TITLE: Given a real function $g$ satisfying certain conditions, can we construct a convex $h$ with $h \le g$?
QUESTION [7 upvotes]: The following is Exercise 8 from Chapter 3 of Rudin's Real and Complex Analysis (not a homework problem, just for fun).

Let $g$ be a positive function on $(0, 1)$ such that $g(x) \to \infty$ as $x \to 0$. Does there exist a convex function $h$ on $(0, 1)$ such that $h \le g$ and $h(x) \to \infty$ as $x \to 0$?

This seems true, but I can't show it. All I've been able to do is to reduce the problem to $g$ being a monotone step function or a strictly monotone piecewise-linear function but this doesn't seem to get me anywhere. I'm not sure how to use the property of $g$ to construct the explicit $h$.

REPLY [2 votes]: Consider the epigraph of $g$, which is the set of all points which lie on or above the graph of $g$; formally ${\rm epi}(g) = \{ (x,y) ~|~ x \in (0,1), y \geq g(x) \}$. Let $h$ be a function whose epigraph is the closure of the convex hull of ${\rm epi}(g)$. Then $h$ does the trick. 
To argue this we need to show: (1) there exists a function $h$ with the above property (2) $ h \leq g$ (3) $h(x) \rightarrow \infty$ as $x \rightarrow 0$. (4) $h$ is convex. 
The easiest are (2) and (4). (2) is equivalent to the assertion that ${\rm epi}(g) \subset {\rm epi}(h)$, which follows from the definition of $h$. As for (4), the convexity of a function is equivalent to the convexity of its epigraph, and the convex hull of any set is convex; moreover, the closure of a convex set is convex. 
Point (3) is not difficult either; simply reformulate the assertion that $g(x) \rightarrow \infty$ as $x \rightarrow 0$ as follows: for any $N>0$, there is a small enough interval $[0,\delta]$ such that if $x$ lies in this interval then $g(x) \geq N$. But by definition of $h$ this implies that $h(x) \geq N$ for $x \in [0,\delta]$, so $h(x)$ goes to infinity as $x \rightarrow 0$ as well. 
To demonstrate (1), define $h(x)$ to be first point on the vertical line going through $(x,0)$ which is in ${\rm cl} ({\rm epi}~ g)$; let us agree to call the latter set $H$. We must now argue that the points in $H$ are precisely those which lie above the graph of $h$. That each point in $H$ lies above the graph of $h$ follows by definition of $h$. To get the other direction - that every point above the graph of $h$ is in $H$  - observe that every point above the graph of $h$ lies above a point in $H$, and it is not too hard to argue it must consequently be in $H$ itself.