-1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

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, $f(x) = x.$

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

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.