CAL1_L18

The mean value theorem

Recall the mean value theorem for functions from $\mathbb{R}$ to $\mathbb{R}$. It is false, in general, for vector valued functions from ${\mathbb{R}}^n$ to ${\mathbb{R}}^m$ when $m>1$, as illustrated here. However, a useful generalization can be obtained by using the dot product:

Theorem 1(Mean Value Theorem).

Let $S$ be an open subset of ${\mathbb{R}}^n$ and assume that $\mathbf{f}:S\to {\mathbb{R}}^m$ is differentiable at each point of $S$. Let $\mathbf{x}$ and $\mathbf{y}$ be two points in $S$ such that $L(\mathbf{x},\mathbf{y})\subseteq S$. Then for every vector $\mathbf{\alpha }\in {\mathbb{R}}^m$ there is a point $\mathbf{z}\in L(\mathbf{x},\mathbf{y})$ such that

$$\mathbf{\alpha }\cdot (\mathbf{f}(\mathbf{y})-\mathbf{f}(\mathbf{x}))=\mathbf{\alpha }\cdot ({\mathbf{f}}^{\prime }(\mathbf{z})(\mathbf{y}-\mathbf{x})).$$

Proof
Let $\mathbf{u}=\mathbf{y}-\mathbf{x}$. Since $S$ is open and $L(\mathbf{x},\mathbf{y})\subseteq S$, there is a $\delta >0$ such that $\mathbf{x}+t\mathbf{u}\in S$ for all real $t$ in the interval $(-\delta ,1+\delta )$. Let $\mathbf{a}$ be a fixed vector in ${\mathbb{R}}^m$ and let $F$ be the real valued function defined on $(-\delta ,1+\delta )$ by $F(t)=\mathbf{a}\cdot \mathbf{f}(\mathbf{x}+t\mathbf{u})$. Then, from the chain rule, $F$ is differentiable on $(-\delta ,1+\delta )$ and its derivative is given by

$$F^{\prime }(t)=\mathbf{a}\cdot {\mathbf{f}}^{\prime }(\mathbf{x}+t\mathbf{u};\mathbf{u})=\mathbf{a}\cdot \{{\mathbf{f}}^{\prime }(\mathbf{x}+t\mathbf{u})(\mathbf{u})\}.$$

By the usual mean value theorem, we have $F(1)-F(0)=F^{\prime }(\theta )$ for some $\theta \in (0,1)$. Now,

$$F^{\prime }(\theta )=\mathbf{a}\cdot \{{\mathbf{f}}^{\prime }(\mathbf{x}+\theta \mathbf{u})(\mathbf{u})\}=\mathbf{a}\cdot \{{\mathbf{f}}^{\prime }(\mathbf{z})(\mathbf{y}-\mathbf{x})\},$$

where $\mathbf{z}\equiv \mathbf{x}+\theta \mathbf{u}\in L(\mathbf{x},\mathbf{y})$. Note that $F(1)-F(0)=\mathbf{a}\cdot \{\mathbf{f}(\mathbf{y})-\mathbf{f}(\mathbf{x})\}$ and we are done.

We can now easily prove the weaker version stated here. Let $\mathbf{f}:S\to {\mathbb{R}}^m$ where $S\subseteq \mathbb{R}$ and $[x,y]\subseteq S$ be differentiable. Take $\mathbf{\alpha }$ to be $\mathbf{f}(y)-\mathbf{f}(x)$:

$$\begin{array}{rl}\Vert \mathbf{f}(y)-\mathbf{f}(x){\Vert }^2& =(\mathbf{f}(y)-\mathbf{f}(x))\cdot ({\mathbf{f}}^{\prime }(z)(y-x))\\ & \le (y-x)\Vert \mathbf{f}(y)-\mathbf{f}(x)\Vert \Vert {\mathbf{f}}^{\prime }(z)\Vert \\ & \\ ⟹\Vert \mathbf{f}(y)-\mathbf{f}(x)\Vert & \le (y-x)\Vert {\mathbf{f}}^{\prime }(z)\Vert ,\end{array}$$

for some $z\in [x,y]$. More generally, if $S\subseteq {\mathbb{R}}^n$, we have

$$\Vert \mathbf{f}(\mathbf{y})-\mathbf{f}(\mathbf{x})\Vert \le \Vert {\mathbf{f}}^{\prime }(\mathbf{z})(\mathbf{y}-\mathbf{x})\Vert .$$

Moreover, using this result, we have

$$\Vert \mathbf{f}(\mathbf{y})-\mathbf{f}(\mathbf{x})\Vert \le M\Vert \mathbf{y}-\mathbf{x}\Vert ,$$

where $M={\sum }_{k=1}^m\Vert \nabla f_k(\mathbf{z})\Vert$. Note that $M$ depends on $\mathbf{z}$ and hence on $\mathbf{x}$ and $\mathbf{y}$. However, if all the partial derivatives $D_j{f}_k$ are bounded on $S$, there exists an upper bound $A$ for $M$, and hence

$$\Vert \mathbf{f}(\mathbf{y})-\mathbf{f}(\mathbf{x})\Vert \le A\Vert \mathbf{y}-\mathbf{x}\Vert$$

for all $\mathbf{x},\mathbf{y}\in S$ (so $\mathbf{f}$ satisfies the Lipschitz condition, with the requirement that $\mathbf{f}$ map between the same spaces being relaxed).

Functions with bounded and zero total derivative

Theorem 2(Theorem).

Suppose $\mathbf{f}$ maps a convex open set $E\subseteq {\mathbb{R}}^n$ in ${\mathbb{R}}^m$, $\mathbf{f}$ is differentiable in $E$, and there is a real number $M$ such that for every $\mathbf{x}\in E$,

$$\Vert {\mathbf{f}}^{\prime }(\mathbf{x})(\mathbf{v})\Vert \le M\Vert \mathbf{v}\Vert$$

for every $\mathbf{v}\in {\mathbb{R}}^n$. Then

$$\Vert \mathbf{f}(\mathbf{b})-\mathbf{f}(\mathbf{a})\Vert \le M\Vert \mathbf{b}-\mathbf{a}\Vert$$

for all $\mathbf{a}\in E$, $\mathbf{b}\in E$.

Proof
Since $E$ is convex, we can define $\mathbf{g}(t)\equiv \mathbf{f}(\mathbf{a}+t(\mathbf{b}-\mathbf{a}))$ for $t\in [0,1]$. Now, ${\mathbf{g}}^{\prime }(t)={\mathbf{f}}^{\prime }(\mathbf{a}+t(\mathbf{b}-\mathbf{a});\mathbf{b}-\mathbf{a})$, so that

$$\Vert {\mathbf{g}}^{\prime }(t)\Vert =\Vert {\mathbf{f}}^{\prime }(\mathbf{a}+t(\mathbf{b}-\mathbf{a}))(\mathbf{b}-\mathbf{a})\Vert \le M\Vert \mathbf{b}-\mathbf{a}\Vert$$

for all $t\in [0,1]$. It follows from the mean value theorem that for some $x\in [0,1]$,

$$\begin{array}{r}\Vert \mathbf{g}(1)-\mathbf{g}(0)\Vert \le \Vert {\mathbf{g}}^{\prime }(x)\Vert \le M\Vert \mathbf{b}-\mathbf{a}\Vert .\end{array}$$

Theorem 3(Theorem).

Let $S$ be an open connected subset of ${\mathbb{R}}^n$, and let $\mathbf{f}:S\to {\mathbb{R}}^m$ be differentiable at each point of $S$. If ${\mathbf{f}}^{\prime }(\mathbf{c})=0$ for each $\mathbf{c}$ in $S$, then $\mathbf{f}$ is constant on $S$.

Follows as a corollary of the previous theorem by plugging $M=0$. Here’s another proof.

Proof
Since $S$ is open and connected, it is polygonally connected, $i$.$e$, every pair of points $\mathbf{x}$ and $\mathbf{y}$ in $S$ can be joined by a polygonal arc lying in $S$. Denote the vertices of this arc by ${\mathbf{p}}_1,\dots ,{\mathbf{p}}_r$, where ${\mathbf{p}}_1=\mathbf{x}$ and ${\mathbf{p}}_r=\mathbf{y}$. Since each segment $L({\mathbf{p}}_{i+1},{\mathbf{p}}_i)\subseteq S$, the mean value theorem shows that

$$\mathbf{a}\cdot (\mathbf{f}({\mathbf{p}}_{i+1})-\mathbf{f}({\mathbf{p}}_i))=0,$$

for every $\mathbf{a}\in {\mathbb{R}}^m$. Adding these equations for $i=1,2,\dots ,r-1$, we get

$$\mathbf{a}\cdot (\mathbf{f}(\mathbf{y})-\mathbf{f}(\mathbf{x}))=0,$$

for every $\mathbf{a}$. Taking $\mathbf{a}=\mathbf{f}(\mathbf{y})-\mathbf{f}(\mathbf{x})$, we find $\mathbf{f}(\mathbf{x})=\mathbf{f}(\mathbf{y})$, so $\mathbf{f}$ is constant on $S$.