Why are monotonicity and convexity so intensely studied in comparison to their higher-order analogies?

Question

I've been recently contemplating what can be said about functions $f:\Bbb R \to \Bbb R$ with positive third-order derivative and their properties:

It puzzles me why monotonicity and convexity, i.e. the properties associate with the first- and second-order derivatives of a function being positive, are frequently used, whilst their higher-order alternatives are so rarely referred to.

I found that the third-order derivative at a point is called jerk in physics and its meaning is discussed here. I found also a discussion on the topic why are third-order concepts so rare. However, what I'm interested in are the uses of the fact that third-order derivative of $f$ is positive in analysis.

For example, whilst strict convexity of $f$ implies that every critical point of $f$ is the unique global minimum of $f$ and that $f$ has at most two roots, having strictly positive third-order derivative would guarantee that every point with $f''(x)=0$ is the unique inflection point of $f$, that the function has at most one local minimum and one local maximum, and no more than 3 roots.

One could argue that it would be informative enough to know that there is $x_2\in \Bbb R$ such that $f''(x_2)=0$ and that $f$ is concave on $(-\infty,x_2]$ and convex on $[x_2,\infty)$. However, analogously one could argue that convexity is not needed for analyzing function's minima because it is enough to know that there is $x_1\in \Bbb R$ such that $f'(x_1)=0$ and that $f$ is decreasing on $(-\infty,x_1]$ and increasing on $[x_1,\infty)$.

The concept of $n$-th order convexity (having positive $n$-th order derivative) is essential for the analysis of the number of roots of one-variable functions: A function with positive $n$-th derivative has at most $n$ roots – an inequality version of the Fundamental theorem of Algebra.

Consider the question on what is the maximum number of strict local minima that a degree $k$ polynomial $p(x,y)$ in two variables can have?

In case of quadratic polynomial one can readily answer: The polynomial has a strict local minima only if it is strictly convex, and then the minimum must be unique.
In case of cubic polynomial, there is also at most one strict local minimum because if there were two, say at points $a$ and $b$, then the third degree polynomial $q(t)=p\big((1-t)a+tb\big)$ would need to have two strict local minima – impossible. [1]
In case of quartic polynomial, none of the above arguments apply and the only ready estimate follows from applying Bézout's theorem to the partial derivatives of $p$, and so we can be sure that $p$ has is no more than $3\times 3$ isolated critical points. [2]

I expected that the above analysis would be trivial for the case of quadratic polynomials thanks to the concept of convexity. However, the analysis is equally trivial in case of cubic polynomial, the only difference is that there is no name for the "third-order convexity" that gives the result. In fact, the analysis becomes difficult as late as in the case of quartic polynomials. This suggests that third-order convexity has practical applications, only the higher-order alternatives would be less practical for the analysis of functions in two or more variables.

Does importance of $n$-th derivative drop?

Let me compare the count of search results of terms

a) convex
b) monotone or monotonic (sum up the count)

\begin{array} {|r|r|r|}\hline & \text{Google} & \text{site:SME} & \text{site:mathoverflow} & \text{G Scholar} \\ \hline \text{a)} & 148M & 79K & 229 & 4.3M \\ \hline \text{b)} & 98M & 77K & 54 & 2.5M \\ \hline \end{array}

Why is the second-order property (convexity) more prevalent than the first-oder one (monotonicity) and yet the third-order one is almost never heard of?

Questions:

Is there a standardized name for functions with $f^{(3)}>0$?
Is the main reason that $f^{(3)}$ is so rarely analyzed the fact that for most problems it is enough to determine on which regions $f$ is convex/concave and the effort needed to analyze $f^{(3)}$ would typically not be justified?
Is there a known set property of $\mathop{epi}(f)$ for $f$ with $f^{(3)}>0$, akin to strict convexity of $\mathop{epi}(f)$ for $f$ with $f^{(2)}>0$?

Related Posts:

Let me share some observations and conjectures that I came up with when contemplating on this topic:

Functions not necessarily differentiable that behave like those with $f^{(3)}\geq 0$: Geometric characterization of functions with positive third derivative.
"Quasi" generalization of the condition that $f^{(3)}>0$: Is there a third-order analogy of quasi-convexity?
An attempt to find the geometric property whose special case $\mathop{epi}(f)$ of $f$ with $f^{(3)}\geq 0$ satisfies: Second-order star convex set: A set whose intersection with any conics passing through two given points consists of at most two connected curves.

convexity and monotonicity are independent of the differentiability of a function, so your question doesn't make sense. The concept of convexity is related to subsets of vector spaces, where to concept applied to functions comes, and the word monotonicity is related to functions between ordered spaces — Masacroso, Jan 17 '23 at 07:07
@Masacroso Thanks for sharing your view, however, I must disagree. I restricted attention to three times differentiable functions $f:\Bbb R \to \Bbb R$ for the sake of simplicity. 1) As shown in https://math.stackexchange.com/questions/4616713/geometric-characterization-of-the-n-th-derivative-of-f-being-positive-conve the property $f^{(n)}\geq 0$ has an analogy that does not require the function to be differentiable. 2) In one of the attached links I also discuss that the property $f^{(3)}>0$ has an ordinal analogy the same way quasi-convexity is an ordinal analogy of convexity. — Pavel Kocourek, Jan 17 '23 at 08:37
interesting... I didn't knew this relation of "convexity" related to polynomials of $n$-degree — Masacroso, Jan 17 '23 at 09:13
@MarianoSuárez-Álvarez But why is that the case? It is understandable that not much is taught about cubic functions in a high-school: There is a practical formula for the roots of a quadratic function, and the concept of convexity is easy to interpret as one only needs to interpolate two points by a line to imagine its meaning. In contrast to interpret "3-convexity" one needs to interpolate 3 points by a quadratic function to imagine its meaning. Such a concept is perhaps too challenging for most high-school students, but how about graduates? — Pavel Kocourek, Jan 19 '23 at 02:53
I must admit that it struck me when I realized how easy it is to see that an $n$ variable cubic polynomial has at most one strict minimum. It is trivial to prove it, but it is not a result that I would be readily aware of – in contrast to being aware of that a quadratic polynomial has at most one strict minimum. — Pavel Kocourek, Jan 19 '23 at 03:15
In some vague sense, I would say that the importance of a $n$-th order property decreases with $n$, while at the same time its complexity increases with $n$. This is why only the first few of such properties do appear in practice. — Giuseppe Negro, Jan 24 '23 at 18:11
I guess an economist would call it *diminishing marginal returns. However, it still does not explain why there is such a radical drop of importance at $n=3$, why not as late as at $n=4$? Both 3rd and 4th order properties are useful in determining the number of minima of a single-variable function, what is more, the 3rd order property is useful in doing so in case of a multi-variate function. So why the drop of importance is not as late as at the 4th order?* — Pavel Kocourek, Jan 25 '23 at 06:10
I added some google search statistics in the question to question the argument that the importance of $n$-th order property drops with $n$. — Pavel Kocourek, Jan 25 '23 at 06:49
@Pavel for some reason many things drops at degree $3$, by example, there is no special reason to define a fundamental physical entity as force based in a second order element as the acceleration. I think that a second order captures the idea of "something that is not linear", what is very different from linearity. However third order, at least visually, doesn't seem to show a big different from second order, at least not compared with the difference between second and first order — Masacroso, Jan 30 '23 at 20:21

user7777777 · Answer 1 · 2023-01-25T03:51:18.540

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The thing with considering how higher-order derivatives behave and what that implies is that we lose information in the process. Our definition of the function becomes more broad or arbitrary. It's similar to how higher-order differential equations have more arbitrary solutions, with one more linearly independent family if we increase the order by one for linear ones for example.

To picture this, we will consider the intervals at which the function is increasing and at which it's decreasing given that an $n^\text{th}$ derivative, $D^n$ is positive everywhere for example and consider what we need to do to be able to get a sketch for it, it's like trying to expand the function in some sense.

For simplicity, let $C_k\subseteq\mathbb{R}$ be a continuous interval in the domain, where $C_1\cup C_2\cup\cdots\cup C_n=\mathbb{R}$ given $D^n>0.$

\begin{align} &D^1>0 \begin{cases} D^1>0&\mathbb R \end{cases}\\ &D^2>0 \begin{cases} D^1>0&C_1\\ D^1<0&C_2\\ \end{cases}\\ &D^3>0 \begin{cases} D^2<0&C_1\\ D^2>0&C_2 \end{cases}\\ \implies&D^3>0 \begin{cases} D^1<0&C_1\cup C_2\\ D^1>0&C_3\cup C_4 \end{cases}\\ &\qquad\qquad\vdots \end{align} It's like a tree diagram where going up an order adds a new branch, or a better analogy is a piecewise function that gets choppier the higher the order gets.

For the first derivative, it's increasing everywhere; to graph it, we need to evaluate $D^0$ at a point. For the second we need to solve $D^1=0$ and evaluate $D^0$ at $2$ points. For the third derivative, we need to solve $D^2=0$ and $D^1=0$ and evaluate $D^0$ at $4$ points and so on.

It can also get increasingly more complex as the order increases to solve for the turning points at which the derivatives are $0,$ but that's not what the main reason we don't consider higher-order derivatives.

edited Jan 25 '23 at 03:51

answered Jan 25 '23 at 03:40

user7777777

556

What you describe could be seen as an advantage as well: Suppose that I have a third degree polynomial with a positive leading coefficient. It is neither convex nor concave. However, its third-order derivative is positive, so I can use that property when analyzing it. Imagine that you asked about why are only quadratic polynomials and not higher-degree ones used to interpolate points of a graph, and I answered that it is because higher-degree polynomials have more complex behavior. Well, more degrees of freedom come with more complexity, that is natural. – Pavel Kocourek Jan 25 '23 at 04:03
Well, I don't think I agree that it's an advantage. What I am trying to point out is that, for example, if I only give you that a cubic polynomial has positive second derivative and that another one has positive third derivative, which one has less possibilities? Which one is more useful for analysing a particular function, which I think is what most of the problems in analysis are? For your arguement about quadratics, I think that saying that higher-degree polynomials are more complex is a fair and sensible answer. – user7777777 Jan 25 '23 at 04:26
When I Google a) convex I receive 148M results; b) for monotone and monotonic the total is only 97.7M. In Google Scholar these counts are a) 4.27M and b) 2.48M. So the second order property (convexity) is studied / used more than the first-order property (monotonicity). How could this be consistent with your argument? – Pavel Kocourek Jan 25 '23 at 06:22
Because convexity is more advanced. You wouldn't search for solutions of seperable differential equations, but you would search for that of homogeneous and second-order ones, right? You are still not getting my point. A monotonic function is a very strict definition that's very exact, at least more than a concave/convex one which encompasses more possibilities. Monotonicity is stronger than convexity, but for example, some theorems might only require the weaker condition which makes the theorems themselves stronger etc. – user7777777 Jan 25 '23 at 07:52
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I think looking for Google results is misleading since convexity is a geometric property that has meaning in many situations outside of derivatives (see functional analysis and convex hulls, elementary geometry and many others etc), while monotonicity depends on order so generally speaking, is just a real-valued property; – Conrad Jan 25 '23 at 16:37

Why are monotonicity and convexity so intensely studied in comparison to their higher-order analogies?

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