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.

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    $\begingroup$ 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
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    $\begingroup$ @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

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.


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