The Mean Value Theorem for Differentiable Functions from Rn to Rm

# The Mean Value Theorem for Differentiable Functions from Rn to Rm

Recall that if $f$ is a continuous function on the closed interval $[x, y]$ and differentiable on the open interval $(x, y)$ (where we assume $x < y$) then there exists a number $c \in (x, y)$ for which:

(1)
\begin{align} \quad f'(c) = \frac{f(y) - f(x)}{y - x} \quad \Leftrightarrow \quad f(y) - f(x) = f'(c)(y - x) \end{align}

We would like to generalize this extremely important result to differentiable functions from $\mathbb{R}^n$ to $\mathbb{R}^m$. Doing so is actually not that straightforward though. The equation above does not immediately generalize to differentiable functions from $\mathbb{R}^n$ to $\mathbb{R}^m$ and we will need to do some more work in order to make a meaningful generalization.

To emphasize this, consider the function $\mathbf{f} : \mathbb{R} \to \mathbb{R}^2$ defined for all $x \in \mathbb{R}$ by:

(2)
\begin{align} \quad \mathbf{f}(x) = (\cos x, \sin x) \end{align}

Then the total derivative of $\mathbf{f}$ at $x$ evaluated at any $h \in \mathbb{R}$ is:

(3)
\begin{align} \quad \mathbf{f}'(x)(h) = (-\sin x, \cos x)h \end{align}

Therefore we have that:

(4)
\begin{align} \quad \mathbf{f}(y) - \mathbf{f}(x) = (\cos y, \sin y) - (\cos x, \sin x) = (\cos y - \cos x, \sin y - \sin x) \quad (*) \end{align}

And also:

(5)
\begin{align} \quad \mathbf{f}'(z)(y - x) = (-\sin z, \cos z)(y - x) \quad (**) \end{align}

Now set $x = 0$ and $y = 2\pi$. Then $(*)$ will always equal the zero vector, $\mathbf{0}$, and $(**)$ will never equal the zero vector for any choice of $z$ between $0$ and $2\pi$. Therefore we see that $\mathbf{f}(y) - \mathbf{f}(x) \neq \mathbf{f}'(z)(y - x)$ in general.

 Theorem 1 (The Mean Value Theorem): Let $S \subseteq \mathbb{R}^n$ be open and let $\mathbf{f} : S \to \mathbb{R}^m$ be differentiable on all of $S$. Let $\mathbf{x}, \mathbf{y} \in S$ be such that the line segment connecting these two points is contained in $S$, i.e., $L(\mathbf{x}, \mathbf{y}) \subset S$. Then for every $\mathbf{a} \in \mathbb{R}^m$ there exists a point $\mathbf{z} \in L(\mathbf{x}, \mathbf{y})$ such that $\mathbf{a} \cdot [\mathbf{f}(\mathbf{y}) - \mathbf{f}(\mathbf{x})] = \mathbf{a} \cdot [\mathbf{f}'(\mathbf{z})(\mathbf{y} - \mathbf{x})]$.

In the following Theorem we use the notation "$L(\mathbf{x}, \mathbf{y})$" to denote the line segment that joints the point $\mathbf{x}$ to $\mathbf{y}$. This line segment can be parameterized as $L(\mathbf{x}, \mathbf{y}) = \{ (1 - t)\mathbf{x} + t \mathbf{y} : t \in [0, 1] \}$.

• Proof: Let $\mathbf{a} \in \mathbb{R}^m$ and define a new function $F : [0, 1] \to \mathbb{R}$ for all $t \in [0, 1]$ by:
(6)
\begin{align} \quad F(t) = \mathbf{a} \cdot \mathbf{f}(\mathbf{x} + t(\mathbf{y} - \mathbf{x})) \end{align}
• Since $\mathbf{f}$ is differentiable on $S$ we have from the Differentiable Functions from Rn to Rm are Continuous page that $\mathbf{f}$ is continuous on $S$ and so $F$ must continuous on $[0, 1]$. Furthermore, $F$ is differentiable on $(0, 1)$ by the chain rule:
(7)
\begin{align} \quad F'(t) = \mathbf{a} \cdot \mathbf{f}'(\mathbf{x} + t(\mathbf{y} - \mathbf{x})) (\mathbf{y} - \mathbf{x}) \end{align}
• So by the Mean Value Theorem for single-variable real-valued functions, for $x = 0$ and $y = 1$ there exists a number $h \in (0, 1)$ for which:
(8)
\begin{align} \quad F(1) - F(0) &= F'(h)(1 - 0) \quad (*)\\ \end{align}
• The lefthand side of $(*)$ is:
(9)
\begin{align} \quad F(1) - F(0) = \mathbf{a} \cdot \mathbf{f}(\mathbf{y}) - \mathbf{a} \cdot \mathbf{f}(\mathbf{x}) = \mathbf{a} \cdot (\mathbf{f}(\mathbf{y}) - \mathbf{f}(\mathbf{x})) \end{align}
• The righthand side of $(*)$ is:
(10)
\begin{align} \quad \mathbf{F}'(h)(1 - 0) = \mathbf{F}'(h) = \mathbf{a} \cdot \mathbf{f}'(\mathbf{x} + h(\mathbf{y} - \mathbf{x})) (\mathbf{y} - \mathbf{x}) \end{align}
• Set $\mathbf{z} = \mathbf{x} + h(\mathbf{y} - \mathbf{x})$. Then $\mathbf{z} \in L(\mathbf{x}, \mathbf{y})$ and we have from the equality at $(*)$ that:
(11)
\begin{align} \quad \mathbf{a} \cdot [\mathbf{f}(\mathbf{y}) - \mathbf{f}(\mathbf{x})] = \mathbf{a} \cdot \mathbf{f}'(\mathbf{z})(\mathbf{y} - \mathbf{x}) \quad \blacksquare \end{align}