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Let $\phi(x)$ be a convex polynomial of degree $m$ at least two. Note that for $x,q \in \mathbb{R}$ $$\phi(x) + \phi(q) - 2\phi(\frac{x+q}{2}) = \sum_{l=1}^{m/2}\frac{\phi^{(2l)}(\frac{x+q}{2})}{2^{2l-1}(2l)!}|x-q|^{2l}$$ is strictly positive unless $x=q$, because the slopes of secant lines to $\phi$ are increasing.

I have proven using naive calculus-type estimates that there is some $C > 0$ such that $$\sum_{k=2}^{m}\frac{\bigl|\phi^{(k)}(\frac{x+q}{2})\bigr|}{k!}|x-q|^{k} \leq C \sum_{l=1}^{m/2}\frac{\phi^{(2l)}(\frac{x+q}{2})}{2^{2l-1}(2l)!}|x-q|^{2l}$$ uniformly in $x$ and $q$. I now need to show that $C \leq 2^{m}$ suffices.

But my approach of splitting $\mathbb{R} \times \mathbb{R}^{+}$ into various $(\frac{x+q}{2},|x-q|)$ regions and appealing to either asymptotics or compactness no longer seems good enough. Has anyone seen such an estimate before, or does anyone know of general theory that makes this easier?

$Edited\ to\ add$ - for example, the estimate is easy if the following is true: suppose $p(x)$ and $q(x)$ are convex polynomials of degree $m$ which both vanish at $x=0$. Suppose also that there are $M_{0}, M_{1}$ where $0 < M_{0} < M_{1} < \infty$ such that $$p \geq q \text{ on both } [0,M_{0}] \text{ and } [M_{1},\infty)$$ Then does $p \geq q$ hold everywhere?

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Crossposted at MSE:… – Noah Stein Sep 10 '12 at 13:57
up vote 0 down vote accepted

This is just an answer to the new question added in the edit: no, $p\geq q$ need not hold everywhere. Let $p(x) = x^2 + (1+\alpha) x^6$ for some $\alpha\in (0,\frac{1}{4})$ and $q(x) = x^4 + x^6$. Then $p(x)\geq q(x)$ if and only if $\lvert x\rvert\in [0,M_0]\cup[M_1,\infty)$ where \[ M_0 = \sqrt{\frac{1-\sqrt{1-4\alpha}}{2\alpha}}\qquad\text{and}\qquad M_1 = \sqrt{\frac{1+\sqrt{1-4\alpha}}{2\alpha}}. \]

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Thanks, Noah! Maybe neither $2^m$ nor any constant independent of $\phi$ will work. (I cannot find a way to quantify the size of the negative coefficients in a convex polynomial. This is my main problem. It seems as if it should be part of some standard theory, but...maybe not.) – Michael Tinker Sep 10 '12 at 13:17

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