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Rudin ask:

If $g$ is a positive function on $(0,1)$ such that $g(x)\rightarrow \infty$ as $x\rightarrow 0$. Then there is a convex function $h$ on $(0,1)$ such that $h\le g$ and $h(x)\rightarrow \infty$ as $x\rightarrow 0$. True or False? Is the problem changed if $(0,1)$ is replaced by $(0,\infty)$ and $x\rightarrow 0$ replaced by $x\rightarrow \infty$?

I thought about using piece-wise linear functions $f_{n}$ to approximate $g$, and use convex functions $h_{ni}$ to approximate $f_{n}$. The selecting $h_{nn}$ we would be able to approximate $g$. But this missed a point; $g$ can be arbitrally strange. For example if we work with a basis for $\mathbb{R}/\mathbb{Q}$, then $g$ can even be dense in $\mathbb{R}$ in any subset of $(0,1)$. So a piecewise approximation is impossible. But is an approximation by convex functions still possible?

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    I think this is impossible.2012-12-25
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    $g$ cannot have a dense range on every subinterval because it tends to $\infty $ . Begin by defining $f(x)$ to be the infimum of $g$ on $ (0,x)$ - this is a monotone function.2012-12-25
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    This is a duplicate of http://math.stackexchange.com/questions/50549/given-a-real-function-g-satisfying-certain-conditions-can-we-construct-a-conv?rq=12012-12-25

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