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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1Yes, for example $g(x)=0$ – 2012-02-05
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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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0I think it's only true that if a convex function takes on the same value at three collinear points, it's constant on the *segment* connecting them (consider for example $f(x)=\max\{0,x\}$). So it might take a bit more work to make sure you can build a bridge connecting any two points to guarantee they have the same value. – 2012-02-05
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0Oops, I meant to say 'line segment'; edited, thanks. It takes some work to see that every point is connected by such line segments, but it's not extremely difficult (and you can sort of just see it intuitively) because of the sheer number of line segments that work. – 2012-02-05
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0For instance, take any line of the form x1+x2=C. That line passes through 4 points on some level curve, so just taking the segment of the line in the first quadrant, the function is constant on that whole segment. But we can also rotate the segment just a little bit and it'll still work, so the function must be constant everywhere. – 2012-02-05
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0Thanks @Lopsy for the nice answer. – 2012-02-06