I am looking for proofs of the (Poincare-) Wirtinger inequality which states that if $f:[0,\pi]\to \mathbb{C}$ is $C^1$ and $f(0)=f(\pi)=0$ then \begin{equation} \int_0^\pi |f(t)|^2 dt \leq \int_0^\pi |f'(t)|^2 dt. \end{equation} See link.

The proof that I know starts by proving that if $$ \int_0^{2\pi} F(t) dt =0 $$ then $$ \int_0^{2\pi} |F(t)|^2 dt \leq \int_0^{2\pi} |F'(t)|^2 dt. $$ using Parseval's identity. From this, one proves the desired inequality for $f$ on $[0,\pi]$ by "extending" $f$ to an odd $C^1$ function on $[-\pi,\pi]$.

Are there other proofs? (Straightforward or otherwise...)


If you are willing to get a non-sharp constant, here's another proof found in many differential geometry texts. Without loss of generality assume $f \geq 0$. (Replacing $f$ by $|f|$ doesn't change the integrals on either side, if $f$ is assumed to be $C^1$.)

Let $2M = \sup f$, and let $t_0 \in (0,\pi)$ attain this maximum.

Let $X(t) = f(t) - M$ and $Y(t) = \sqrt{M^2 - X(t)^2}$ if $t \leq t_0$ and $-\sqrt{M^2 - X(t)^2}$ if $t \geq t_0$.

We have that $(X(t),Y(t))$ lies on the circle of radius $M$, and goes around the circle exactly once as $t$ goes from $0$ to $\pi$. We thus can use a well-known formula to conclude that

$$ -\int_0^\pi Y(t) X'(t) \mathrm{d}t = \text{Area of disk} = \pi M^2 $$

By Schwarz inequality, however, we have

$$ \int_0^\pi Y(t) X'(t) \mathrm{d}t \leq \sqrt{ \int_0^\pi Y^2\mathrm{d}t \int_0^\pi X'^2\mathrm{d}t} = \sqrt{ \left(\pi M^2 - \int_0^\pi X^2\mathrm{d}t \right) \int_0^\pi X'(t)^2\mathrm{d}t }$$

Squaring we get

$$ \pi^2 M^4 \leq \left(\pi M^2 - \int_0^\pi X^2 \mathrm{d}t\right) \int_0^\pi f'^2\mathrm{d}t $$

Now, notice that $$ \int_0^\pi f^2 ~\mathrm{d}t = \int_0^\pi (X + M)^2 ~\mathrm{d}t = \pi M^2 + \int_0^\pi X^2 ~\mathrm{d}t + 2M \int_0^{\pi} X ~\mathrm{d}t \leq \pi M^2 (1+A)^2 $$ where $$ A: = \left[ \frac{1}{\pi M^2} \int_0^\pi X^2 ~\mathrm{d}t \right] < 1. $$ This implies $$ \int_0^\pi f^2 ~\mathrm{d}t \leq (1 + A)^2(1-A^2) \int_0^\pi |f'|^2 ~\mathrm{d}t$$ The coefficient has a maximum when $A = 1/2$ or that $$ \int_0^\pi f^2 ~\mathrm{d}t \leq \frac{27}{16} \int_0^\pi |f'|^2~\mathrm{d}t $$

If $\int_0^\pi X ~\mathrm{d}t = 0$, we can sharpen the coefficient to $(1 + A^2)(1-A^2) = 1 - A^4 \leq 1$. This can be achieved by extending $f$ to a function $g$ on $(-\pi,\pi)$ with an odd extension, exactly as you have described for the Fourier proof.

  • $\begingroup$ you let $2M=$sup $f$ at first, but why you say $2f$ is bounded by $2M$ at last? $\endgroup$ – user360777 Jun 1 '17 at 8:06
  • $\begingroup$ @user360777: you are right! let me see how to fix this. $\endgroup$ – Willie Wong Jun 1 '17 at 13:51

The following proof is found in section 7.7 of Hardy-Littlewood-Polya Inequalities, it is motivated by Hilbert's investigations into calculus of variations, especially Hilbert's method of invariant integrals.

Consider the expression $$ (y'^2 - y^2) - (y' - y\cot x)^2 = -(1+ \cot^2 x) y^2 + 2y y' \cot x $$ So $$ \left[(y'^2 - y^2) - (y' - y\cot x)^2 \right]\mathrm{d}x = -(\csc^2 x)y^2 \mathrm{d}x + 2y \mathrm{d}y \cot x = \mathrm{d} ( y^2 \cot x )$$

Now, since $y' \in L^2$, we have that $$ y^2(x) = \left(\int_0^x y'(s) \mathrm{d}s\right)^2 \leq \int_0^x y'(s)^2 \mathrm{d}s \int_0^x 1\mathrm{d}s \leq x \int_0^x y'(s)^2 \mathrm{d}s $$ So we have that $$ \frac{y^2(x)}{x} = o(1) $$ and hence $y = o(\sqrt{x})$. Similarly we have that $y^2$ approaches 0 superlinearly at $\pi$. This implies that $\lim_{x\to \{0,\pi\}} y^2 \cot x = 0$. Hence the exact integral

$$ \int_0^\pi (y'^2 - y^2) - (y' - y\cot x)^2 \mathrm{d}x = \int \mathrm{d}\left( y^2 \cot x\right) = 0 - 0 = 0 $$

Therefore we have

$$ \int_0^\pi (y'^2 - y^2) \mathrm{d}x = \int_0^\pi (y' - y\cot x)^2 \mathrm{d}x \geq 0 $$

with equality only if

$$ y' = y \cot x $$

which is when $y = k \sin x$.

  • 2
    $\begingroup$ Interesting to note that similar proofs using hypergeometric functions can be used to give analogous statements for the $L^{2n}$ versions of the inequality when $n$ is a positive integer. The general arguments are given in section 7.6 of the above mentioned book. $\endgroup$ – Willie Wong Jun 12 '12 at 15:46

You can find different proof in the book of B. Dacorogna 'Introduction to the calculus of variations'.

  • 3
    $\begingroup$ Could you give for example the chapter and the number of the theorem? If the book is not available online, could you write a sketech of proof? $\endgroup$ – Davide Giraudo Dec 29 '12 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.