Define an inner product and show the sum of the inner product is bounded

Question

Here is the problem:

For f $\in$ C([0,2$\pi$]), define $a_k$ = $<f,sin(kx)>$, and show that for any integer N$\ge$1, $$\sum_{k=1}^N |a_k|^2\le||f||^2$$

Hint:

In class, I saw that {$\sin x, \sin2x, \sin3x, \dots$} is an orthonormal set in C([$0,2\pi$]) with respect to the inner product $<f,g>= \frac{1}{\pi}\int_{0}^{2\pi}f(x)g(x)dx$.

How would you prove the corresponding statement for an orthonormal set in a finite-dimensional vector space? Can the idea of that proof be adapted to the current situation?

Thank!

Answer 1

This is known as the Bessel Inequality. Let $F_n$ be the span generated by $(1,\sin x, \sin (2x), \dots , \sin (nx))$ then the orthogonal projection of $f$ on $F_n$ is the function $$P_{F_n}(f)=\sum_{k=1}^n \langle f|\sin(kx) \rangle \sin(kx)$$ moreover $P_{F_n}(f)$ is orthogonal to $f-P_{F_n}(f)$ for the product you defined therefore $$\|f\|^2=\|f-P_{F_n}(f)\|^2+\|P_{F_n}(f)\|^2 \geq \|P_{F_n}(f)\|^2$$ and because the family $(1,\sin x, \sin (2x), \dots , \sin (nx))$ is orthonormal $$\|P_{F_n}(f)\|^2=\sum_{k=1}^n \langle f|\sin(kx) \rangle^2$$ so the inequality holds

Answer 2

You can use the Cauchy-Schwarz inequality : you have $\langle f| \sin(kx) \rangle ^2 \leq \|f\|^2 \left(\frac{1}{\pi}\int_0^{2\pi}sin(kx)^2dx\right)$ and the last integral is in fact $\pi-\frac{\sin(4\pi k)}{4k}$ according to wolphram so by summing these inequalities for $k=1, \ 2, \ \dots \ , \ n$ I assume you get the inequality you wanted. Yet this inequality is more general and regards prehilbertian spaces I advice you to go and see the article on wikipedia on this subject, projection stuff is not very complicated if you studied classic linear algebra before. Have a look to the article on bessel inequality too ;) (also oon wikipedia)