$S \subset R^n,~~f : S \rightarrow R $ is a concave function.
$S^{'}= \{ x \in S: f(x)>0 \}. $
Prove that $\frac{1}{f}$ is a convex function on $S^{'}.$
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Jack D'Aurizio
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user107723
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Since concavity implies continuity, $S'$ is an open set and $\frac{1}{f}$ is a continuous function over $S'$, the convexity of $\frac{1}{f}$ follows from the midpoint-convexity, so we only have to prove $$\frac{1}{f\left(\frac{x+y}{2}\right)}\leq\frac{1}{2}\left(\frac{1}{f(x)}+\frac{1}{f(y)}\right)\tag{1}$$ for any $x,y\in S'$ such that $\frac{x+y}{2}\in S'$ (we can remove this assumption since $S'$ is a convex set by the concavity of $f$). We can write $(1)$ as: $$ f\left(\frac{x+y}{2}\right)\geq HM(f(x),f(y)), \tag{2}$$ but the concavity of $f$ over $S'$ gives: $$ f\left(\frac{x+y}{2}\right)\geq AM(f(x),f(y)),\tag{3} $$ so the claim follows by the AM-HM inequality - equivalent to the convexity of the $\frac{1}{x}$ function over $\mathbb{R}^+$.
Provided that $S'$ is a convex set, another approach is to consider that $\log f$ is concave on $S'$ since it is the composition of two concave functions, then $-\log f$ is convex, then $\frac{1}{f}=\exp(-\log f)$ is convex since it is the composition of two convex functions.
Jack D'Aurizio
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I think that the important point first is to show that $S'$ is convex. Plus, the fact that concavity implies continuity is much more complicated than the original statement... – user37238 Jan 27 '14 at 12:04
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2The fact that concavity implies continuity was only used to reduce the problem to a simple midpoint-convexity proof, but you could re-insert the $\lambda$ parameter and follow the same lines, then use the weigthed AM-HM inequality. Just a matter of taste. – Jack D'Aurizio Jan 27 '14 at 12:08
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I'll assume that $f$ is already positive, since you know that $S'$ is a convex set. So, $\log \circ f$ is concave: indeed, for $1/p+1/q=1$, $$ \log f\left( \frac{x}{p}+\frac{y}{q} \right) \geq \frac{\log f(x)}{p} + \frac{\log f(y)}{q} $$ is equivalent to $$ \log f\left( \frac{x}{p}+\frac{y}{q} \right) \geq \log \left( f(x)^{\frac{1}{p}}f(y)^{\frac{1}{q}} \right), $$ or equivalently $$ f\left( \frac{x}{p}+\frac{y}{q} \right) \geq f(x)^{\frac{1}{p}}f(y)^{\frac{1}{q}}. $$ But $$ f(x)^{\frac{1}{p}}f(y)^{\frac{1}{q}} \leq \frac{f(x)}{p} + \frac{f(y)}{q} $$ by some elementary inequality for positive numbers that you usually prove before Hölder's inequality. You conclude by the concavity of $f$. Now, consider $$ \log \frac{1}{f} = -\log f, $$ which is convex. Then $1/f = \exp \left( -\log f \right)$ is convex because the exponential is a convex increasing function.
Siminore
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The claim holds since the exponential is a convex function (the composition of two convex functions gives a convex function), not an increasing function: for istance, $\log x$ is increasing, but $\log(-\log f)$ is not always convex. – Jack D'Aurizio Jan 27 '14 at 18:59
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$x_1, x_2 \in S^{'}$ implies $f(x_1)>0$ and $f(x_2)>0$
$f(\lambda x_1 + (1- \lambda) x_2) >= \lambda f(x_1) + (1- \lambda) f(x_2) >0$ for $\lambda \in [0,1] $
which implies, $S^{'}$ is convex
– user107723 Jan 27 '14 at 11:51