# What is wrong with my proof of Wirtinger inequality.

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 $$\int |f|^2 = \sum_{k\in{\mathbb{Z}\setminus \{0\}}} |\hat f(k)|^2\leq \left[ \frac{1}{4\pi^2}\sum_{k\in{\mathbb{Z}\setminus \{0\}}} \frac{1}{k^2}\right]\int |f'|^2 = \frac{1}{12}\int |f'|^2$$

• I think there is a problem in third line. we only have \begin{align} |\hat{f}(k)|^{2}= \frac{1}{4\pi^{2}k^{2}} \left|\int f'(x)e^{-i2\pi kx}dx\right|^{2}\leq \frac{1}{4\pi^{2}k^{2}}\left(\int |f'(x)|dx\right)^{2} \end{align} – Seewoo Lee Aug 11 '16 at 23:05
• I am using Jensen's inequality to bring the square from outside of the integral to the inside. – Xiao Aug 11 '16 at 23:09
• It is not true that $\hat{f}(0) = 0$ when $f$ is even. This requires $\int_{[-1/2,1/2]} f = 0$ (which is true if $f$ is odd). For example if $f(x) = \sin(\pi(x-1/2))$ then $f(-1/2) = f(1/2) = 0$ but $\hat{f}(0) = -\frac{2}{\pi} \not = 0$. – Winther Aug 12 '16 at 0:22
• @Winther nice catch, it is a typo, I meant odd function. The original idea was for a functions on $[a,b]$, we can work with $[0,\frac{1}{2}]$ and I want to extend the function to $[ - \frac{1}{2}, \frac{1}{2}]$ by $f(-x) = -f(x)$, so that $\hat f(0) = 0$. The "even" extension $f(-x) = f(x)$ does not guarantee $f\in C^1 [-\frac{1}{2}, \frac{1}{2}]$, – Xiao Aug 12 '16 at 0:27
• I can't find any mistakes in your proof. If I repeat the proof of Wirtinger in Jack's answer below under the assumptions that $f$ is odd with $f(-1/2) = f(1/2) = 0$ (so $f(x) = \sum a_n \sin(2\pi n x)$) then I get $\int f^2\leq \frac{1}{4\pi^2}\int_{-1/2}^{1/2}f'^2$ with equality for $f = \sin(2\pi x)$. If this is correct then your result using CS is weaker than this ($\frac{1}{4\pi^2} < \frac{1}{12}$) so the derivation could be fine. – Winther Aug 12 '16 at 1:06

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 $f(x)=\lambda\sin(x)$.