A polynomial $\sum\limits_{k=0}^n a_kx^k$ is log-concave if $a_0,\ldots,a_n$ constitute a log-concave sequence. I wonder whether the log-concave polynomial is also a log-concave function with respect to $x$ if $a_k\ge 0$ for all $k$ and $x\ge0$. I have tested some log-concave sequences. The results support my guess. If this proposition is true, how to prove it rigorously?
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8$\begingroup$ This doesn't seem to be true. The sequence $(1,1,1)$ is log concave, but $$\frac{d^2}{(dx)^2} \log (1+x+x^2) = \frac{1-2x-2x^2}{(1+x+x^2)^2}$$ which is positive for $0 \leq x < \frac{\sqrt{3}-1}{2}$. Did I misunderstand something? $\endgroup$– David E SpeyerCommented Jan 22, 2016 at 14:49
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2$\begingroup$ A stronger condition than log-concave is having only real zeros. Such polynomials are log-concave functions. $\endgroup$– Richard StanleyCommented Jan 22, 2016 at 19:50
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$\begingroup$ A good counter example. The log-concave condition is not sufficient. The original problem acctually is from a probability computation. I attempt to make the polynomial become a log-concave function by using the transform $x=1-\exp(y)$. The sequence $a_k$ is a decreasing and log-concave sequence. More than that, $0<a_k<1$ and $0<x<1$. After the transform, does the polynomial become a log-concave function? I can prove it is log-concave when $a_k=a^k, 0<a<1$. Is it still log-concave in other cases? $\endgroup$– Johnny YinCommented Jan 23, 2016 at 4:01
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$\begingroup$ Is the transformed function is log-concave with respect to $y$？ $\endgroup$– Johnny YinCommented Jan 23, 2016 at 4:13
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