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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...)

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3 Answers

up vote 3 down vote accepted

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

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

So we get,

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

Using that $2f$ is bounded by $2M$, we see that the product of the first two terms on the RHS is bounded by $\pi^2 M^4$. And this implies the desired inequality after canceling the factor.

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I am going to accept this answer. Thanks for the proofs. –  Jim Jun 14 '12 at 14:20
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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 $y = k \sin x$.

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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. –  Willie Wong Jun 12 '12 at 15:46
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You can find different proof in the book of B. Dacorogna 'Introduction to the calculus of variations'.

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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? –  Davide Giraudo Dec 29 '12 at 13:44
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