MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

3 deleted 13 characters in body

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 from convex analysis that makes this easier?<3

$Edited\ to\ add$ - 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?

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.

Unfortunately

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 makes this more delicate task look...hardno longer seems good enough.

I'm curious if Has anyone has seen such an estimate before, or is aware know of general theory from convex analysis that could make makes this more tractable. easier? <3

$Edited\ to\ add$ - 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?

1

# Best constant in a convex polynomial inequality.

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 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.

Unfortunately 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 makes this more delicate task look...hard.

I'm curious if anyone has seen such an estimate before, or is aware of general theory from convex analysis that could make this more tractable. <3