mixedmath

This is joint work with Paul Carter.We completed this while on a cross-country drive as we moved the newly minted Dr. Carter from Brown to Arizona.

I've had a longtime fascination with the standard mean value theorem of calculus.

Mean Value Theorem Suppose $f$ is a differentiable function. Then there is some $c \in (a,b)$ such that \begin{equation} \frac{f(b) - f(a)}{b-a} = f'(c). \end{equation}

The idea for this project started with a simple question: what happens when we interpret the mean value theorem as a differential equation and try to solve it? As stated, this is too broad. To narrow it down, we might specify some restriction on the $c$, which we refer to as the mean value abscissa, guaranteed by the Mean Value Theorem.

So I thought to try to find functions satisfying \begin{equation} \frac{f(b) - f(a)}{b-a} = f' \left( \frac{a + b}{2} \right) \end{equation} for all $a$ and $b$ as a differential equation. In other words, let's try to find all functions whose mean value abscissas are midpoints.

This looks like a differential equation, which I only know some things about. But my friend and colleague Paul Carter knows a lot about them, so I thought it would be fun to ask him about it.

He very quickly told me that it's essentially impossible to solve this from the perspective of differential equations. But like a proper mathematician with applied math leanings, he thought we should explore some potential solutions in terms of their Taylor expansions. Proceeding naively in this way very quickly leads to the answer that those (assumed smooth) solutions are precisely quadratic polynomials.

It turns out that was too simple. It was later pointed out to us that verifying that quadratic polynomials satisfy the midpoint mean value property is a common exercise in calculus textbooks, including the one we use to teach from at Brown. Digging around a bit reveals that this was even known (in geometric terms) to Archimedes.

So I thought we might try to go one step higher, and see what's up with \begin{equation}\label{eq:original_midpoint} \frac{f(b) - f(a)}{b-a} = f' (\lambda a + (1-\lambda) b), \tag{1} \end{equation} where $\lambda \in (0,1)$ is a weight. So let's find all functions whose mean value abscissas are weighted averages. A quick analysis with Taylor expansions show that (assumed smooth) solutions are precisely linear polynomials, except when $\lambda = \frac{1}{2}$ (in which case we're looking back at the original question).

That's a bit odd. It turns out that the midpoint itself is distinguished in this way. Why might that be the case?

It is beneficial to look at the mean value property as an integral property instead of a differential property, \begin{equation} \frac{1}{b-a} \int_a^b f'(t) dt = f'\big(c(a,b)\big). \end{equation} We are currently examining cases when $c = c_\lambda(a,b) = \lambda a + (1-\lambda b)$. We can see the right-hand side is differentiable by differentiating the left-hand side directly. Since any point can be a weighted midpoint, one sees that $f$ is at least twice-differentiable. One can actually iterate this argument to show that any $f$ satisfying one of the weighted mean value properties is actually smooth, justifying the Taylor expansion analysis indicated above.

An attentive eye might notice that the midpoint mean value theorem, written as the integral property \begin{equation} \frac{1}{b-a} \int_a^b f'(t) dt = f' \left( \frac{a + b}{2} \right) \end{equation} is exactly the one-dimensional case of the harmonic mean value property, usually written \begin{equation} \frac{1}{\lvert B_h \rvert} = \int_{B_h(x)} g(t) dV = g(x). \end{equation} Here, $B_h(x)$ is the ball of radius $h$ and center $x$. Any harmonic function satisfies this mean value property, and any function satisfying this mean value property is harmonic.

From this viewpoint, functions satisfying our original midpoint mean value property~\eqref{eq:original_midpoint} have harmonic derivatives. But the only one-dimensional harmonic functions are affine functions $g(x) = cx + d$. This gives immediately that the set of solutions to~\eqref{eq:original_midpoint} are quadratic polynomials.

The weighted mean value property can also be written as an integral property. Trying to connect it similarly to harmonic functions led us to consider functions satisfying \begin{equation} \frac{1}{\lvert B_h \rvert} = \int_{B_h(x)} g(t) dV = g(c_\lambda(x,h)), \end{equation} where $c_\lambda(x,h)$ should be thought of as some distinguished point in the ball $B_h(x)$ with a weight parameter $\lambda$. More specifically,

Are there weighted harmonic functions corresponding to a weighted harmonic mean value property? In one dimension, the answer is no, as seen above. But there are many more multivariable harmonic functions [in fact, I've never thought of harmonic functions on $\mathbb{R}^1$ until this project, as they're too trivial]. So maybe there are weighted harmonic functions in higher dimensions?

This ends up being the focus of the latter half of our paper. Unexpectedly (to us), an analogous methodology to our approach in the one-dimensional case works, with only a few differences.

It turns out that no, there are no weighted harmonic functions on $\mathbb{R}^n$ other than trivial extensions of harmonic functions from $\mathbb{R}^{n-1}$.

Harmonic functions are very special, and even more special than we had thought. The paper is a fun read, and can be found on the arxiv now. It has been accepted and will appear in American Mathematical Monthly.

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