# The Unapologetic Mathematician

## The Mean Value Theorem

Here’s a nice technical result we may have call for from time to time: a higher-dimensional version of the differential mean value theorem. Remember that this says that if we’ve got a function $f$ continuous on the closed interval $\left[a,b\right]\subseteq\mathbb{R}$ and differentiable on its interior, there is some point $\xi$ in the middle where the derivative of the function is the same as the average — the mean — rate of change of the function over the interval. In more than one dimension we’re going to modify this a bit to make it clearer what it means.

First of all, instead of talking about the closed interval $\left[a,b\right]$, we’re going to use the closed straight line segment. That is, the collection of all the points between $a$ and $b$ in a straight line, and including the endpoints. We first look at the total displacement $b-a$ from one point to the other. Then we start at $a$ and move some portion of this displacement towards $b$. That is, the closed line segment $\left[a,b\right]$ consists of all points of the form $a+t(b-a)$ for $t$ in the closed interval $\left[0,1\right]$. Setting $t=0$ gives us the point $a$, and $t=1$ gives us the point $b$. Similarly, the open line segment $\left(a,b\right)$ consists of all points of the form $a+t(b-a)$ for $t$ in the open interval $\left(0,1\right)$.

Next, we have to be clear about the average rate of change. As we move from $a$ to $b$, the value of the function $f$ changes by $f(b)-f(a)$. It takes a displacement of $\lVert b-a\rVert$ to get there, so on average the rate of change is

$\displaystyle\frac{f(b)-f(a)}{\lVert b-a\rVert}$

Finally, we don’t just have a single value for the instantaneous rate of change, we have a differential $df(\xi)$. But we can use it to find directional derivatives. Specifically, we’ll consider the derivative of $f$ in the direction pointing from $a$ to $b$. We’ll pick out this direction with the unit vector we get by normalizing the displacement

$\displaystyle\frac{b-a}{\lVert b-a\rVert}$

So the mean value theorem will tell us that if $f$ is differentiable in some open region $S$ that contains the whole closed line segment $\left[a,b\right]$. Then there is some point $\xi$ in the open line segment $\left(a,b\right)$ so that the average rate of change of $f$ from $a$ to $b$ is equal to the directional derivative of $f$ at $\xi$ in the direction pointing from $a$ to $b$:

$\displaystyle\frac{f(b)-f(a)}{\lVert b-a\rVert}=df(\xi)\left(\frac{b-a}{\lVert b-a\rVert}\right)$

or, more simply

$\displaystyle f(b)-f(a)=df(\xi)\left(b-a\right)$

We’ll get at this by changing to a function of one variable so we can bring the one-dimensional version to bear. To that end, we define $h(t)=f(a+t(b-a))$ for $t$ in the closed interval $\left[0,1\right]$. Then $f(b)-f(a)=h(1)-h(0)$, and we can also show that $h$ is differentiable everywhere inside the interval. Indeed, we can evaluate the difference quotient

$\displaystyle\frac{h(s)-h(t)}{s-t}=\frac{f((a+t(b-a))+(s-t)(b-a))-f(a+t(b-a))}{s-t}$

Taking the limit as $s$ approaches $t$, we find

$\displaystyle h'(t)=\left[D_{b-a}f\right](a+t(b-a))=df(a+t(b-a))(b-a)$

which exists since $f$ is differentiable.

So our old differential mean-value theorem tells us that there is some $\tau\in\left(0,1\right)$ so that

\displaystyle\begin{aligned}f(b)-f(a)&=h(1)-h(0)\\&=\frac{h(1)-h(0)}{1-0}\\&=h'(\tau)\\&=df(a+\tau(b-a))(b-a)\\&=df(\xi)(b-a)\end{aligned}

where $\xi=a+\tau(b-a)$ is a point in the open line segment $(a,b)$.

October 13, 2009 Posted by | Analysis, Calculus | 4 Comments