Undergrad Regarding Wirtinger's inequality

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The discussion centers on proving Wirtinger's inequality for a continuous function f on the interval (0, π) with boundary conditions f(0) = f(π) = 0 and f' in L²([0, π]). Participants explore the implications of f' being in L², questioning its necessity for integrability and the validity of their Fourier series approach. The solution involves using Fourier series to establish the inequality and identifying conditions for equality, specifically when f(x) = C sin(x). The conversation highlights the importance of square integrability and the Cauchy-Schwarz inequality in this context. Overall, the participants clarify the mathematical foundations necessary for the problem's solution.
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I have a doubt about about an assumption in Wirtinger's inequality.
Consider the following problem:
Let ##f## be a continuous real-valued function on ##0<x<\pi## such that ##f(0)=f(\pi)=0## and ##f' \in L^2([0,\pi])##.
a) Prove that $$\int_0^{\pi}(f(x))^2 dx \leq \int_0^{\pi} (f'(x))^2 dx.$$
b) For what functions does equality hold?
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?
 
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psie said:
In order for this solution to work, we need ##f'## to be integrable, right? Does ##f'\in L^2([0,\pi])## guarantee this?
We need square integrability of ##f'## to even phrase the statement and ##f'\in L^2## is just that.
 
fresh_42 said:
We need square integrability of ##f'## to even phrase the statement and ##f'\in L^2## is just that.
I oversaw something there :smile: 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?
 
psie said:
I oversaw something there :smile: 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?
Cauchy-Schwarz
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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