We know that the function $f(x)=(|x_1|^p+\dots+|x_n|^p)^{1/p}$ on $\mathbb{R}^n$ is not convex for $0 . I am wondering whether there is some non constant continuously differentiable function $g$ on $\mathbb{R}$ such that $g \circ f$ is convex.
$\ell_p$ space for p<1
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real-analysis
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0Yes Norbert, in fact, any constant function will do. I should have been more careful when I asked this. Sorry for that. I am looking for a non constant continuously differentiable $g$. I will change the question accordingly. – 2012-02-05
1 Answers
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Nope.
For simplicity, let's say we're working in $R^2$. For some constant $c$, graph the curve where $f=c$.
It'll look something like this:
There are three collinear points on this curve. $f$ takes on the same value for those three points, so $g ∘f$ must also take on the same value for those three points.
But for any convex function, if it takes on the same value at three collinear points, then it's constant on the line segment containing those three points. This argument works for any line segment passing through at least three points on any level curve of $f$, so there is no nonconstant $g$ which fits your specifications.
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0Thanks @Lopsy for the nice answer. – 2012-02-06