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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 positive real numbers $s \neq t$. Can one show that $$\frac{d^2}{dx^2}(\log f(e^x)) > 0$$ for all $x \in \mathbb{R}$?

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

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This is false. Take $f(x) = x^4+bx^3+cx^2+bx+1$ where $-0.5<b<0.5$ and $c$ is the unique root of $$216 b^2 + 108 b^4 - 324 b^2 c - 54 b^2 c^2 + 128 c^3 + b^2 c^3=0 \quad (\ast)$$ between $-1$ and $0$. The specific example I used to check this is $b=0.1$, $c\approx -0.2880429933200467$.

Then $\tfrac{d^2}{(dz)^2} \log f(e^z) = g(e^z)/f(e^z)^2$ where $$g(x) = x(b + 4 c x + 9 b x^2 + b c x^2 + 16 x^3 + 4 b^2 x^3 + 9 b x^4 + b c x^4 + 4 c x^5 + b x^6).$$ We rewrite this as $g(x) = x^4 h(x+1/x)$ where $$h(y) = b y^3 + 4 c y^2 + b(c+6) y + 4 (b^2-2 c+4).$$ The discriminant of $h$ is $$-4 (8 + b^2 - 4 c) (216 b^2 + 108 b^4 - 324 b^2 c - 54 b^2 c^2 + 128 c^3 + b^2 c^3)$$ so $(\ast)$ implies that $h$ has a double root. Some numerical verification shows that $h$ is positive on $[0,\infty)$ except at the double root, which occurs at some $y_0>2$. For example, with the values of $(b,c)$ above, the double root is at $y_0 \approx 7.42471$. Therefore, $$\tfrac{d^2}{(dz)^2} \log f(e^z) = e^{4z} h(e^z+e^{-z})/f(e^z)^2$$ is also positive except for vanishing at the two solutions to $e^{z_0}+e^{-z_0} = y_0$. In our example, $z_0 = \pm 1.98616$.

Finally, it is also east to check that $f(x)$ has no real roots. (This is the part that fails if we let $b$ get bigger than $0.5$.)


In case you prefer exact solutions, take $b = \tfrac{8 \sqrt{2}}{17 \sqrt{17}}$ and $c = -7/17$. Then $g(x)$ has a double roots at $\tfrac{19 \sqrt{34} \pm \sqrt{11118}}{34} \approx 0.15724, 6.35971$ and is $\geq 0$ for all $x \geq 0$. Thus, $g(e^z)$ is $\geq 0$ for all real $z$, but has two double roots.

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  • $\begingroup$ Is there a way to see that your constructed $f$ has the property that $\log f(e^x)$ is strictly convex on $\mathbb{R}$, i.e. $f(s^2) \cdot f(t^2) > f(st)^2$ whenever $s \neq t$ are positive reals? $\endgroup$
    – user17982
    Jul 24, 2016 at 12:39
  • $\begingroup$ Because $\tfrac{d^2}{(dx)^2} \log f(e^z)$ has isolated zeroes. For any smooth function $h$, we have $h(2s)-2 h(s+t) + h(2t) = \int_{u=s}^{s+t} \int_{v=u}^{u+t-s} h''(z) dz$. So if $h''(z) \geq 0$ and has only isolated zeroes, then $h(2s)-2 h(s+t) + h(2t)>0$ for any $s<t$. $\endgroup$ Jul 24, 2016 at 13:50
  • $\begingroup$ Great! I learnt something from your integral expression of $h(2s) - 2h(s + t) + h(2t)$. $\endgroup$
    – user17982
    Jul 25, 2016 at 1:23

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