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?