# This finite sum involving roots of unity is bounded, but why?

For $$N \in \mathbb{N}$$, let $$\omega := \exp(\frac{2 \pi}{N} \sqrt{-1} )$$. For $$k,l \in \{1, \dots, N \}$$, define

$$d_{k,l}^{(N)} := \begin{cases} \frac{\sin\left(\frac{2\pi}{N}(k-1)\right)}{\sin\left(\frac{2\pi}{N}(k-1)\right)^2 + \sin\left(\frac{2\pi}{N}(l-1)\right)^2}&\mbox{ if } \sin\left(\frac{2\pi}{N}(k-1)\right) \neq 0\\ 0 & \mbox{ otherwise } \end{cases}$$

For $$i,j,m,n \in \{1, \dots, N\}$$, let

$$a_{i,j,m,n}^{(N)} :=\frac{1}{N^2} \sum_{l=1}^N \sum_{k=1}^N (\omega^{i-j})^{k-1} (\omega^{m-n})^{l-1} d^{(N)}_{k,l}$$

Numerical experiments very strongly suggest that there exists $$C>0$$ independent of $$N$$ ($$C \approx 0.25$$) such that $$\forall N$$, $$\max_{1 \le i,j,m,n \le N} | a^{(N)}_{i,j,m,n} | \le C$$

How can one formally show that this is true?

I eventually found a (fairly lengthy) proof of this fact. I will not write it in detail, but I sketch the ideas, in hopes somebody might eventually find them helpful -- I would be happy to clarify any question in the comments. Let for convenience $$p := i-j$$, $$q := m-n$$.

1. Using Euler's formula $$e^{ix} = \cos(x) + i\sin(x)$$, expand the terms $$\omega^{p(k-1)}$$ and $$\omega^{q(l-1)}$$ as sines and cosines.

2. The sum can then be identified with four Riemann sums, for integrals $$\int_{y=0}^{2\pi}\int_{x=0}^{2\pi} \frac{\sin(x)}{\sin(x)^2 + \sin(y)^2 }h^x(px)h^y(qy) \,dx\, dy$$ for $$h^x, h^y \in \{ \sin, \cos\}$$, up to multiplicative constants.

3. A simple argument of parity shows that three of these four integrals are zero, and the only nonzero is: $$I_{p,q} := \int_{y=0}^{2\pi}\int_{x=0}^{2\pi} \frac{\sin(x)}{\sin(x)^2 + \sin(y)^2 }\sin(px)\cos(qy) \,dx\, dy$$

4. I refer to user TheSimpliFire's excellent proof here that this integral tends to zero as $$q\to\infty$$. A completely analogous argument shows that it also tends to zero as $$p\to\infty$$.

5. A careful application of Riemann-Lebesgue's lemma to the integrals in the proof above determines that the rate of convergence to zero is $$I_{p,q} = O(p^{-1}q^{-2})$$, and $$I_{p,0} = O(p^{-1})$$. The asymmetry with respect to $$p$$ and $$q$$ is caused by the following fact: if one takes the innermost integral to be the one with respect to $$x$$, as in the aforementioned proof, then eventually one arrives to an integral of the form $$\int_0^{2\pi} g(y)\cos(qy)\,dy$$, with $$g$$ smooth and such that $$g(0) = g(2\pi)$$. Riemann-Lebesgue thus gives a bound of $$O(q^{-2})$$. Conversely, integrating first by $$y$$, one reaches an integrand of the form $$\int_0^{2\pi} f(x)\sin(px) \,dx$$ with $$f$$ smooth but $$f(0)\neq f(2\pi)$$; Riemann-Lebesgue only gives a $$O(p^{-1})$$ bound.

6. The integrands $$\frac{\sin(x)}{\sin(x)^2 + \sin(y)^2 }h^x(px)h^y(qy)$$ are (smooth) periodic functions, integrated over one period. Periodic functions display accelerated convergence of its Riemann sums (e.g. see Guillemin and Stroock's "Some Riemann sums are better than others"); so the error of the Riemann sums, $$\epsilon_N$$, decays faster than any polynomial of $$N (\ge |p|,|q|)$$, and can effectively be ignored.

7. In conclusion, if $$m\neq n$$, $$a_{i,j,m,n}^{(N)} = \begin{cases}\epsilon_N + O(|i-j|^{-1}|m-n|^{-2}) &\text{ if }m\neq n \\ \epsilon_N + O(|i-j|^{-1}) &\text{ if } m=n,\,i\neq j\\ \epsilon_N &\text{ if } m=n,\,i=j\end{cases}$$

Which is obviously bounded for all $$i,j,m,n$$.