In the book Controllability of Partial Differential Equations by Enrique Zuazua page 110 I have found a problem to prove the last equality in the last line formula $$\sum\limits_{n,m} {{a_n}{{\bar a}_m}K({\mu _n} - {\mu _m})}. $$ I tried to use the Fourier transform of $e^{ixt}$ in $S'(R)$ to prove it but I found $$\sum\limits_{n,m} {{a_n}{a_m}K({\mu _n} + {\mu _m})}. $$ (With plus sign). Any Ideas? Thank you.
Define the function $$ h \colon \mathbb{R} \to \mathbb{R}, \quad h(t) = \begin{cases} \cos(t/2) & \text{if $|t| \leq \pi$}, \\ 0 & \text{if $|t| > \pi$}. \end{cases} $$ and let us compute its Fourier transform $K(\varphi)$, $$ K(\varphi) = \int_{-\pi}^{\pi} h(t) e^{i t \varphi} \,\mathrm{d}t = \int_{-\infty}^\infty h(t) e^{i t \varphi} \,\mathrm{d}t = \frac{4 \cos \pi \varphi}{1 - 4 \varphi^2}. $$ On the other hand, since $0 \leq h(t) \leq 1$ for any $t \in [-\pi, \pi]$, we have that $$ \int_{-\pi}^\pi \left| \sum_n a_n e^{i \mu_n t} \right|^2 \,\mathrm{d}t \geq \int_{-\pi}^\pi h(t) \left| \sum_n a_n e^{i \mu_n t} \right|^2 \,\mathrm{d}t = \sum_{n,m} a_n \overline{a}_m K(\mu_n - \mu_m) = $$
(Original picture of the problem here.)
We have :
$$\int_{-\pi}^\pi h(t) \left\vert \sum_n a_n e^{i\phi_nt} \right\vert^2 dt=\int_{-\pi}^\pi h(t) \sum_n a_n e^{i\phi_nt} \overline{\sum_m a_m e^{i\phi_mt}}dt.$$ Since :
$$\sum_n a_n e^{i\phi_nt} \overline{\sum_m a_m e^{i\phi_mt}}=\sum_{n,m} a_n \overline{a_m} e^{i(\phi_n-\phi_m)t}.$$
Then :
$$\int_{-\pi}^\pi h(t) \left\vert \sum_n a_n e^{i\phi_nt} \right\vert^2 dt=\sum_{n,m} a_n \overline{a_m}\int_{-\pi}^\pi h(t) e^{i(\phi_n-\phi_m)t}dt=\sum_{n,m} a_n \overline{a_m}K(\phi_n-\phi_m).$$