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 John Armstrong | Analysis, Calculus

4 Comments »

[…] theorem that resembles the mean value theorem. And we’ll even use an approach like we did for the extension of that result to higher […]

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[…] the ball is convex, the line segment is completely contained within , and so we can bring the mean value theorem to bear. For each component function we can […]

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[…] also in for sufficiently small . Then writing and we find . The mean value theorem then tells us […]

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[…] identical to, but is not to be confused with, the closed line segment from to we used in the mean value theorem. I’ll try to keep them separate by always referring to this rectangular parallelepiped as an […]

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