$\newcommand{\R}{\mathbb R }$ $\newcommand{\N}{\mathbb N }$ $\newcommand{\Z}{\mathbb Z }$ $\newcommand{\bfa}{\mathbf a}$ $\newcommand{\bfb}{\mathbf b}$ $\newcommand{\bfc}{\mathbf c}$ $\newcommand{\bff}{\mathbf f}$ $\newcommand{\bfg}{\mathbf g}$ $\newcommand{\bfG}{\mathbf G}$ $\newcommand{\bfh}{\mathbf h}$ $\newcommand{\bfu}{\mathbf u}$ $\newcommand{\bfx}{\mathbf x}$ $\newcommand{\bfp}{\mathbf p}$ $\newcommand{\bfy}{\mathbf y}$ $\newcommand{\ep}{\varepsilon}$
Theorem 1. the Mean Value Theorem. Assume that $f$ is a real-valued function of class $C^1$ defined on an open set $S\subset\R^n$. For two points $\bfa,\bfb\in S$, let $L_{\bfa, \bfb}$ denote the line segment that connects them. If $L_{\bfa,\bfb}\subset S$, then there exists $\bfc\in L_{\bfa,\bfb}$ such that $$ f(\bfb)-f(\bfa) = (\bfb-\bfa)\cdot \nabla f(\bfc) $$
Proof.
For $s\in [0,1]$, let $\gamma(s) := s\bfb+(1-s)\bfa$. Note that $$ L_{\bfa,\bfb} = \{ \gamma(s) : 0\le s \le 1 \}. $$ Next, define $$ \phi(s) := f(\gamma(s)). $$ According to the single-variable Mean Value Theorem from MAT137, there exist $\sigma\in (0,1)$ such that $$ \frac{\phi(1) - \phi(0)}{1-0} = \phi'(\sigma). $$ Also, the Chain Rule implies that $\phi'(\sigma) = \nabla f(\gamma(\sigma))\cdot (\bfb-\bfa)$. So if we define $\bfc := \gamma(\sigma)$, then $\bfc\in L_{\bfa,\bfb}$, and since $\phi(0)=f(\bfa)$ and $\phi(1) = f(\bfb)$, the above identity becomes $$ f(\bfb) - f(\bfa) = \nabla f(\bfc)\cdot(\bfb-\bfa). \qquad \qquad\Box $$
The $1d$ Mean Value Theorem, familar from MAT137, is used to prove things like
if a function $f$ has the property that $f'(t)=0$ for all $t$ in an interval $(a,b)$, then $f$ is constant on $(a,b)$.
if a function $f$ has the property that $|f'(t)|\le M$ for all $t$ in an interval $(a,b)$, then the slope of $f$ between any two points is at most $M$. More precisely, $$ |f'(t)|\le M\mbox{ for all }t\in (a,b) \quad\Rightarrow \quad |f(t) - f(s)| \le M|t-s| \mbox{ for all }s,t\in (a,b). $$
In this section we will show how the Mean Value Theorem can be used to prove similar facts in higher dimensions.
First, we introduce a class of sets on which the Mean Value Theorem is particularly useful.
Definition 1: A set $S\subset \R^n$ is said to be convex if, for every $\bfa, \bfb\in S$, the line segment $L_{\bfa,\bfb}$ is contained in $S$. That is, $$ \forall \bfa,\bfb\in S,\forall s\in [0,1], \qquad s\bfb+ (1-s)\bfa \in S. $$
In other words, if $S$ is convex, then the geometric assumption in the Mean Value Theorem is satisfied for every pair of points $\bfa$ and $\bfb$ in $S$.
Example 1. A ball $B(r,\bfp)$ is convex.
The proof below is essentially copied from Section 1.5, where we proved that $B(r,\bfp)$ is path-connected. As you can see, the proof we gave there actually shows that it is convex.
Proof.
Let's write
$$
\gamma(s) = \bfa + s(\bfb - \bfa) = (1-s) \bfa + s\bfb .
$$
We have to show that $|\gamma(s) - \bfp|<r$ for all $s\in [0,1]$.
In fact this is the case, because for $s\in [0,1]$,
\begin{align*}
|\gamma(s) - \bfp|
&=
|(1-s) \bfa + s\bfb - \bfp| &\mbox{(definiton of $\gamma(s)$)}\\
&=
|(1-s) (\bfa-\bfp) + s(\bfb - \bfp)| &\mbox{(rewrite)}\\
&\le
|(1-s) (\bfa-\bfp)| + |s(\bfb - \bfp)| &\mbox{(triangle ineq.)}\\
%&=
%|1-s|\, |\bfa-\bfp| + |s|\ |\bfb - \bfp| &\mbox{(triangle ineq.)}\\
&<
(1-s) r + s r = r &\mbox{ since $\bfa, \bfb \in B(r,\bfp)$}.
\end{align*}
Thus $B(r,\bfp)$ is convex. $\qquad\qquad \Box$
Examples 2. Here are a number of other examples of convex sets. The proofs are execises.
A solid ellipsoid is convex. By this we mean a set of the form $$ S = \{ \bfx \in \R^n : (x_1/a_1)^2+ \cdots + (x_n/a_n)^2 \le 1\} $$ where $a_1,\ldots, a_n$ are nonzero constants.
An intersection of convex sets is convex. (A union of convex sets need not be convex; you can easily convince yourself of this by drawing a picture.)
A subspace of $\R^n$ is convex. In particular, the range and the nullspace of a matrix are both convex.
If $L:\R^n \to \R^m$ is a function of the form $$ L(\bfx) = A\bfx + \bfb\qquad\mbox{ where }A\mbox{ is an }m\times n\mbox{ matrix, and }\bfb\in \R^m $$ and if $S$ is a convex subset of $\R^n$, then $$ L(S) := \{L(\bfx) : \bfx\in S\}\quad\mbox{ is convex}. $$
Theorem 2. Assume that $S$ is an open, convex subset of $\R^n$ and that $f:\R^n\to \R$ is a function that is differentiable in $S$, and moreover that there exists $M\ge 0$ such that $|\nabla f(\bfx)|\le M$ for all $\bfx\in S$. Then for every $\bfa, \bfb\in S$, $$ |f(\bfb)- f(\bfa)| \le M |\bfb - \bfa|. $$
This is very similar to a standard application of the 1d-mean value theorem.
Proof.
Fix any $\bfa,\bfb\in S$. The Mean Value Theorem implies that there exists some $\bfc\in L_{\bfa,\bfb}\subset S$ such that $$ f(\bfb) - f(\bfa) = \nabla f(\bfc)\cdot (\bfb - \bfa). $$ Then Cauchy's inequality implies that $$ |f(\bfb) - f(\bfa)| = |\nabla f(\bfc)\cdot (\bfb - \bfa)| \le |\nabla f(\bfc)| \ |\bfb - \bfa|. $$ Our hypotheses imply that $|\nabla f(\bfc)|\le M$, so the conclusion of the theore follows.
Theorem 3. Assume that $S$ is an open, convex subset of $\R^n$ and that $f:\R^n\to \R$ is a function that is differentiable in $S$. If $\nabla f(\bfx )={\bf 0}$ for every $\bfx\in S$, then $f$ is constant on $S$.
This is the multi-variable version of a familiar theorem from first-year calculus: if $f'=0$ everywhere on an interval, then $f$ is constant on that interval. (Recall, the proof of that theorem uses the 1d version of the the mean value theorem.)
Proof. Fix some $\bfa \in S$ and let $c = f(\bfa)$. Apply Theorem 2 with $M=0$ to find that for every $\bfb\in S$, $$ |f(\bfb) - c| = |f(\bfb)-f(\bfa)| \le 0\cdot |\bfb-\bfa| = 0. $$ Thus $f(\bfb) = c$ for every $\bfb\in S$. $\quad \Box$
In fact, the hypothesis of convexity is much stronger than necessary, and it can be replaced by a much weaker geometric condition.
Theorem 4. Assume that $S$ is an open, path-connected subset of $\R^n$ and that $f:\R^n\to \R$ is a function that is differentiable in $S$. If $\nabla f(\bfx )={\bf 0}$ for every $\bfx\in S$, then $f$ is constant on $S$.
The proof is not very difficult, but it is a slightly sneaky.
Proof.
We need to show that if $\bfa, \bfb$ are any two points in $S$, then $f(\bfa) = f(\bfb)$. So, fix any $\bfa,\bfb$. By the hypothesis of path-connectedness, there exists $\gamma:[0,1]\to S$ that is continuous such that $\gamma(0)=\bfa$ and $\gamma(1)=\bfb$.
Define $\phi(s) = f(\gamma(s))$. We will show that $\phi'(s)=0$ for every $s\in (0,1)$. Note that we cannot use the chain rule, since we only know that $\gamma$ is continuous, not differentiable.
To do this, fix $s\in (0,1)$. Since $S$ is open, there exists $\ep>0$ such that $B(\ep,\gamma(s))\subset S$. Since $\gamma$ is continuous, there exists $\delta>0$ such that if $|h|<\delta$, then $s+h\in (0,1)$ and $|\gamma(s+h)-\gamma(s)|<\ep$. In other words, $$ |h|<\delta \quad\Rightarrow\quad \gamma(s+h)\in B(\ep,\gamma(s)) $$ However, $B(\ep,\gamma(s))$ is a convex open set on which $\nabla f = {\bf 0}$ everywhere, so Theorem 3 implies that $f(\bfx) = f(\gamma(s))$ for every $\bfx\in B(\ep, \gamma(s))$. In particular, it follows that $$ |h|<\delta \quad \Rightarrow \quad \phi(s+h) - \phi(s) = f(\gamma(s+h)) - f(\gamma(s)) = 0. $$ It easily follows that $\phi'(s)=0$. Since $s$ was arbitrary, we conclude that $\phi'=0$ everywhere in $(0,1)$. Finally, the $1$-d Mean Value Theorem implies that $$ f(\bfb) - f(\bfa) = \phi(1)-\phi(0) = 0. $$
There are not really any Basic Skills connected to the material in this section.
(This question was discussed in Tutorial 5.) Assume that $S$ is an open subset of $\R^2$, and that $f:S\to \R$ is a differentiable function such that $\partial_1 f = 0$ everywhere in $S$.
If $S$ is convex, is it true that $f$ depends only on the $y$ variable, in other words, that $f(x,y)= f(x', y)$ whenever $(x,y)$ and $(x',y)$ belong to $S$?
Same question if $S$ is not convex. For concreteness, assume that $$ S = \{ (x,y)\in \R^2 : 2x^2< y <1+x^2 \}. $$
Assume that $f:\R^n\to \R$ is a $C^1$ function and that there exists a vector ${\bf v}\in \R^n$ such that $$ {\bf v}\cdot \nabla f(\bfx) = 0\qquad\mbox{ for all }\bfx\in \R^n. $$ Prove that for every $\bfx \in \R^n$ and every $t\in \R$, $$ f(\bfx + t{\bf v}) = f(\bfx). $$
Prove that every convex set is path-connected. (This should be easy).
Draw a picture of the following sets and determine whether they are convex
$S = \{ (x,y)\in \R^2 : (x/2)^2+ (y/3)^2 \le 1\}$.
$S = \{ (x,y)\in \R^2 : (x/2)^2- (y/3)^2 \le 1\}$.
$S = \{ (x,y)\in \R^2 : y \ge e^{x} \}$.
$S = \{ (x,y)\in \R^2 : x < e^{-y^2} \}$.
$S = \{ (x,y)\in \R^2 : xy <1 \}$.
$S = \{ (x,y)\in \R^2 : y> k - x/k^2 \mbox{ for all }k\in \mathbb N \}$.
(In this question, the issue of convexity is really just an excuse to
ask you to draw pictures of various subsets of $\R^2$, something you shoud be able to do without the assistance of an electronic device. If you find it difficult, put away your phone and practice.)
Assume that $S$ is a convex subset of $\R^n$ and that
$f:\R^n\to \R^m$ is a function of the form
$$
f(\bfx) = A \bfx + \bfb
$$
for some $m\times n$ matrix $A$ and some $b\in \R^m$.
Prove that
$$
f(S) := \{ f(\bfx) : \bfx \in S\}
$$
is convex.
Prove that if $S_1, S_2, \ldots, $ are convex sets, then
Prove that a set $S$ of the form $$ S = \{ \bfx \in \R^n : (x_1/a_1)^2+ \cdots + (x_n/a_n)^2 \le 1\} $$ is convex, where $a_1,\ldots, a_n$ are nonzero constants. Hint: a relatively easy way to do this is by combining one of the exercises above and the fact that the unit ball in $\R^n$ is convex,
Let $g:\R^n\to [0,\infty)$ be a function that is homogeneous of degree
$1$, and such that $g(\bfx+\bfy) \le g(\bfx) + g(\bfy)$
for all $\bfx, \bfy\in \R^n$. Prove that
$$
\{ \bfx\in \R^n : g(\bfx) < 1 \}
$$
is convex.
Recall, homogenoeus of degree $1$
means that $f(\lambda \bfx) = \lambda f(\bfx)$ for all $\bfx\in \R^n$ and $\lambda>0$.