Given a real function $g$ satisfying certain conditions, can we construct a convex $h$ with $h \le g$?

Solution 1:

I think the answer is yes. Define a sequence $\{\delta_n\}$ as follows: $0<\delta_{n+1}<\delta_n<1$, $\delta_{n}-\delta_{n+1}\geq \delta_{n+1}-\delta_{n+2}$,$^{\dagger}$ and $g(x)>n$ on $(0,\delta_n)$; we can also arrange it so that $\delta_n\to0$. Define lines $\ell_n(x)$ by $$ \ell_n(x)=\frac{1}{\delta_{n+1}-\delta_n}(x-\delta_n)+n-1, $$ so that $g(x)>\ell_n(x)$ on $(0,\delta_n)$, and the slope of $\ell_{n+1}(x)$ is not greater than that of $\ell_n(x)$. Define $h(x)$ by $h(x)=0$ on $[\delta_1,1)$ and $$ h(x)=\ell_n(x)\qquad(x\in[\delta_{n+1},\delta_n)). $$ Then $h(x)$ is convex because its slope is clearly nondecreasing, and it is continuous. But also $$ \lim_{x\to0}{h(x)}=\lim_{n\to\infty}h(\delta_{n+1})=\lim_{n\to\infty}\left(\frac{1}{\delta_{n+1}-\delta_n}(\delta_{n+1}-\delta_n)+n-1\right) =\infty. $$

$^{\dagger}$ To see that we can get $\delta_n-\delta_{n+1}\geq\delta_{n+1}-\delta_{n+2}$, let $\{\epsilon_n\}$ be any sequence satisfying the other conditions, let $n$ be the least integer for which $\epsilon_n-\epsilon_{n+1}<\epsilon_{n+1}-\epsilon_{n+2}$, and choose $k$ large enough so that $\epsilon_{n}-\epsilon_{n+k}\geq\epsilon_{n+k}-\epsilon_{n+k+1}$ (such $k$ exists because $2\epsilon_{n+k}-\epsilon_{n+k+1}\to0$); take $\delta_i=\epsilon_i$ for $i\leq n$, $\delta_{n+1}=\epsilon_{n+k}$, $\delta_{n+2}=\epsilon_{n+k+1}$, and proceed (e.g., by induction).

Solution 2:

I am going to give you a hint toward an argument, and then en explanation of a) intuition about convex functions, and b) why constructing the most optimal convex function $h$ might not be possible.

First, the hint. For any non-empty family of functions $\{f_\alpha\}_{\alpha\in I}$ bounded above by some function $g$, we can construct a supreme function $f=\sup\{f_\alpha\}$ given by $f(x)=\sup_{\alpha\in I}\{f_\alpha(x)\}$. This is a generalization of the least upper bounder property from sets of numbers to sets of real-valued functions.

A property which you might not know about is that if the family of functions $\{f_\alpha\}$ consists of convex functions, then the supreme function $f=\sup\{f_\alpha\}$ is also convex. This allows you sufficient mileage to solve your problem with the following strategy:

1. Show that the supreme function of bounded above convex functions is convex.
2. Show that there exists a convex function $h$ such that $h\leq g$ when $g$ is bounded below. This shows that the family of convex functions bounded above by $g$ has a supreme convex function $f$.
3. Show that if a convex function $h$ such that $h\leq g$ is bounded above by some number $M\leq g(x)$ for some $x$, then there exists a convex function $h'$ such that $h<h'\leq g$ and $\sup_x\{h'(x)\}=M$.
4. (non-constructive) The previous step establishes that the supreme convex function $f$ (whose existence you know of non-constructively) must be not bounded above by any number $M\leq g(x)$ for some $x$. Since $f$ is in fact dominated by $g$, show that for every $a$, $g(x)\to\infty$ as $x\to a$ implies $f(x)\to\infty$ as $x\to a$ (i.e. that every pole of $g$ demands a pole of $f$; you can do this by clever domain restriction to domains where $g$ would have only one pole).
5. (constructive) Use the construction of 2. to define an infinite sequence of bounded convex functions $h_1<h_2<\dots$ with supremums increasing to infinity which will converge to an unbounded convex function dominated by $g$.

Second, the intuition. A good way to think about convex functions is that $h$ is convex on $(0,1)$ if and only if the epigraph $E_h=\{(x,y)\colon y\geq h(x)\}$ (the set of all points in $(0,1)\times\mathbb R$ that lie on or above the graph of $h$) is convex.

Epigraphs are pretty cool, and useful. For example, we can express the statement $h\leq g$ as the statement that the epigraph of $h$ contains the epigraph of $g$. To see this, note that $(x,y)\in E_g$ if and only if $y\geq g(x)$, and the assumption that $g(x)\geq h(x)$ then implies that $y\geq h(x)$ and hence $(x,y)\in E_h$.

For another example, suppose that $f=\sup\{f_\alpha\}$ is the supreme function of some bounded above family. In terms of epigraphs, we have $E_f=\bigcap_{\alpha\in I}E_{f_\alpha}$, i.e. taking supremums of functions is the same as intersecting their epigraphs. This also shows why we need the condition that the family of functions is bounded above by some other function -- if we don't have that condition, then the intersection of the epigraphs could end up missing certain lines. For example, if our family of functions was $0$ everywhere except at $a$ where $f_i(a)=i$, then the intersection of the epigraphs would be everything above $y=0$ except for the line $x=a$.

Using epigraphs we can also easily see that the supreme of a family of bounded above convex functions is a convex functions, since then we are simply intersecting closed convex sets (epigraphs are always closed sets), and convexity theory tells us that arbitrary intersection of convex sets gives us a convex set.

Additionally, we can almost construct the supreme convex function $f$ bounded above by $g$: it is the function whose epigraph is the closed convex hull of the epigraph of $g$. This then is the problem with determining $f$ explicitly, namely determining the boundary of the closed convex hull of a closed set is (or seems to me) difficult to do so in a useful way (you could just look at how horizontal lines intersect the epigraph of $g$, fill in segments between points of intersection, and keep track of data, but that seems painful and difficult to use to show that $f$ is unbounded in a way different from what I sketched above).