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Does there exist a $\gamma > 0$ and functions $g$ and $f$ s.t. we have the following for all $x \in [0, 1]$: $$ g(x) \in [0, 1] \\ f(x) \in [0, x] \\ x \frac{d}{dx}g(x) - \frac{d}{dx}(g(x)f(x)) = 0 \\ g(x)f(x) \geq \gamma x $$ In general, dealing with the inequality constraints in this differential equation setting is troublesome to deal with. I've tried constructing various such $\gamma$, $g$, and $f$ but found nothing and I have very little knowledge on how to prove non-existence.

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This is not possible. We can view the ODE as an equation for $f$ (given $g$). This gives $f=x-G/g$, with $G'=g$. The second condition now forces us to take $G\ge 0$, and in particular $G(0)\ge 0$. If $G(0)>0$, then $gf=xg-G<0$ near $x=0$, and if $G(0)=0$, then $gf=xg-G=O(x^2)$ will again be $<\gamma x$ for any $\gamma>0$ for sufficiently small $x>0$.

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  • $\begingroup$ Wait I realized there was a mistake here now. Mainly when $G(0) = 0$, it's not clear that $gf = xg - G = O(x^2)$ as $g \in [0, 1]$ and not necessarily in $[0, x]$. $\endgroup$ – Daishisan Jul 11 '16 at 17:18
  • $\begingroup$ @Daishisan: Yes, it is clear, as $G(x)=xg(0)+O(x^2)$ by a Taylor expansion. $\endgroup$ – Christian Remling Jul 11 '16 at 17:19

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