A difficulty in understanding a solution .

Question

I have question(1) in the following picture:

and the solution to it is given in the following picture:

but I do not understand why $||f^{'}||_{2}^{2} \geq \sum |(e^{inx}/\sqrt{2\pi},f^{`})|^{2}$

Could anyone explain this for me?

Answer 1

This is called Bessel's inequality. A straightforward proof is given in the Wikipedia article, but it essentially hinges on $$ 0 \le \left\| x - \sum_{k=1}^n \langle x, e_k \rangle e_k\right\|^2 = \|x\|^2 - 2 \sum_{k=1}^n |\langle x, e_k \rangle |^2 + \sum_{k=1}^n | \langle x, e_k \rangle |^2 = \|x\|^2 - \sum_{k=1}^n | \langle x, e_k \rangle |^2, $$ where $e_k$ are orthonormal. Taking $n \to \infty$ gives the general result.

Answer 2

Let the space $S$ spanned by the given orthonormal set $V=\{v_n\}$, and $g$ be an element in $L^2$. Then $g=g_S+g_\perp$, where $g_S\in S$ and $g_\perp$ which can be obtained by the Gramm-Schmidt process. With some algebra with the inner product, we obtain $$\|g\|_2^2=\|g_S\|_2^2+\|g_\perp\|_2^2$$ and $$\|g\|_2^2\ge\|g_S\|_2^2=\sum \big|(v_n,g)\big|^2,$$ where the last step could be deduced easily by the othornormality of $V$.