Any help? I thought about solving the differential equation $f(x) = x f'(x)$ but I don't think that does anything useful.
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Try $f(x)=2$, then $f'(x)=0$. – vadim123 Apr 27 '14 at 19:10
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Any non-constant functions? – Brian Apr 27 '14 at 19:12
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Try $f(x)=\begin{cases}3 & x<1\4-x & 1<x<2 \ 2 & x>2\end{cases}$. Smooth at the corners if you like. – vadim123 Apr 27 '14 at 19:17
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Say $f(0)=0$ and $f$ is concave, $f^{\prime\prime}<0$ what can you say? – Sergio Parreiras Apr 27 '14 at 19:17
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f'(x) tends to 0? Not following you Sergio, sorry. – Brian Apr 27 '14 at 19:23
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Well solving the differential equation isn't entirely useless. The solutions of it are of the form $$f(x) = a x, \qquad a \; \text{a constant}$$
Now add any positive constant $b$ to that function to get $$f(x) = ax + b$$ motivated by the fact that differentiation on the right hand side will kill the constant. Now we have $$f(x) = ax + b, \qquad x f'(x) = ax$$ and $ax + b > ax$ follows as long as $b > 0$
Squid
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Solving the ODE:
$f(x)=xf'(x)\Rightarrow \ln f(x)=\ln x+C\rightarrow f(x)=x+C$ let $C=1$
so $f(x)=x+1\gt x(f'(x))=x$
Ellya
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