Schur's proof of Hilbert's inequality: streamlining? TL;DR: Is there a way to make Schur's (elegant) proof of Hilbert's inequality feel like
less of a trick/miracle?
Longer version: Let me go quickly over Schur's proof to show what I mean. Actually, let me do it for Montgomery and Vaughan's generalized Hilbert inequality, since the idea is the same, and let me also change it a little, so as to have fewer sums floating around. (Perhaps I've already taken the first step towards answering my question by asking it.)

Let $\{r_i\}_{i\in I}$, $r_i$ real, $I$ an index set, be given. Our task is to bound the norm of the operator $A = (a_{i,j})_{i,j\in I}$ given by
$$a_{i,j} = \frac{1}{r_i-r_j}$$
for $i\ne j$, $a_{i,i}=0$, under the assumption that $|r_i-r_j|\geq \delta$ for all distinct $i,j\in I$ and some $\delta>0$.
Write $A^2 = (b_{i,j})_{i,j\in I}$. Of course $b_{i,j} = \sum_{k\ne i,j} a_{i,k} a_{k,j}$, and so, for $i\ne j$,
$$\begin{aligned}b_{i,j} &= \sum_{k\ne i,j} \frac{1}{(r_i-r_k) (r_k - r_j)} =
\frac{1}{r_i-r_j} \sum_{k\ne i,j} \left(\frac{1}{r_i-r_k} + \frac{1}{r_k - r_j}\right)\\
&=\frac{1}{r_i-r_j} \left(\sum_{k\ne i} \frac{1}{r_i-r_k} + \sum_{k\ne j}\frac{1}{r_k - r_j}\right) - \frac{2}{(r_i-r_j)^2}
\end{aligned}$$
(the second equality here is the tiny miracle) whereas $b_{i,i} = - \sum_k
1/(r_i-r_k)^2$. In other words,
$$A^2 = J A - A J - D - S,$$
where $J$ and $D$ are diagonal matrices with $J_{i,i}=\sum_{k\ne i} 1/(r_i-r_k)$ and $D_{i,i} = \sum_{k\ne i} 1/(r_i-r_k)^2$, and $S$ is a matrix with $S_{i,j} = 1/(r_i-r_j)^2$ for $i\ne j$ and $S_{i,i}=0$.
Because $A$ is real antisymmetric, the spectral theorem holds. Hence, to bound the norm of $A$,
it is enough to bound $|A v|_2/|v|_2$ for all eigenvectors $v$ of $A$. Again because $A$ is real antisymmetric, all of its eigenvalues are pure imaginary. Hence, for $v$ an eigenvector with eigenvalue $\lambda$,
$$\langle v, J A v\rangle  - \langle v, A J v\rangle =
\lambda \langle v, J v\rangle + \langle A v, J v\rangle =
(\lambda + \overline{\lambda}) \langle v, J v\rangle = 0,$$
and so $$|A v|_2^2 = -\langle v, A^2 v\rangle = \langle v, D v\rangle + \langle v, S v\rangle.$$
The norm of $D$ is clearly $\max_i |D_{i,i}| =  \max_i \sum_{k\ne i} 1/(r_i-r_k)^2
\leq 2 \zeta(2)/\delta^2 = \pi^2/3 \delta^2$, whereas the norm of $S$ is
$\leq \max_i \sum_j |S_{i,j}| = \max_i \sum_{j\ne i} 1/(r_i-r_j)^2\leq  \pi^2/3 \delta^2$. Hence,
$$|A v|_2^2  \leq \frac{\pi^2}{\delta^2} |v|_2^2$$
for any eigenvector $v$ of $A$. We conclude that the norm of $A$ is
$$\leq \frac{\pi}{\delta}.$$

So, questions:
(a) Can this variant of the proof be shortened further, or made more natural?
(b) The above was closely inspired by Montgomery-Vaughan and the exposition in Iwaniec-Kowalski. I take the above may literally appear elsewhere?
(c) Are there other interesting matrices for which we have expressions such as $A^2 = J A - A J - D - S,$ with similar consequences?
 A: When I was a graduate student, I worked out a more simple direct variant of this argument avoiding any use of matrics, bilinear forms, etc. It's so simple I can't imagine I was the first to do this.
https://lewko.wordpress.com/2009/08/31/the-sharp-large-sieve-inquality
A: One can even attempt the same argument to attempt to derive a bound of the form
$$\left|\sum_{i\ne j} \frac{\overline{v_i} v_j}{r_i-r_j}
\right| \leq C \sum_i \delta_i^{-1} |v_i|^2,$$
where $\delta_i = \max_{j\ne i} |r_j-r_i|$. Bounds of this form are known from Montgomery-Vaughan (with $C = (3/2) \pi$) and Preissmann (with $C = (4/3) \pi$; still suboptimal). The optimal value is conjectured to be $C=\pi$; there were rumors of an unpublished proof with $C=3.2$, but apparently they are unsubstantiated so far (see Does a proof of Selberg's 3.2 inequality exist?).

Our task is to bound the norm of the operator $A = (a_{i,j})_{i,j\in I}$ given by
$$a_{i,j} = \frac{\delta_i}{r_i-r_j}$$
for $i\ne j$, $a_{i,i}=0$, under the assumption that $|r_i-r_j|\geq \delta_i$ for all distinct $i,j\in I$. The norm here is to be understood with respect to the inner product $\langle v,w\rangle = \sum_{i} \delta_i^{-1} \overline{v_i} w_i$, both for the domain and for the image.
Write $A^2 = (b_{i,j})_{i,j\in I}$. Of course $b_{i,j} = \sum_{k\ne i,j} a_{i,k} a_{k,j}$, and so, for $i\ne j$,
$$\begin{aligned}b_{i,j} &= \sum_{k\ne i,j} \frac{\delta_i \delta_k}{(r_i-r_k) (r_k - r_j)} =
\frac{\delta_i}{r_i-r_j} \sum_{k\ne i,j} \left(\frac{\delta_k}{r_i-r_k} + \frac{\delta_k}{r_k - r_j}\right)\\
&=\frac{\delta_i}{r_i-r_j} \left(\sum_{k\ne i} \frac{\delta_k}{r_i-r_k} + \sum_{k\ne j}\frac{\delta_k}{r_k - r_j}\right) - \frac{\delta_j^2 + \delta_i \delta_j}{(r_i-r_j)^2}
\end{aligned}$$
(the second equality here is the tiny miracle) whereas $b_{i,i} = - \sum_k \delta_i \delta_k/(r_i-r_k)^2$. In other words,
$$A^2 = J A - A J - D - S,$$
where $J$ and $D$ are diagonal matrices with $J_{i,i}=\sum_{k\ne i} \delta_k/(r_i-r_k)$ and $D_{i,i} = \sum_{k\ne i} \delta_i \delta_k/(r_i-r_k)^2$, and $S$ is a matrix with $S_{i,j} = (\delta_i \delta_j + \delta_i^2)/(r_i-r_j)^2$ for $i\ne j$ and $S_{i,i}=0$.
Perhaps contrary to appearances, $A$ is real antisymmetric with respect to our inner product:
$$\langle v, A w\rangle = 
\sum_{i,j} \delta_i^{-1} \frac{\overline{v_i} \delta_i w_j}{r_i-r_j} = -
\sum_{i,j} \delta_j^{-1} \frac{\overline{\delta_j v_i} w_j}{\overline{r_j-r_i}} = -\langle A v, w\rangle,
$$
and so the spectral theorem holds. Hence, to bound the norm of $A$, it is enough to bound $|A v|_2/|v|_2$ for all eigenvectors $v$ of $A$. Again because $A$ is real antisymmetric, all of its eigenvalues are pure imaginary. Hence, for $v$ an eigenvector with eigenvalue $\lambda$,
$$\langle v, J A v\rangle  - \langle v, A J v\rangle =
\lambda \langle v, J v\rangle + \langle A v, J v\rangle =
(\lambda + \overline{\lambda}) \langle v, J v\rangle = 0,$$
and so $$|A v|_2^2 = -\langle v, A^2 v\rangle = \langle v, D v\rangle + \langle v, S v\rangle.$$
The norm of $D$ is clearly $$\max_i |D_{i,i}| =  \max_i \sum_{k\ne i} \frac{\delta_i \delta_k}{(r_i-r_k)^2}
\leq 2\cdot \left(1 + \max_{1<x_1<x_2<\dotsc} \sum_{n=1}^\infty \frac{x_{n+1}-x_n}{x_{n+1}^2}\right)\leq 4,$$ since $1/x^2 = - (1/x)'$ and $x\mapsto 1/x$ is convex.
It only remains to bound $\langle v,S v\rangle$. Here I am a bit stuck. It is not as if there weren't plenty to work with -- $S$ is symmetric and positive definite with respect to $\langle \cdot,\cdot\rangle$, and so we can assume that $v$ is an eigenvector of $S$. (We can assume instead that $v$ is an eigenvalue of $A$, for that matter.)
If only we could bound the norm of $S$ by $2 \max_i \sum_{k\ne i} \frac{\delta_i \delta_k}{(r_i-r_k)^2}$, we would have a total bound of $|A v|_2^2\leq \frac{12}{\delta^2} |v|_2^2$, and so we would be able to conclude that the norm of $A$ is $\leq \sqrt{12}/\delta$; in other words, we would have
$$\left|\sum_{i\ne j} \frac{\overline{v_i} v_j}{r_i-r_j}
\right| \leq C \sum_i \delta_i^{-1} |v_i|^2$$
with $C = \sqrt{12} < 4\pi/3$.
