Best constant in a convex polynomial inequality. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T07:51:15Z http://mathoverflow.net/feeds/question/106696 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106696/best-constant-in-a-convex-polynomial-inequality Best constant in a convex polynomial inequality. Michael Tinker 2012-09-09T00:26:05Z 2012-09-10T12:36:24Z <p>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. </p> <p>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.</p> <p>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?</p> <p>$Edited\ to\ add$ - 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 $0 &lt; M_{0} &lt; M_{1} &lt; \infty$ such that $$p \geq q \text{ on both } [0,M_{0}] \text{ and } [M_{1},\infty)$$ Then does $p \geq q$ hold everywhere?</p> http://mathoverflow.net/questions/106696/best-constant-in-a-convex-polynomial-inequality/106811#106811 Answer by Noah Stein for Best constant in a convex polynomial inequality. Noah Stein 2012-09-10T12:36:24Z 2012-09-10T12:36:24Z <p>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 $M_0 = \sqrt{\frac{1-\sqrt{1-4\alpha}}{2\alpha}}\qquad\text{and}\qquad M_1 = \sqrt{\frac{1+\sqrt{1-4\alpha}}{2\alpha}}.$</p>