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While the set of log-convex functions is closed under addition, the set of log-concave functions is not. Yet if $f$ is log-concave, $\ln(k f) = \ln(k)+\ln(f)$, with $k \in \mathbb{R}^+$ constant, is concave. This suggests one can find a set of log-concave functions whose sums are still log-concave (possibly closed under addition).

I am interested in the concavity/convexity of $\ln(b+s)$, where $b,s$ are functions on $\mathbb{R}^+$, increasing, positive, convex and log-concave.

Example: I found $\ln(x^2+x^\beta)''<0, \forall x>0$ for $\beta<6$ (numerical result, perhaps not exact).

Are there any results that could help delineate this set of functions? Perhaps for polynomials?

Some suggestions and more detailed description here

The initial motivation for this problem comes from population dynamics.

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I guess you are aware of log-concavity being closed under convolution, if not addition.... –  Suvrit Mar 22 '13 at 16:31
    
I've seen this - do you think I could use it? –  Fred B Mar 22 '13 at 18:23
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In fact $\ln(x^2 + x^\beta)$ is concave for $x > 0$ iff $3-2\sqrt{2} \le \beta \le 3 + 2 \sqrt{2}$. This comes from writing $$ \dfrac{d^2}{dx^2} \ln(x^2 + x^\beta) = \dfrac{x^2}{(x^2 + x^\beta)^2} \left(-\beta x^{2\beta - 4} + (\beta^2 - 5 \beta + 2) x^{\beta - 2} -2\right)$$ and noting that the discriminant of $-\beta t^2 + (\beta^2 - 5 \beta + 2) t - 2$ with respect to $t$ is $(\beta^2 - 6 \beta + 1) (\beta - 2)^2$.

EDIT: Somewhat more generally, $\ln(x^\alpha + x^\beta)$ is concave for $x > 0$ iff $\alpha=\beta$ or $(\alpha-\beta)^2 - 2 (\alpha + \beta) + 1 \le 0$.

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Thanks! The condition for log-concavity is very neat. Assuming $\beta > \alpha \geq 1$, $\gamma=\beta-\alpha$, the condition further simplify into $\gamma<1+2\sqrt{\alpha}$. In the general case, maybe I should seek similar relationships for elasticities ($\frac{x}{f}\frac{df}{dx}$). Writing down $\frac{d^2}{dx^2}\ln(x^\alpha+x^\beta)$, I obtained $\frac{x^{2\alpha-2}}{x^\alpha+x^\beta} (-\beta x^{(\beta-\alpha)2}+(\alpha^2+\beta^2-\alpha-\beta-2\alpha \beta)x^{(\beta-\alpha)}-\alpha)$ I suppose you used $t=x^{\beta-\alpha}$ and factored the discriminant as in the case $\alpha=2$? –  Fred B Mar 26 '13 at 15:00
    
Intermediate step: polynomial discriminant can be factored into $(\alpha-\beta)^2 (\alpha^2 + \beta^2 -2 \alpha \beta - 2 \alpha - 2 \beta +1)$. –  Fred B Apr 2 '13 at 9:06
    
Just checked: the criterion holds for sums of the form $\left(\frac{x}{A}\right)^\alpha + \left(\frac{x}{B}\right)^\beta$, with $A,B>0$ constants. –  Fred B Apr 2 '13 at 9:27
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