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- TL;DR Summary
- I have a doubt about about an assumption in Wirtinger's inequality.
Consider the following problem:
In order for this solution to work, we need ##f'## to be integrable, right? Does ##f'\in L^2([0,\pi])## guarantee this?
What I struggle with in this exercise is why ##f'## is in ##L^2([0,\pi])##. Does this make sense? The way I prove this inequality is that I extend ##f## to an odd function on ##[-\pi,\pi]##. I find its Fourier series, namely ##f(x)\sim\sum_{n=1}^\infty b_n\sin nx##. Then, since ##f'## will be even, its Fourier coefficients are $$a_n=\frac{2}{\pi}\int_0^\pi f'(x)\cos(nx)dx=\frac{2n}{\pi }\int_0^\pi f(x)\sin(nx)=nb_n,$$ since ##f(0)=f(\pi)=0## by partial integration. Then I use Parseval's formula $$\int_0^\pi f^2= \frac{\pi}{2}\sum_{n=1}^\infty b_n^2 \quad \text{and} \quad \int_0^\pi (f')^2= \frac{\pi}{2}\sum_{n=1}^\infty n^2b_n^2.$$ And the inequality follows from this. Equality holds for ##a_n=0## for ##n\neq 1##, i.e. when ##f(x)=C\sin x##.
In order for this solution to work, we need ##f'## to be integrable, right? Does ##f'\in L^2([0,\pi])## guarantee this?
In my solution, however, I also need ##a_n=\frac2{\pi}\int_0^\pi f’(x)\cos(nt)dt## to be well-defined. Does ##f’\in L^2## ensure this?