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Problem says:

Suppose that continuous $f(x)$ is such that:

$\int_0^x f(t) dt= [f(x)]^2$

and $f(x)$ is never zero. Then find $f(x)$.

I do the question thinking it say "$f(x)$ is not the zero function" I find $f(x) = \frac12 x$ as solution.

But, I see from again reading again $f(x)$ can never be zero. Now my solution is not correct. $f(0) = 0$ for my solution.

From how I have came to solution I wonder:

Is this possible even to find?

In the way I have solved it seemed this was the only way possible solution.

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    (Hmm, I suppose I should amend my previous comment to have "any *non-horizontal* line"...)2011-05-16

2 Answers 2

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If you put $x=0$ then you immediately get $f(x) = 0$. I guess what they meant was "$f$ is not identically zero", since $f \equiv 0$ is another solution.

Also, if you follow J.M.'s advice then it's quite helpful to know that $f(x) \neq 0$, at least for $x \neq 0$. You have to assume that $f$ is differentiable, though.

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    It suffices to assume that $f$ is continuous, as this implies $\int_0^x f(t)\, dt$ is differentiable. In fact it suffices to assume $f$ is locally $L_1$, as that implies $\int_0^x f(t)\, dt$ is continuous.2011-05-16
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EDIT: Expanding on J. M.'s advice.

From $\int_0^x {f(t)dt} = f^2 (x)$, we get $f(x) = \frac{d}{{dx}}f^2 (x)$ (using that $f$ is continuous). From this, one wants to write f(x)=2f(x)f'(x). In general, however, differentiability of $f^2$ does not imply that of $f$ (consider, for example, the functions $|x-1|$ and $(x-1)^2$, $x \geq 0$; the former is not differentiable at $x=1$), and we are not given that $f$ is differentiable. Nevertheless, in our case, for any $x > 0$ we have $ f(x) = \frac{d}{{dx}}f^2 (x) = \mathop {\lim }\limits_{h \to 0} \frac{{f^2 (x + h) - f^2 (x)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{[f(x + h) + f(x)][f(x + h) - f(x)]}}{h}, $ leading to $ 1 = \mathop {\lim }\limits_{h \to 0} \bigg[\frac{{f(x + h) + f(x)}}{{f(x)}}\frac{{f(x + h) - f(x)}}{h}\bigg], $ where we have used the assumption $f(x)>0$. Since $f$ is continuous at $x$, it thus follows that $ 1 = 2\mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}, $ and hence the conclusion that f'(x) = \frac{1}{2}. (The rest is straightforward.)

Original answer (see first comment below).

If $f$ is continuous on $[0,\infty)$, then, on the one hand, $ \mathop {\lim }\limits_{x \to 0^ + } \int_0^x {f(t)dt} = 0, $ and, on the other hand, $ \mathop {\lim }\limits_{x \to 0^ + } f^2 (x) = f^2 (0). $ So the condition $\int_0^x {f(t)dt} = f^2 (x)$ implies that $f^2 (0) = 0$, hence also $f(0)=0$.

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    Rather than deleting it, I'll edit it later on (but not soon), expanding on J. M.'s advice (since there is some delicate point there).2011-05-16