A convex function's epigraph is convex. What property has that of an increasing function?

Question

In some sense, convex functions are the second-order version of increasing functions: under suitable hypotheses (for instance if the function is $C^2$), an analytic characterization of convexity is that the second derivative is positive, while a characterization of being increasing is that the first derivative is positive.

Convex real functions have the property that their epigraph is a convex subset of the plane (in fact that property is equivalent to convexity). Is there an analogue geometric property of epigraphs of increasing functions? I suspect that the answer is no, and I suspect also that the reason for it is that convexity is a much more general (in some sense) kind of property, that for instance generalizes to real functions of a real vector space.

Is there then an intuition for why such a general property reduces, in the one dimensional case, to something so related the property of being increasing?

Answer 1

I'm not sure if this would answer your questions but I hope it helps.

Regarding if there is any analogous geometric property of the epigraph of an increasing function to that of a convex function, such a property must necessarily depends on the direction in $\Bbb R$. Indeed, let $f:\Bbb R\to \Bbb R$ be an increasing function. We can say that $(x,r)\in\text{epi}\ f$ if and only if $(-\infty,x]\times \{r\} \subset \text{epi}\ f$. This means that $\text{epi}\ f$ looks like a set with infinity long "tails" to the left of any point in it.

Yes, this property seems artificial unlike the property of being a convex set which doesn't come with "direction". However, this is to be expected since if $f$ is increasing, then the function $\hat f(x):=f(-x)$, whose epigraph is the mirror image of the epigraph of $f$, is decreasing instead. This shows that any geometric property that characterize $\text{epi}\ f$ involves direction.

Now, let's consider a convex function $g$. Why doesn't the geometric property of $\text{epi}\ g$ depend on direction, you may ask. For simplicity, let's assume that $g\in C^2$ and $f\in C^1$. Think of it like this, the defining property of convexity is $g''(x)\ge 0$ for all $x\in \Bbb R$ so it's mirror image $\hat g$ satisfies $$ \frac {d^2\hat g(x)}{dx^2} = \frac {d^2 g(-x)}{dx^2} = (-1)^2g''(-x) = g''(-x)\ge 0. $$ This simply means that $\hat g$ is also convex. On the other hand, the property of being an increasing function is $f'(x)\ge 0$ for all $x\in \Bbb R$ and similar computation gives $\hat f'(x)\le 0$. The fact that convexity is a property about $2^{\text{nd}}$ order derivative gives the factor $(-1)$ twice so the direction doesn't matter.

As to why convexity of $g$ reduces to "$g'$ is increasing" in the one dimension case, this follows from the Mean Value Theorem. Intuitively, it shouldn't be hard to imagine that if the slope of $g$ is decreasing in some small interval $[a,b]$ then the straight line from $a$ to $b$ would lie under the graph of $g$ (i.e. outside the epigraph).

PS. The relatinship between convexity and "first order derivative is increasing" actually goes beyond $\Bbb R$, provided that you generalize the first order derivative to subgradient. It is a well-known fact that the subgradient of a convex function is a monotone operator, so it is increasing in some sense.

Answer 2

Informal Answer

The epigraph of an increasing function $f$ has the property that $\mathop{epi}(f) = \mathop{epi}(f) + \Bbb R_-\times \{0\}$, i.e. it contains its own translation leftwards. In other terms:

A function $f:\Bbb R \to \Bbb R$ is increasing iff $\mathop{epi}(f)$ is star-convex with the vantage point "$(-\infty,0)$".

Definitions:

• Intervals $\Bbb R_- = (-\infty,0]$, $\ \Bbb R_+ = [0,\infty)\ $ and $\ \Bbb R_{++} = (0,\infty)$.
• Minkowski sum $S + \Bbb R_-\times \{0\} = \{(x-\alpha,y)\in \Bbb R^2: (x,y)\in S, \alpha \in \mathbb R_-\}$
• A set $S\subset \Bbb R^n$ is star-convex if there is a point $s_0$ such that for any point $s\in S$ the line segment between $s$ and $s_0$ is contained in $S$. The point $s_0$ is then called the vantage point and $S$ is said to be star-convex at $s_0$.

Note that a set might have multiple vantage points, convex sets provide an extreme example of it (a related post):

A set $S$ is convex iff it is star-convex at every point $s_0\in S$.

Usually the point $s_0$ is required to be a point in $S$. However, in order to be able to characterize the property $S=S+\Bbb R_-\times \{0\}$ as that $S$ is star-convex, we need to augment the definition of star-convexity by allowing for a vantage point to be a "point at infinity" in a certain directions. I will formalize this below.

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Oriented extended Euclidean plane

Let me define an "oriented" version of the extended Euclidean plane:

Let equivalence relation $\sim$ be defined on the set $H = \Bbb R^2 \times \Bbb R_+ \setminus \{(0,0,0)\}$ so that points $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ form $H$ are in relation iff $(a_1,b_1,c_1)=(a_2t,b_2t,c_2t)$ for some $t\in \Bbb R_{++}$. We say that the quotient space $H/_{\sim}$ is the oriented extended Euclidean plane. For every point $(a,b,c)\in H$ we denote its quotient class as $[a,b,c]\in H/_\sim$. We distinguish two types of points in $H/_\sim$:

• A point $[x,y,1]$ represents the Euclidean space point $(x,y)\in\Bbb R^2$, we write then $[x,y,1]\sim (x,y)$;
• A point $[a,b,0]$, where $(a,b)\in \Bbb R^2 \setminus \{(0,0)\}$ represents an oriented point at infinity.

Note: Unlike in the in the conventional extended Euclidean plane, here $[a,b,0]$ and $[-a,-b,0]$ represent two different points (that is why I chose the adjective "oriented"). Hence, every line has two oriented points at infinity.

Generalized star-convexity:

For a set $S\subset \mathbb R^2$ we denote $[S]\subset \Bbb R^3$ the set of all points $(a,b,c)\in H$ such that $[a,b,c]=[x,y,1]$ for some $(x,y)\in S$; and $S^\infty\subset \Bbb R^3$ the set of all the points $(a,b,0) \in \mathop{cl}[S]$ (meaning that $(a,b,0)$ is a limit point of $[S]$ in the Euclidean space $\Bbb R^3$).

We say that $S\subset \Bbb R^2$ is star-convex at $[a,b,0]\in H/_\sim$ if and only if $[S] \cup S^\infty \subset \Bbb R^3$ is star-convex at $(a,b,0)\in\Bbb R^3$.

Or equivalently:

$S\subset \Bbb R^2$ is star-convex at $[a,b,0]\in H/_\sim$ iff for any $u \in [S]$ the half-open line segment connecting $u$ (included) and $(a,b,0)$ (excluded) is contained in $[S]$.

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Formal Answer

A function $f:\Bbb R \to \Bbb R$ is increasing iff $\mathop{epi}(f)$ is star-convex at the oriented point at infinity $[-1,0,0]$.

Note: In fact $\mathop{epi}(f)$ of an increasing function is generalized star convex at any oriented point at infinity $[-a,b,0]$ with $(a,b)\in \Bbb R_+^2$.

Discussion: Formalizing in which sense $\mathop{epi}(f)$ of an increasing function is star-convex is somewhat cumbersome. The reason why I find it meaningful is thanks to the relation between star-convex sets and convex sets. I also believe that it can shed light on how epigraph of functions with positive third-order derivative could be characterized, as I'm trying to do: Geometric characterization of functions with positive third derivative.