Details on a proof of Ingham Theorem In the proof of Ingham Theorem in the book "Control of Wave and Beam PDEs: The Riesz Basis Approach", page 61:

$$ \int_{-\pi}^{\pi}|f(t)|^{2} d t \geq \sum_{m, n}
 K\left(\lambda_{m}-\lambda_{n}\right) a_{m} \overline{a_{n}} $$ Let
$S_{1}$ be part of the sum for which $m=n$ and $S_{2}$ the remaining
part. Then $$ S_{1}=4 \sum_{n}\left|a_{n}\right|^{2} $$ since $K(x)$
is even and $2\left|a_{m} \overline{a_{n}}\right|
 \leq\left|a_{m}\right|^{2}+\left|a_{n}\right|^{2},$ there is a
constant $\theta$ such that $|\theta| \leq 1$ and $$ S_{2}=\theta
 \sum_{m \neq n}
 \frac{\left|a_{m}\right|^{2}+\left|a_{n}\right|^{2}}{2}\left|K\left(\lambda_{m}-\lambda_{n}\right)\right|=\theta
 \sum_{n}\left|a_{n}\right|^{2} \sum_{m, m \neq n}\left|K\left(\lambda_{m}-\lambda_{n}\right)\right| $$

I was wondering if someone could help me in the following two issues concerns the estimate of the $S_{2}$ part:
(1). At the end of the page it says that " there is a constant $\theta$ such that $|\theta| \leq 1$ and...", I don't understand how this comes true?
(2). How the sum $\sum\limits_{m\neq n}$ is replaced by
$\sum\limits_{m,m\neq n}$? It is like fixing $n$ and sum over all $m\neq n$, where we have a double sum!
 A: For the first question about $\theta$, we can think about the following sort of general principle: if $A$ and $B$ are quantities for which $|A| \leq B$, then there is a $\theta$ with $|\theta| \leq 1$ such that $A = \theta B$. In particular, as long as $B \neq 0$ the value of $\theta$ is just $A/B$. (If $B = 0$, we can take $\theta$ to be any complex number with magnitude not more than $1$.)
For the specific situation here, we have $A = S_2$, and
$$B = \sum_{m \neq n} \frac{|a_m|^2 + |a_n|^2}{2} | K(\lambda_m - \lambda_n) |.$$
The statement in the proof that $2 | a_m \overline{a_n}| \leq |a_m|^2 + |a_n|^2$ then justifies (together with the triangle inequality) why $|A| \leq B$.
As for the second question relating to the sum, we can work through the simplification as follows:
\begin{align}
\sum_{m \neq n} &\frac{|a_m|^2 + |a_n|^2}{2} | K(\lambda_m - \lambda_n) | \\\\
&= \sum_m \sum_{n : n \neq m} \frac{|a_m|^2}{2} | K(\lambda_m - \lambda_n) |
+ \sum_n \sum_{m : m \neq n} \frac{|a_n|^2}{2} | K(\lambda_m - \lambda_n) | \\\\
&= \sum_n \sum_{m : m \neq n} \frac{|a_n|^2}{2} | K(\lambda_n - \lambda_m) |
+ \sum_n \sum_{m : m \neq n} \frac{|a_n|^2}{2} | K(\lambda_m - \lambda_n) || \\\\
&= \sum_n \sum_{m : m \neq n} \frac{|a_n|^2}{2} | K(\lambda_m - \lambda_n) |
+ \sum_n \sum_{m : m \neq n} \frac{|a_n|^2}{2} | K(\lambda_m - \lambda_n) || \\\\
&= \sum_n \sum_{m : m \neq n} |a_n|^2 | K(\lambda_m - \lambda_n) | \\\\
&= \sum_n |a_n|^2 \sum_{m : m \neq n} | K(\lambda_m - \lambda_n) |.
\end{align}
At the second step we just renamed the indices, and at the third step we used the fact that $K$ is even.
