \in \mathbb Z} \hat{f} ( 2 \pi m + \alpha)$$

By the formula for the derivative of the Fourier transform:

$$= \sum_{m \in \mathbb Z} \frac{1}{ (2\pi i m + \alpha i)^k}\widehat{f^{(k)}} ( 2 \pi m + \alpha) \leq \sum_{m \in \mathbb Z} \frac{1} {\left| 2 \pi m + \alpha \right|^k}\left|\widehat{f^{(k)}}\right|_{\infty}$$

if $k >1$ then this sum converges and we may replace your constant by $\sum_{m \in \mathbb Z} \frac{1} {\left| 2 \pi m + \alpha \right|^k}$. I'm not sure if there's an analytic formula for this for various values of $k$, we can certainly estimate.

If $k=1$ then this sum diverges and this method fails to prove a bound. Does this mean that no bound is posssible? Yes. Simply take a Schwartz function $g$ that does something nice and smooth on $[-1,1]$, behaves like $1/|\xi|$ on $[1,n]$ and $[-n,-1]$, and then declines rapidly to $0$ outside $[-n,n]$. Then let $f$ be the inverse Fourier transform, also a Schwartz function, so $\hat{f}=g$. Then the Poisson summation formula is justified but:

$$\sum_{m \in \mathbb Z} \hat{f} ( 2 \pi m + \alpha) \approx \log n$$

$$|\widehat{ f'} |_\infty \approx 1$$

So no bound in terms of $|\widehat{ f'} |_\infty$ is possible.

• On $k=1$: nice! – H A Helfgott Sep 14 '15 at 21:52
• On $k\geq 2$: to obtain the bound I gave for $k=2$, I start as you do (I think we are normalizing the Fourier transform differently) and then I use the identity $\sum_{n=-\infty}^\infty \frac{1}{(n+s)^2} = \frac{\pi^2}{(\sin s\pi)^2}$ (a consequence of a formula of Euler's). – H A Helfgott Sep 14 '15 at 21:54
• Seems that Harald is working on binary Goldbach's conjecture now :-) – Sylvain JULIEN Sep 14 '15 at 21:56
• Ha-ha. Just to complete what I was saying: taking derivatives, we obtain $\sum_{n=-\infty}^{n=\infty} \frac{1}{(n+s)^4} =\frac{\pi^4}{(\sin s\pi)^4} (1 -\frac{2}{3} \sin^2 s \pi)$, and so on; hence, for $k=4$, $\frac{|\widehat{f^{(4)}}|_1}{(2 \sin \pi \alpha)^4}$ can be replaced by $\frac{1 - 2 (\sin \pi \alpha)^2/3}{(2 \sin \pi \alpha)^4} |\widehat{f^{(4)}}|_1$, and so on. – H A Helfgott Sep 14 '15 at 22:00
• If we drop the absolute values, Will's constant can be written in terms of the Hurwitz zeta function $\zeta(s,q)$ as $(2 \pi)^{-k} (\zeta(k,\alpha/2\pi)+(-1)^k\zeta(k,-\alpha/2\pi))-\alpha^{-k}$, and there are plenty of identities and estimates in the literature for this function. – Emanuele Tron Sep 14 '15 at 22:11