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I have encountered the following interesting technical relation.

$$ \pi^2 = \inf_{x \in \mathcal{D}(0,1) \setminus\{0\}} \frac{\int_0^1 |x'(s)|^2 \, \text{d}s}{\int_0^1 |x(s)|^2 \, \text{d}s}$$

where $\mathcal{D}(0,1)$ is the set of all smooth functions in $(0,1)$ with a compact support.

Amazing. Can anyone please give a hint why this is true? Thank you.

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  • 2
    $\begingroup$ When you say "encountered," what do you mean? $\endgroup$ – Thomas Andrews Nov 29 '14 at 5:12
  • $\begingroup$ That should be inf instead of sup. $\endgroup$ – user99914 Nov 29 '14 at 5:17
  • $\begingroup$ Note, you can restrict the problem to when $\int_0^1 |x(s)|^2ds=1$. That also elminates the case $x(s)=0$ for all $s$... $\endgroup$ – Thomas Andrews Nov 29 '14 at 5:25
  • $\begingroup$ @John Thanks for your reminder. $\endgroup$ – Empiricist Nov 29 '14 at 5:25
  • $\begingroup$ If you want to, I can write an answer, but do you just want a hint? $\endgroup$ – user99914 Nov 29 '14 at 5:26
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First of all, we write down the fourier expansion of $x(t)$:

$$x(t) = \sum_{n=1}^\infty a_n \sin(n\pi t) + \sum_{n=1}^\infty b_n \cos (n\pi t)$$

(the main point is that $x(t)$ has no constant term, do you know why?). Thus

$$\int_0^1 |x(t)|^2 dt = \sum_{n=1}^\infty a_n^2 + b_n^2$$

and

$$\int_0^1 |x'(t)|^2 dt = \pi ^2 \sum_{n=1}^\infty n^2(a_n^2 + b_n^2)$$

$$\Rightarrow \frac{\int_0^1|x'(t)|^2 dt}{\int_0^1 |x(t)|^2 dt} = \frac{\pi^2 \sum_{n=1}^\infty n^2 (a^2_n + b_n^2)}{ \sum_{n=1}^\infty a_n^2 + b_n^2} \geq \pi^2\frac{\sum_{n=1}^\infty a^2_n + b_n^2}{\sum_{n=1}^\infty a_n^2 + b_n^2} = \pi^2 $$

With equality as $f(t)= \sin(\pi t)$. (Strictly speaking the infimum is not attained as this function does not have compact support).

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  • $\begingroup$ That first equality on the last line is totally wrong. $\frac{a+b}{c+d}\neq \frac{a}{b}+\frac{c}{d}$. $\endgroup$ – Thomas Andrews Nov 29 '14 at 5:36
  • $\begingroup$ @ThomasAndrews: Thanks, let me fix that. $\endgroup$ – user99914 Nov 29 '14 at 5:39
  • $\begingroup$ You've still got the fraction error in the last line, only on the inequality step, and you also lost the square from $\pi^2$. $\endgroup$ – Thomas Andrews Nov 29 '14 at 5:44
  • $\begingroup$ @ThomasAndrews: Sorry, my bad. $\endgroup$ – user99914 Nov 29 '14 at 5:45
  • $\begingroup$ Thanks! I had been thinking about this for several hours but failed to figure out it is done simply by Fourier series expansion. $\endgroup$ – Empiricist Nov 29 '14 at 5:48
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I can prove that if $\pi^2$ is a lower bound then it must be the greatest one. Simply consider uniform approximations of $x(s)=\sin(\pi s)$ by functions with compact support. Then $x'$ will be approximated in the sense of $L^2$ and so you recover the result. I am not sure why $\pi^2$ is a lower bound, however.

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  • $\begingroup$ I am sorry I made a mistake in posing the question. $\endgroup$ – Empiricist Nov 29 '14 at 5:28
  • $\begingroup$ Thanks! Your answer is very useful indeed. $\endgroup$ – Empiricist Nov 29 '14 at 5:56

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