# Tag Info

The inequality you wrote down is a special case of a general principle about higher-order convexity: any "simple enough" linear inequality of the type you wrote down will be true as long as ...
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Accepted

### Why the sequence of Bernstein polynomials of $\sqrt x$ is increasing?

As noted by Paata Ivanishvili, if $f$ is concave on $[0,1]$, then the Bernstein polynomials $B_n(f,p)$ are increasing in $n$. Here is a probabilistic proof: Let $I_j$ for $j \ge 1$ be independent ...
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### Nonconvexity and discretization

I was asked to answer this question. The people one really needs to ask here are the people in high-dimensional convex geometry/probability/statistics/computer science and metric geometry, but in any ...
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Accepted

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### The missing link: an inequality

I sketch a method that should show $F_n$ is convex for $n$ sufficiently large. A Taylor expansion gives that $F_n(x) = x^{2n} (1-x)^2 + O(x^{4n-1})$, and so F_n''(x) = 2x^{2n-2}[2n^2(1-x)^2 + x^2 - ...
• 4,652
Yes. At first, if $h=\min(h_A,h_B)$ is convex (note that it is also 1-homogeneous), it is a support function of the body $C:=\{x:\forall\xi\in \mathbb{R}^n,\langle \xi,x\rangle\leqslant h(\xi)\}$. ...