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

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

1 Answer 1

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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.

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  • $\begingroup$ (+1) All too easy. $\endgroup$
    – Mark Viola
    Sep 24, 2020 at 22:40

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