# An inner product on $\mathcal{C}[a,b]$

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 $$\langle f,f\rangle \geq 0.$$

• I'd be inclined to try a Fourier transform here. Sep 24, 2020 at 13:51
• @CameronWilliams we don't know anything about Fourier in this course I'm taking Sep 24, 2020 at 14:37
• @supinf I think OP means they don't know how to do that part. Sep 24, 2020 at 14:54
• Here's a suggestion: try polynomials. By Stone-Weierstrass, we know that they are linearly dense in $C([a,b])$. Sep 24, 2020 at 14:55

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.

• (+1) All too easy. Sep 24, 2020 at 22:40