# Proof of Wirtinger inequality

Quoting from Ana Cannas da Silva's book on Symplectic Geometry: "As an exercise in Fourier series, show the Wirtinger inequality: for $f\in C^1([a,b])$, with $f(a)=f(b)=0$ we have $$\int_a^b\Big|\frac{\mathrm{d}f}{\mathrm{d}t}\Big|^2\mathrm{d}t \ge\frac{\pi^2}{(b-a)^2} \int_a^b\left|\ f\right|^2\mathrm{d}t."$$ I already found a few questions about this topic in the site, but I couldn't actually grasp what's happening here. Also, I would very much like you to show me where I go wrong with my try, which is sketched below.

I know that, for $f\in \mathcal{L}^2([0,2\pi])\supset C^1([0,2\pi])$, I can expand: $$f(t)=\sum_{n=-\infty}^{+\infty}c_n e^{int},\ \ \ c_n=\frac{1}{2\pi}\int_0^{2\pi}\mathrm{d}t\ e^{-int}f(t).$$ Rescaling $t \to \omega (t - a)$, where $\omega = 2\pi/(b-a)$, we can get a more general form for $f\in C([a,b])$: $$f(t)=\sum_{n=-\infty}^{+\infty}c_n e^{in\omega t},\ \ \ c_n=\frac{1}{b-a}\int_a^{b}\mathrm{d}t\ e^{-in\omega t}f(t).$$ Now, having: $$\frac{\mathrm{d}}{\mathrm{d}t}f(t) = \sum_{n=-\infty}^{+\infty}\tilde{c}_ne^{i\omega nt},\ \ \tilde{c}_n=\frac{1}{b-a}\int_a^b\mathrm{d}t\ e^{-i\omega nt}\frac{\mathrm{d}}{\mathrm{d}t}f(t).$$ Using the fact that $\,f(a)=f(b)=0$ we get: $$\tilde{c}_0 = \frac{1}{b-a}\int_a^b\mathrm{d}t \frac{\mathrm{d}}{\mathrm{d}t} f(t) = \frac{f(b)-f(a)}{b-a} =0 \longrightarrow \frac{\mathrm{d}}{\mathrm{d}t}f(t) = \sum_{n\in\mathbb Z\setminus\{0\}}\tilde{c}_n e^{i\omega nt}$$ Now deriving the series expansion of $f$ yields: $$\frac{\mathrm{d}}{\mathrm{d}t}f(t) = \frac{\mathrm{d}}{\mathrm{d}t} \sum_{n=-\infty}^{+\infty}c_n e^{in\omega t} = \sum_{n=-\infty}^{+\infty}(in\omega)c_n e^{in\omega t}=\sum_{n\in\mathbb Z\setminus\{0\}}(in\omega)c_n e^{in\omega t}$$ Comparing the two expressions we establish: $\tilde{c}_n = i\omega n c_n$, for $n\not= 0$. Parseval's Equality reads here for $\mathrm{d}f/\mathrm{d}t$: $$\int_a^b\left|\frac{\mathrm{d}f}{\mathrm{d}t}\right|^2\mathrm{d}t= \sum_{n\in\mathbb Z\setminus\{0\}} |\tilde{c_n}|^2 = \omega^2\sum_{n\in\mathbb Z\setminus\{0\}} n^2 |c_n|^2 \ge \omega^2\sum_{n\in\mathbb Z\setminus\{0\}} |c_n|^2 = \omega^2 \left(\int_a^b\mathrm{d}t\left|f(t)\right|^2-|c_0|^2\right)$$ where in the last passage we used Parseval's Equality for $f$: $\int_a^b|f|^2\mathrm{d}t=\sum_{n=-\infty}^{+\infty}|c_n|^2$.

We are now left with finding a suitable way of estimating $|c_0|^2$: $$|c_0|^2 = \left|\frac{1}{b-a}\int_a^b\mathrm{d}t\ f(t)\right|^2,$$ thus $$\int_a^b\left|\frac{\mathrm{d}f}{\mathrm{d}t}\right|^2\mathrm{d}t \ge \frac{4\pi^2}{(b-a)^2}\left(\int_a^b\left|f\right|^2\mathrm{d}t - \frac{1}{(b-a)^2}\left|\int_a^bf\mathrm{d}t\right|^2\right).$$ Which is not exactly what I wanted.

• do you know the Parseval formula? Commented Mar 6, 2014 at 20:17
• Yes, that's the equality corresponding to Bessel Inequality when considering a complete orthonormal set as a basis. Commented Mar 6, 2014 at 20:19
• and express the Fourier coefficients of the derivative in function of those of the function, and you are done. Commented Mar 6, 2014 at 20:29
• I get $\sum |c'_n|^2=\omega^2\sum n^2 |c_n|^2$ but I'm still getting an additional factor 4. Commented Mar 6, 2014 at 20:56
• @mookid please, could you have a look at how I tried to apply your hint? :) (I edited my question) Commented Mar 7, 2014 at 9:42

The standard Wirtinger's inequality requires that $$\int_a^b f\,dx=0$$, which implies that $$c_0=0$$, and hence the difficulty you are encounter does not exist.

If instead we assume that $$f(a)=f(b)=0$$, then we exploit this by expanding $$f$$ in a sine series, i.e., for $$a=0$$ and $$b=\pi$$, $$f(x)=\sum_{n=1}^\infty a_n\sin nx,$$ and hence $$\int_0^\pi f^2= \frac{\pi}{2}\sum_{n=1}^\infty a_n^2 \qquad \text{while} \qquad \int_0^\pi (f')^2= \frac{\pi}{2}\sum_{n=1}^\infty n^2a_n^2,$$ and therefore $$\int_0^\pi (f')^2\ge \int_0^\pi f^2,$$ with best constant $$c=1$$.

For arbitrary $$b>a$$, $$f(x)=\sum_{n=1}^\infty a_n\sin \left(\frac{n\pi(x-a)}{b-a}\right),$$ and hence $$\int_a^b\!\! f^2= \frac{b-a}{2}\sum_{n=1}^\infty a_n^2 \quad \text{while} \quad \int_a^b\!\! (f')^2= \frac{b-a}{2}\sum_{n=1}^\infty \left(\frac{\pi}{b-a}\right)^2n^2a_n^2,$$ and hence $$\int_a^b\! (f')^2\ge \frac{\pi^2}{(b-a)^2}\int_0^\pi f^2.$$

• Thanks a lot. Only...I don't think I get exactly where all the extra factors come from! Commented Mar 7, 2014 at 10:24
• @Yiorgos, why $\int_a^b f = 0$ is equivalent to $f(a) = f(b) = 0$ and where you used that $f(a) = f(b) = 0$ in your proof? Commented Oct 25, 2017 at 3:27
• @George It is not equivalent, of course! Commented Oct 25, 2017 at 4:22
• @YiorgosS.Smyrlis, but why you assume that $f(a) = f(b) = 0$ in place of $\int_a^b f = 0$ and where you used that $f(a) = f(b) = 0$ in your proof? Commented Oct 25, 2017 at 6:36
• @Brightsun A function in $[0,\pi]$ with $f(0)=f(\pi)=0$ can be always viewed as a $2\pi-$periodic function of odd parity Commented May 31, 2018 at 12:03

Here is a proof not involving Fourier Series at all.

Since $f(a) = 0$ we can write, using the fundamental theorem of calculus: $$f(t) = \int_a^t \frac{\mathrm{d}f(\tau)}{\mathrm{d}\tau}\mathrm{d}\tau,\ \ \ t\in[a,b],$$ hence $$\left|f(t)\right| \le \int_a^t \left|\frac{\mathrm{d}f(\tau)}{\mathrm{d}\tau}\right|\mathrm{d}{\tau}.$$ Cauchy-Schwarz inequality ensures: $$\int_a^t \left|\frac{\mathrm{d}f(\tau)}{\mathrm{d}\tau}\right|\mathrm{d}{\tau} \le \left(\int_a^t \left|\frac{\mathrm{d}f(\tau)}{\mathrm{d}\tau}\right|^2\mathrm{d}{\tau}\right)^{1/2}\cdot \left(\int_a^t\mathrm{d}\tau\right)^{1/2}\\ \hspace{2.7cm}\le \left(\int_a^b \left|\frac{\mathrm{d}f(\tau)}{\mathrm{d}\tau}\right|^2\mathrm{d}{\tau}\right)^{1/2}\cdot \left(\int_a^b\mathrm{d}\tau\right)^{1/2} \\ \hspace{2cm} =(b-a)^{1/2}\left(\int_a^b \left|\frac{\mathrm{d}f(\tau)}{\mathrm{d}\tau}\right|^2\mathrm{d}{\tau}\right)^{1/2}.$$ Thus $$\left|f(t)\right|^2 \le (b-a)\int_a^b \left|\frac{\mathrm{d}f(\tau)}{\mathrm{d}\tau}\right|^2\mathrm{d}{\tau}$$

$$\int_a^b\left|f(t)\right|^2\mathrm{d}t \le (b-a)^2 \int_a^b \left|\frac{\mathrm{d}f(\tau)}{\mathrm{d}\tau}\right|^2\mathrm{d}{\tau}$$ $$\int_a^b \left|\frac{\mathrm{d}f(\tau)}{\mathrm{d}\tau}\right|^2\mathrm{d}{\tau} \ge \frac{1}{(b-a)^2}\int_a^b \left|f(t)\right|^2\mathrm{d}t,$$ which is a slightly less strong result.

EDIT: The fact that $f(a)=0=f(b)$ can be incorporated as follows. Let us first assume for simplicity that $a=0$ and $b=1$ and denote $||\varphi||^2=\int_0^1 \varphi(t)^2 dt$. $$\varphi(t)=\sum_{\mathbb Z} e^{i2\pi n t} c_n\,,\qquad c_n=\int_0^1 \varphi(t) e^{-i2\pi nt}dt\,,$$ then $$||\varphi||^2=\sum_{\mathbb Z} |c_n|^2= |c_0|^2 + \sum_{\mathbb Z\setminus \{0\}} |c_n|^2\,,\qquad ||\dot \varphi||^2=4\pi^2 \sum_{\mathbb Z\setminus\{0\}} n^2 |c_n|^2\ge 4\pi^2 \sum_{\mathbb Z\setminus \{0\}} |c_n|^2\,.$$ On the other hand $\varphi(0)=0=\varphi(1)$ requires $$0=\sum_{\mathbb Z} c_n = c_0 + \sum_{\mathbb Z\setminus\{0\}} c_n$$ and using this equation to get rid of $c_0$ we obtain $$||\varphi||^2 = \Big|\sum_{\mathbb Z\setminus\{0\}} c_n \Big|^2+\sum_{\mathbb Z\setminus \{0\}} |c_n|^2 \le 2 \sum_{\mathbb Z\setminus \{0\}} |c_n|^2\,.$$ Comparing the two inequalities $$||\dot\varphi||^2 \ge 2\pi^2 ||\varphi||^2\,.$$ More generally, reinstating the dependence on the interval, $$\int_a^b \dot f(t)^2 dt \ge \frac{2\pi^2}{(b-a)^2}\int_a^b f(t)^2dt\,.$$ Curiously enough, this approach gives an extra factor of two, unless I am missing something.

• This is not weaker. It does not require $f(b) = 0$. Commented Jan 13, 2015 at 1:12
• Can anyone enlighten me as to why, on the second to last inequality he multiplies by (b-a) on both sides and on the LHS he puts the |f(t)|² inside the integral of d/tau from a to b? Why can one do this? Commented Nov 27, 2017 at 20:07
• In the next-to-last line I integrated both sides w.r. to $dt$ from $a$ to $b$, but since the RHS does not depend on $t$, this amounts to multiplying it by $b-a$ Commented Nov 28, 2017 at 22:21