Mean-Value Theorems for Vector Valued Functions

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:

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

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:

${\begin{aligned}\quad \mathbf {f} (x)=(\cos x,\sin x)\end{aligned}}$

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

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

Therefore we have that:

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

And also:

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

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:

${\begin{aligned}\quad F(t)=\mathbf {a} \cdot \mathbf {f} (\mathbf {x} +t(\mathbf {y} -\mathbf {x} ))\end{aligned}}$

• 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:

${\begin{aligned}\quad F'(t)=\mathbf {a} \cdot \mathbf {f} '(\mathbf {x} +t(\mathbf {y} -\mathbf {x} ))(\mathbf {y} -\mathbf {x} )\end{aligned}}$

• 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:
  ${\begin{aligned}\quad F(1)-F(0)=F'(h)(1-0)\quad (*)\\\end{aligned}}$

• The lefthand side of $(*)$ is:

${\begin{aligned}\quad F(1)-F(0)=\mathbf {a} \cdot \mathbf {f} (\mathbf {y} )-\mathbf {a} \cdot \mathbf {f} (\mathbf {x} )=\mathbf {a} \cdot (\mathbf {f} (\mathbf {y} )-\mathbf {f} (\mathbf {x} ))\end{aligned}}$

• The righthand side of $(*)$ is:

${\begin{aligned}\quad \mathbf {F} '(h)(1-0)=\mathbf {F} '(h)=\mathbf {a} \cdot \mathbf {f} '(\mathbf {x} +h(\mathbf {y} -\mathbf {x} ))(\mathbf {y} -\mathbf {x} )\end{aligned}}$

• 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:

${\begin{aligned}\quad \mathbf {a} \cdot [\mathbf {f} (\mathbf {y} )-\mathbf {f} (\mathbf {x} )]=\mathbf {a} \cdot \mathbf {f} '(\mathbf {z} )(\mathbf {y} -\mathbf {x} )\quad \blacksquare \end{aligned}}$

Licensing

Content obtained and/or adapted from:

• The Mean Value Theorem for Differentiable Functions from Rn to Rm, mathonline.wikidot.com under a CC BY-SA license