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Suppose that $f$ is continuous on $[0,\infty)$ such that $f(x)>0$ for all $x>0$ and that $f^2(x)=2\int_0^x f(t)\,dt$ for all $x>0$. Prove that $f(x)=x$ for all $x\geq 0$.

Attempt at a proof: Let $f(t)=t$ for all $t\geq 0$. Since $f$ is continuous on $[0,\infty)$, $f(t)=t$ is integrable. Then \begin{align*} f^2(x)&=2\int_0^x f(t)\, dt \\ &=2\int_0^x t\, dt \\ &=2\frac{t^2}{2}\Big|_0^x \\ &=2\frac{x^2}{2} \\ &=x^2 \end{align*} Therefore $f(x)=x$ for all $x\geq 0$.

J126
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user113009
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    In the first sentence of your attempted proof you are assuming what you intend to prove. – Jay Dec 03 '13 at 00:55
  • You've shown that $f(x) = x$ satisfies the integral equation, not that the integral equation implies $f(x) = x$. –  Dec 03 '13 at 00:55
  • I am not quite sure how I can get the integral equation to implies that f(x)=x then. Should I take the derivative of both sides and solve it that way? – user113009 Dec 03 '13 at 00:58
  • Yes, that will give it to you quickly. – André Nicolas Dec 03 '13 at 01:01

2 Answers2

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derivative of both sides gives you $2f(x)*f'(x)=2f(x) $ so $f'(x)=1$ imples that $f(x)=x$

Eli Elizirov
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Your guess is right. Take the derivative of both sides using the chain rule and FTC. Divide by f (which is not zero) and integrate.

Brian Rushton
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