3
$\begingroup$

Claim: For every $y \in C^1[0,1]$ with $y(0) = y(1) = 0$, the inequality $$\int_0^1y'(x)^2dx \geqslant\pi^2\int_0^1y(x)^2dx$$ holds.

How may this be shown? The direction is giving me trouble as most inequality proofs go in the opposite direction.

  • 1
    If you could be bothered, what does $y \in C^1[0,1]$ mean? – 2017-02-27
  • 1
    @AnotherJohnDoe: It means that $y$ is continuously differentiable (which in turn means that $y$ is differentiable and that $y'$ is continuous). – 2017-02-27
  • 1
    https://en.wikipedia.org/wiki/Wirtinger's_inequality_for_functions – 2017-02-27

2 Answers 2

3

Some details in my answer may require additional mathematical validation, but I think the flow of the proof should be as follows.

The function $y$ could be continued on the interval $[-1,0]$ so that the resulting function will be odd. For this function $(y\in C^1[0,1])$ we have pointwise convergent Fourrier series. The function is odd, so we have only sinus functions

$$ y(x) = \sum_{n=1}^\infty b_n\sin\left(\pi nx\right).$$

If we take the derivative

$$ y^\prime(x) = \sum_{n=1}^\infty \pi nb_n\cos\left(\pi nx\right).$$

According to Parseval's identity

$$\int_0^1y'(x)^2dx = \sum_{n=1}^\infty (\pi nb_n)^2 = \pi^2\sum_{n=1}^\infty ( nb_n)^2\geqslant \pi^2\sum_{n=1}^\infty b_n^2 = \pi^2\int_0^1y(x)^2dx$$

Edit.

As JeanMarie and uniquesolution have stated in the comments to the post, we should justify our term-by-term differentiation by the condition that $y\in C^1[0,1]$.

  • 1
    Looks nice, but how is the differentiation term-by-term justified? – 2017-02-27
  • 1
    @uniquesolution, it is valid for continous + piecewise continously differentiable function. Our assumtion is even stronger than that – 2017-02-27
  • 1
    What is valid? The series converges point-wise. As far as I know, you cannot automatically differentiate convergent series. You don't know that the series converges uniformly. – 2017-02-27
  • 1
    @uniquesolution, according to https://www.math24.net/differentiation-integration-fourier-series/ we can differentiate the series, given $y\in C^1[0,1]$ – 2017-02-27
  • 1
    Ok, thanks for the reference. – 2017-02-27
  • 4
    @Andrei Kulunchakov An important thing is that the series $\sum_{n=1}^\infty (\pi nb_n)^2$ is convergent because a $C^1$ function has Fourier coefficients $O(\frac{1}{n^e})$ with $e\geq2$. – 2017-02-27
2

This is called Wirtinger's inequality. The ordinary proof uses Fourier series, but a nice proof of this can by done by hubristically assuming that we can write $$ \int_0^1 (y'^2-\pi^2y^2) \, dx = \int_0^1 (y'-\psi y)^2 \, dx $$ for some function $\psi$. When will this be the case? Expanding out, $$ \int_0^1 (y'-\psi y)^2 \, dx = \int_0^1 (y'^2-2\psi yy'+\psi^2 y^2) \, dx, $$ and integrating the middle term by parts, $$ \int_0^1 -2\psi yy' \, dx = [-\psi y^2]_0^1 + \int_0^1 \psi' y^2 \, dx. $$ Hence we can make this identification if $\psi'-\psi^2 = -\pi^2$, which has solutions $$ \psi = \pi\cot{(\pi x+c)}. $$ We also need $\psi y^2$ to vanish at the endpoints, and $\psi$ to be continuous elsewhere. Since $y\in C^1[0,1]$, $y^2 = O(x^2)$ at the endpoints, so we can get away with choosing $c=0$, and we conclude that $$ \int_0^1 (y'^2-\pi^2y^2) \, dx = \int_0^1 (y'-y\pi\cot{\pi x})^2 \, dx \geqslant 0, $$ with equality if and only if $y'-y\pi\cot{\pi x} = 0$, or $y=k\sin{\pi x}$.

(The above can be found in more detail in the classic Hardy, Littlewood and Pólya, Inequalities, p. 184f.)

  • 1
    [+1] very very interesting. It is to be noticed that this inequality was found at the beginning of XXth century at the same time the young Fejer established quadratic convergence of Fourier series (a striking result that the great XIXth century mathematicians hadn't found). – 2017-02-27