7
$\begingroup$

I've been asked to prove multi-dimensional Mean Value Theorem. I'd be grateful if someone could give me feedback if it is okay.

Proof of Mean Value Theorem:

Let $f: [a,b]\rightarrow \mathbb{R}$ be a continuous on $[a,b]$ and differentiable on $(a,b)$. Consider the function:

$$g(x)=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a) \mbox{.}$$

This function is continuous on $[a,b]$, differentiable on $(a,b)$ and $g(a)=g(b)$. Thus there is $c\in (a,b)$ such that $g'(c)=0$. But this means that there is $c\in (a,b)$ such that

$$f'(c)=\frac{f(b)-f(a)}{b-a}\mbox{.}$$

Proof of multi-dimensional Mean Value Theorem:

Let $f:U\rightarrow\mathbb{R}$ be a differentiable function ($U$ is an open subset of $\mathbb{R}^n)$. Let $\mathbf{a}$ and $\mathbf{b}$ be points in $U$ such that the entire line segment between them is contained in $U$. Define $h:[0,1]\rightarrow U$ in the following way:

$$h(t)=(a_1+(b_1-a_1)t,\ldots,a_n+(b_n-a_n)t) \mbox{.}$$

This function is differentiable on $(0,1)$ and continuous on $[0,1]$, so is $f \circ h$. If we apply Mean Value Theorem to $f\circ h$ we get

$$(f \circ h )'(c)=(f \circ h )(1)-(f \circ h )(0)$$

where $c\in (0,1)$ and

$$f '(h(c))(\mathbf{b}-\mathbf{a})=f(\mathbf{b})-f(\mathbf{a}) \mbox{.}$$

If we set $\zeta=h(c)$ we get

$$f '(\zeta)=\frac{f(\mathbf{b})-f(\mathbf{a})}{\mathbf{b}-\mathbf{a}} \mbox{.}$$

(Obviously $f '(\zeta)$ is a gradient vector.)

$\endgroup$
9
$\begingroup$

Everything is fine, except the last formula: You cannot divide by a vector. Leave it at $$f({\bf b})-f({\bf a})=\nabla f({\bf z})\cdot({\bf b}-{\bf a})$$ for some ${\bf z}\in[{\bf a},{\bf b}]$.

$\endgroup$
  • $\begingroup$ Thank you, very good point! $\endgroup$ – luka5z Dec 29 '15 at 8:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.