TITLE: An inner product on $\mathcal{C}[a,b]$
QUESTION [11 upvotes]: I've to prove that the functional
$$\langle f,g\rangle = \int_{a}^{b} \int_{a}^{b} \frac{\sin(\pi(t-s))}{\pi (t-s)} f(s) \overline{g(t)}dsdt$$
is an inner product on $\mathcal{C}[a,b]$ (complex continuous functions).
I've already made a similar exercise where I shown that
$$\int_{a}^{b} f(t) \overline{g(t)} dt$$
is an inner product. And most of the properties follow the same reason. But I can't see how
$$\langle f,f\rangle = \int_{a}^{b} \int_{a}^{b} \frac{\sin(\pi(t-s))}{\pi (t-s)} f(s) \overline{f(t)}dsdt = 0 \iff f \equiv 0.$$
The implication $f=0 \Longrightarrow \langle f,f\rangle=0$ is clear from of the definition of the functional, but since $\frac{\sin(\pi(t-s))}{\pi (t-s)} = 0$ for $\pi(t-s) = n\pi$ I can't get the other implication. Same with $\langle f,f\rangle \geq 0.$

REPLY [8 votes]: Write
$$\langle f,f\rangle = \int_{a}^{b} \int_{a}^{b} \int_0^1 \cos(\pi u(t-s))du\,  f(s)\overline{f(t)}ds\,dt  \\
= \int_0^1 \left(\int_a^b  \cos(\pi u s) f(s) ds \int_a^b  \cos(\pi u t) \overline{f(t)} ds\right. \\ +\left. \int_a^b  \sin(\pi u s) f(s) ds \int_a^b  \sin(\pi u t) \overline{f(t)} ds \right)du \\
= \int_0^1 \left(\left|\int_a^b  \cos(\pi u s) f(s) ds\right|^2 + \left|\int_a^b  \sin(\pi u s) f(s) ds\right|^2 \right)du.$$
Then, $\langle f,f\rangle \ge 0$, and for $f\in C[a,b]$, $\langle f,f\rangle = 0$ iff
$$
\int_a^b  \cos(\pi u s) f(s) ds= \int_a^b  \sin(\pi u s) f(s) ds = 0 
$$
for all $u\in [0,1]$ (observing that both integrals are continuous in $u$). In other words, the Fourier transform $\hat g$ of $g(x) = f(x) \mathbf{1}_{[a,b]}$ vanishes on $[0,1]$. Since $\hat g$ is analytic, it vanishes everywhere, therefore, $f = 0$ a.e., whence $f \equiv 0$ due to continuity.