3
$\begingroup$

Let $f(x)$ be a smooth function satisfying $$f(0)=f(2)=0$$ and $$\int_0^2 (f(x))^2 dx=1$$ and $$\int_0^2 (f'(x))^2 dx=1$$ Does such an f exist? Why? I'm (perhaps stupidly) presuming that this function doesn't exist but I can't intuitively think why, and I don't know how to go about the question mathematically. I know there must be a stationary point between 2 and 0 by definition. Any help would be greatly appreciated.

$\endgroup$
  • 1
    $\begingroup$ I would bet that such a function does exist. One possibility to construct it would be taking two free parameters $0<a<b<2$ and then setting $$f(x)=\begin{cases} 0 & x<0 \\ \frac{b}{a}x & 0\le x < a \\ b & a\le x \le 1 \end{cases}$$ and then extend symmetrically along the axis $x=1$. (Visualize it, it is a trapezoid). You have two free parameters, so imposing your two conditions you will determine $f$ uniquely. Of course there is a problem: this is not smooth. But this is only a technical issue. $\endgroup$ – Giuseppe Negro Dec 9 '13 at 17:55
  • $\begingroup$ P.S.: If Norbert is right, which is most probable, then my previous comment is wrong. I am leaving it there though, perhaps it might be of some help. I wonder where the flaw is. $\endgroup$ – Giuseppe Negro Dec 9 '13 at 17:59
  • 2
    $\begingroup$ @GiuseppeNegro the problem is that $f(0)=f(2)=0$ $\endgroup$ – Norbert Dec 9 '13 at 18:06
  • $\begingroup$ @Norbert: Yes, I think I understand what you mean. I treated $a$ and $b$ as if they were independent parameters, but they are not because of the condition $f(0)=f(2)=0$. $\endgroup$ – Giuseppe Negro Dec 9 '13 at 18:37
  • $\begingroup$ thanks- interesting, all the same :) $\endgroup$ – Lucy Dec 9 '13 at 18:40
4
$\begingroup$

Such function does not exist. Use Wirtinger's inequality $$ \pi^2\int_0^a|f(x)|^2dx\leq a^2\int_0^a |f'(x)|^2 dx $$ to see this.

$\endgroup$

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.