# Abstract Nonsense

## The Mean Value Theorem for Multivariable Maps

Point of Post: In this post we state and prove the multidimensional analogue of the mean value theorem.

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Motivation

Anyone who has taken a basic analysis course knows that the mean value theorem (MVT) is a very important, and widely used tool. In fact, we’ve had several occasions to use it in our study of multidimensional analysis. So, the obvious question is “is there a multidimensional analogue?” Well, in the strictest sense, there isn’t. By this I mean if one literally transposes the usual MVT to higher dimensions one arrives at something like “Let $f$ be everywhere differentiable on $U$ and let $x,y\in U$. Then, there exists some $\xi\in \overline{xy}$ (where $\overline{xy}$ is the line segment connecting $x$ and $y$) such that $f(x)-f(y)=D_f(\xi)(x-y)$” Unfortunately, this is wildly false as the map $f:\mathbb{R}\to\mathbb{R}^2:t\mapsto (\cos(t),\sin(t))$ clearly shows. So, what then is the correct formulation? We play our old trick of thinking of a map $\mathbb{R}^n\to\mathbb{R}^m$ as secretly $\mathbb{R}\to\mathbb{R}^n\to\mathbb{R}^m$ by first mapping $t\mapsto a+tb$ for some vectors $a,b\in\mathbb{R}^n$ and then evaluating our map there (more explicitly something of the form $t\mapsto a+tb\mapsto f(a+tb)$). From there if we could find some way of going back into $\mathbb{R}$ we’d have a map $\mathbb{R}\to\mathbb{R}^n\to\mathbb{R}^m\to\mathbb{R}$ which, being an honest to god real valued real variable map, can have the one dimensional case of the MVT applied to it. So, the question remains as to what kind of maps $\mathbb{R}^m\to\mathbb{R}$ we want to consider? Well, we know from the above that we’re going to have to apply the chain rule to find the derivative of the map $\mathbb{R}\to\mathbb{R}$ and so we don’t want to pick something so crazy that we are left knowing nothing new. No, we’ll restrict our maps into $\mathbb{R}$ to be the simplest (in terms of derivatives), namely we’ll consider $\varphi\in\text{Hom}\left(\mathbb{R}^m,\mathbb{R}\right)$ and so our map really looks like $\varphi\circ f\circ g$ (where $g(t)=a+tb)$.

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Mean Value Theorem

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We begin with the theorem:

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Theorem: Let $f:U\to\mathbb{R}^m$, with $U\subseteq\mathbb{R}^n$ open, be differentiable and $\varphi\in\text{Hom}\left(\mathbb{R}^m,\mathbb{R}\right)$. Suppose further that the line $\overline{xy}\subseteq U$ for some $x,y\in U$. Then, there exists some $\xi\in \overline{xy}$ such that $\varphi\left(f(x)-f(y)\right)=\varphi\left(D_f(\xi)(x-a)\right)$.

Proof: As indicated in the motivation consider $g:\mathbb{R}\to\mathbb{R}^n$ given by $g(t)=t+t(x-y)$. Note then that since $\overline{xy}\subseteq U$ and $U$ is open we actually have that $g(-\delta,1+\delta)\subseteq U$ for some $\delta>0$. So, consider then the function $\varphi\circ f\circ g:(-\delta,\delta+1)\to\mathbb{R}$. Now, since (recalling that constant maps and multillinear maps are differentiable) each of these functions are differential be know from the chain rule that $\varphi\circ f\circ g$ is differentiable on $(-\delta,\delta+1)$. Thus, by the one-dimenionsal mean value theorem there exists $\eta\in(-\delta,\delta+1)$ such that

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$\varphi(f(g(1)))-\varphi(f(g(0)))=D_{\varphi\circ f\circ g}(\eta)(1)$

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That said, we first note that $\varphi(f(g(1)))-\varphi(f(g(0)))=\varphi(f(x))-\varphi(f(y))=\varphi(f(x)-f(y))$. Next, by the chain rule we have that

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$D_{\varphi\circ f}(g(\eta))\circ D_g(\eta)=D_\varphi(f(g(\eta))\circ D_f(g(\eta))\circ D_g(\eta)$

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Note though that a quick check shows that $D_g(\eta)(t)=t(x-y)$ (this is clear since $g$ is, up to an irrelevant constant, just the linear map $g(t)=t(x-y)$ and thus its derivative is itself) and so $D_g(\eta)(1)=x-y$. Next, we note that since $\varphi$ is linear we know that $D_\varphi(f(g(\eta))=\varphi$. Thus, finally letting $\xi=g(\eta)$ we may conclude that

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$(\varphi\circ f\circ g)'(1)=\varphi\left(D_f(\xi)(x-y)\right)$

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Thus, putting this all together

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$\varphi\left(f(x)-f(y)\right)=\varphi(f(g(1)))-\varphi(f(g(0)))=D_{\varphi\circ f\circ g}(\eta)(1)=\varphi\left(D_f(\xi)(x-y)\right)$

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lastly noting that $\xi\in\text{im}(g)\subseteq \overline{xy}$ finishes the argument. $\blacksquare$

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Remark: I believe it is important to point out precisely why we conclude from the one-dimensional mean value theorem that $\cdots=D_{\varphi\circ f\circ g}(\eta)(1)$. This is because we (thinking of $\varphi\circ f\circ g$ as a real valued function from analysis, i.e. that the derivative is a number) know from the mean value theorem that $\cdots=(\varphi\circ f\circ g)'(\eta)$ that said, we have from first principles that $D_{\varphi\circ f\circ g}(\eta)(t)=(\varphi\circ f\circ g)'(\eta)t$ from where that mysterious inclusion should now seem clear.

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Where this most often comes up (this is actually taken as the usual multidimensional mean value theorem, but if we recall the canonical isomorphism between an inner product space and its dual we realize that these two theorems are actually equivalent):

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Corollary: Let $f:U\to\mathbb{R}^m$, with $U\subseteq\mathbb{R}^n$ open , and $u\in\mathbb{R}^m$. Then, if $\overline{xy}\subseteq U$ for some $x,y\in U$ then there exists $\xi\in\overline{xy}$ such that $u\cdot(f(x)-f(y))=u\cdot D_f(\xi)(x-y)$.

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We now prove a theorem which, in all calculus textbooks, immediately follows the multivariable mean value theorem. Namely, it is the generalization that the only functions with zero derivatives on open connected subsets of the real numbers are the constants. Namely:

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Theorem: Let $f:U\to\mathbb{R}^m$, where $U\subseteq\mathbb{R}^n$ is open and connected, is differentiable everywhere with $D_f(a)=\bold{0}$ for all $a\in U$. Then, $f$ is constant on $U$.

Proof: Let $x,y\in U$ be arbitrary. It is a common fact that every open connected subspace of Euclidean space is polygonally line connected, so let $t_1,\cdots,t_k$ be points $\overline{t_it_{i+1}}\subseteq U$ for $i=1,\cdots,k-1$ and $t_1=x,t_k=y$. We evidently have then by the mean value theorem that there exists some $c_i\in \overline{t_it_{i+1}}$ such that $\left(f(x)-f(y)\right)\cdot f(t_i)-f(t_{i+1})=\left(f(x)-f(y)\right)\cdot D_f(c_i)(t_i-t_{i+1})$ but by assumption this right hand side is zero. Thus,

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$\displaystyle 0=\sum_{i=1}^{k-1}\left(f(x)-f(y)\right)\cdot\left(f(t_i)-f(t_{i+1})\right)=\left(f(x)-f(y)\right)\cdot\left(\sum_{i=1}^{k-1}f(t_i)-f(t_{i+1})\right)$

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but this sum telescopes to $f(x)-f(y)$ and so we see that $\|f(x)-f(y)\|^2=0$ and so $f(x)=f(y)$. Since $x,y\in U$ were arbitrary the conclusion follows. $\blacksquare$

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From this we get the obvious corollary:

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Corollary: Let $f,g:U\to\mathbb{R}^m$, with $U\subseteq\mathbb{R}^n$ open, be such that $D_f(a)=D_g(a)$ for every $a\in U$. Then, there exists $c\in\mathbb{R}^m$ such that $f(a)=g(a)+c$ for every $a\in U$.

Proof: We merely note that $D_f-D_g=D_{f-g}$ is zero everywhere on $U$ and so by the previous theorem $f-g=c$ for some constant $c$. The conclusion follows. $\blacksquare$

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References:

1.  Spivak, Michael. Calculus on Manifolds; a Modern Approach to Classical Theorems of Advanced Calculus. New York: W.A. Benjamin, 1965. Print.

2. Apostol, Tom M. Mathematical Analysis. Reading, MA: Addison-Wesley Pub., 1974. Print.

June 11, 2011 -