What is wrong with my proof of Wirtinger inequality.

Question

Given an odd function in $f\in C^1[-\frac{1}{2}, \frac{1}{2}]$ with $f(-\frac{1}{2})=f(\frac{1}{2})= 0$, I want to show $$\int |f|^2 \leq \frac{1}{\pi^2} \int |f'|^2 .$$ Since $f$ is even, $\hat f(0) = 0$. For $k\neq 0$, from integration by parts we have $$\hat f(k) = \int f(x)e^{-i2\pi kx} dx = -\int f'(x) \frac{e^{-i2\pi kx}}{-i2\pi k} dx $$ and $$|\hat f(k)|^2 = \left| -\int f'(x) \frac{e^{-i2\pi kx}}{-i2\pi k} dx\right|^2\leq \frac{1}{4\pi^2k^2} \int |f'(x)|^2 dx $$ From Parseval, we have $$\int |f|^2 = \sum_{k\in{\mathbb{Z}\setminus \{0\}}} |\hat f(k)|^2\leq \left[ \frac{1}{4\pi^2}\sum_{k\in{\mathbb{Z}\setminus \{0\}}} \frac{1}{k^2}\right]\int |f'|^2 = \frac{1}{12}\int |f'|^2 $$

Answer 1

By translation, we may assume to have a $C^1$ function over the interval $[0,1]$ with the property that $f(0)=f(1)=0$ and $f(x)=f(1-x)$. Such function has an expansion as a trigonometric polynomial of the form $$ f(x)=\sum_{n\geq 1}a_n \sin((2n-1)\pi x) \tag{1}$$ and by Parseval's identity $$ \|f\|_2^2 = \frac{1}{2}\sum_{n\geq 1} a_n^2,\qquad \|f'\|_2^2 = \frac{\pi^2}{2}\sum_{n\geq 1}a_n^2 (2n-1)^2 \tag{2} $$ hence $$ \|f'\|_2^2 \geq \pi^2 \|f\|_2^2\tag{3} $$ with equality attained only by $f(x)=\lambda\sin(x)$.