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Let $f(t)$ be a monic real polynomial such that $f(t) > 0$ for all $t \ge 0$. Suppose that $\log f(e^x)$ is strictly convex on $\mathbb{R}$, i.e. $f(s^2) \cdot f(t^2) > f(st)^2$ for all $s, t \ge 0$. Can one show that in fact $$\frac{d^2}{dx^2}(\log f(e^x)) > 0$$ for all $x \in \mathbb{R}$?

Remark: Since $\log f(e^x)$ is strictly convex (and hence convex), we always have $\frac{d^2}{dx^2}(\log f(e^x)) \ge 0$.

(This question is migrated from Math Stack Exchange, where it has not received a response.)

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    $\begingroup$ For those who think this is trivial -- it doesn't come out of local bounds. If $f(t) = t^3 + (t^2+5t+1)(t-1)^4$, then $\tfrac{d^2}{(dx)^2} \log f(e^x)$ has a double zero at $x=0$ and is positive as $x \to \pm \infty$ (but it is negative on the intervals $\pm [0.849743, 3.87148]$.) $\endgroup$ Jul 7, 2016 at 21:34
  • $\begingroup$ Perhaps a remark is in order to say that the class of polynomials in consideration is not empty. For $f(t) = 1 + t$, then $\frac{d^2}{dx^2} \log f(e^x) = \frac{e^x}{(1 + e^x)^2} > 0$ on $\mathbb{R}$. $\endgroup$
    – user94803
    Jul 8, 2016 at 2:24
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    $\begingroup$ @Suvrit What I wrote is true, but let me fill in the context. Let $f$ be anypolynomial without positive roots. Set $g(x) = \tfrac{d^2}{(dx)^2} \log f(e^x)$. The contrapositive of Colin's statement is that, if $g(x_0)=0$, then $g$ is negative somewhere. This choice of $f$ obeys this claim, but the place where $g$ is $0$ is neither near $x_0=0$, nor near $\pm \infty$, so we need to use global properties of $g$ to find it. $\endgroup$ Jul 8, 2016 at 16:12
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    $\begingroup$ @DavidSpeyer thanks for the added context. $\endgroup$
    – Suvrit
    Jul 8, 2016 at 17:59
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    $\begingroup$ This is false. Counter example is $x^4+bx^3+cx^2+bx+1$ where $-0.5<b<0.5$ and $c$ is the unique root of $216 b^2 + 108 b^4 - 324 b^2 c - 54 b^2 c^2 + 128 c^3 + b^2 c^3=0$ between $-1$ and $0$. More details left at math.SE. $\endgroup$ Jul 23, 2016 at 2:01

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