Let $f:[0,\infty )\rightarrow R$ a real valued continous function such that: $f(x)\neq 0 \quad \forall x>0 \quad$ and

${ (f(x)) }^{ 2 }=2\int _{ 0 }^{ x }{ f(t)dt } \quad \forall x\ge 0$.

Prove that $f(x)=x\quad \forall x\ge 0$.

  • $\begingroup$ I tried to try deriving on both sides and reaching the conclusion that f '(x) =1 but I think it is wrong $\endgroup$ – JuanCamilo Sep 25 '16 at 20:19
  • $\begingroup$ No you are right. Now integrate both sides. $\endgroup$ – Extremal Sep 25 '16 at 20:24
  • $\begingroup$ But as we know that f is indeed differentiable in the domain? $\endgroup$ – JuanCamilo Sep 25 '16 at 20:28
  • $\begingroup$ Can you show that if $f$ has one sign and $f^2$ is differentiable then $f$ is also differentiable? (The bad thing that can happen if all you know is that $f^2$ is differentiable is very rapid switching of signs, for example $f(x)=\begin{cases} 1 & x \in \mathbb{Q} \\ -1 & x \not \in \mathbb{Q} \end{cases}$.) $\endgroup$ – Ian Sep 25 '16 at 20:31

$$f(x)f'(x) = f(x) ~\forall x \ge 0.$$

If $x > 0$ then $f'(x) = 1.$ So, $f(x) = x + c$, where $c$ is a constant.

If $x = 0$ then $f(0) = 0.$

So, $c = 0.$ And then, $f(x) = x.$

  • $\begingroup$ once it is continuous its integral is derivable. Once $f(x)^2$ is equal to one derivable function the claim follows. $\endgroup$ – L.F. Cavenaghi Sep 25 '16 at 20:31
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    $\begingroup$ it's true, thanks $\endgroup$ – JuanCamilo Sep 25 '16 at 20:34
  • $\begingroup$ @frusciante14 It's slightly more complicated than that, because $(f^2)'$ can exist even when $f'$ does not exist, such as in the example I gave in the comments. But this requires sign changes in $f$ which are prohibited by the continuity assumption. $\endgroup$ – Ian Sep 25 '16 at 20:47
  • $\begingroup$ @Ian, the function you stated is not even integrable $\endgroup$ – L.F. Cavenaghi Sep 26 '16 at 0:53
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    $\begingroup$ @frusciante14 I know. But my point is that "if $f^2$ is differentiable then $f$ is differentiable" is not at all true. However, if $f$ is continuous at $x_0$, $f(x_0) \neq 0$ and $g(x)=(f(x))^2$ is differentiable at $x_0$, then $f$ is differentiable at $x_0$, with $f'(x_0)=\frac{g'(x_0)}{2f(x_0)}$. Things break if you drop any of these hypotheses, which is why I was saying it is a bit more complicated than you made it sound. $\endgroup$ – Ian Sep 26 '16 at 1:21

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