9.1 Extreme Values

9.1   Definition (Maximum, Minimum.) Let $f\colon S\to\mbox{{\bf R}}$ be a function from a set $S$ to $\mbox{{\bf R}}$, and let $a\in S$. We say that $f$ has a maximum at $a$ if $f(a)\geq f(x)$ for all $x\in S$, and we say $f$ has a minimum at $a$ if $f(a)\leq f(x)$ for all $x\in S$.

9.2   Definition (Maximizing set.) Let $f\colon S\to\mbox{{\bf R}}$ be a function and let $M$ be a subset of $S$. We say $M$ is a maximizing set for $f$ on $S$ if for each $x\in S$ there is a point $m\in M$ such that $f(m)\geq f(x)$.

9.3   Examples. If $f$ has a maximum at $a$ then $\{a\}$ is a maximizing set for $f$ on $S$.

If $M$ is a maximizing set for $f$ on $S$, and $M\subset B\subset S$, then $B$ is also a maximizing set for $f$ on $S$.

If $f\colon S\to\mbox{{\bf R}}$ is any function (with $S\neq \emptyset$), then $S$ is a maximizing set for $f$ on $S$, so every function with non-empty domain has a maximizing set.

Let

$$f(z)=\cases{ {1\over {\vert z\vert}} &for $z\neq 0$\vspace{1ex}\cr 0 &for $z=0$.\vspace{1ex}\cr}$$

Then every disc $D(0,\varepsilon)$ is a maximizing set for $f$, since if $z\in\mbox{{\bf C}}\backslash\{0\}$ we can find $n\in\mbox{{\bf N}}$ with $${n>\max \left( {1\over \varepsilon},{1\over {\vert z\vert}}\right)}$$; then $${n>{1\over \varepsilon}}$$, so $${{1\over n}<\varepsilon}$$, so $${ {1\over n}\in D(0,\varepsilon)}$$ and $${f\left({1\over n}\right)=n>{1\over {\vert z\vert}}=f(z)}$$. This argument shows that $${\left\{ {1\over {n+1}}\colon n\in\mbox{{\bf N}}\right\}}$$ is also a maximizing set for $f$.

9.4   Remark. Let $S$ be a set, and let $f\colon S\to\mbox{{\bf R}}$, and let $M$ be a subset of $S$. If $M$ is not a maximizing set for $f$ on $S$, then there is some point $x\in S$ such that $f(x)>f(m)$ for all $m\in M$.

9.5   Lemma. Let $S$ be a set, let $f\colon S\to\mbox{{\bf R}}$ be a function, and let $M$ be a maximizing set for $f$ on $S$. If $M= A\cup B$, then at least one of $A,B$ is a maximizing set for $f$ on $S$.

Proof: Suppose $A\cup B$ is a maximizing set for $f$ on $S$, but $A$ is not a maximizing set for $f$ on $S$. Then there is some $s\in S$ such that for all $a\in A$, $f(s)>f(a)$. Since $A\cup B$ is a maximizing set for $f$ on $S$, there is an element $t$ in $A\cup B$ such that $f(t)\geq f(s)$, so $f(t)>f(a)$ for all $a\in A$, so $t\notin A$, so $t\in B$. Now, for every $x\in S$ there is an element $c$ in $A\cup B$ with $f(c) \geq f(x)$. If $c\in A$, then the element $t\in B$ satisfies $f(t)>f(c) \geq f(x)$ so there is some element $u\in B$ with $f(u)\geq f(x)$ (if $c\in A$, take $u=t$; if $c\in B$, take $u= c$.) Hence $B$ is a maximizing set for $f$ on $S$. $\mid\!\mid\!\mid$

9.6   Theorem (Extreme value theorem.) Let $a,b\in\mbox{{\bf R}}$ with $a<b$ and let $f\colon [a,b]\to\mbox{{\bf R}}$ be a continuous function. Then $f$ has a maximum and a minimum on $[a,b]$.

Proof: We will construct a binary search sequence $\{[a_n,b_n]\}$ with $[a_0,b_0]=[a,b]$ such that each interval $[a_n,b_n]$ is a maximizing set for $f$ on $[a,b]$. We put

$$\begin{eqnarray*}[a_0,b_0]&=& [a,b] \cr [a_{n+1},b_{n+1}] &=& \cases{ \left[ a_... ...r } f$\ \cr \left[{{a_n+b_n}\over 2},b_n\right] &otherwise.\cr} \end{eqnarray*}$$

By the preceding lemma (and induction), we see that each interval $[a_n,b_n]$ is a maximizing set for $f$ on $[a,b]$. Let $c$ be the number such that $\{[a_n,b_n]\} \to c$ and let $s\in [a,b]$. Since $[a_n,b_n]$ is a maximizing set for $f$ on $[a,b]$, there is a number $s_n\in[a_n,b_n]$ with $f(s_n)\geq f(s)$. Since

$$a_n\leq c\leq b_n \mbox{ and } a_n\leq s_n\leq b_n,$$

we have $${\vert s_n-c\vert\leq \vert b_n-a_n\vert={{(b-a)}\over {2^n}}}$$, so $\{s_n\}\to c$. By continuity of $f$, $\{f(s_n)\}\to f(c)$. Since $f(s_n)\geq f(s)$, it follows by the inequality theorem for limits that

$$f(c)=\lim\{f(s_n)\}\geq f(s).$$

Hence $c$ is a maximum point for $f$ on $[a,b]$. This shows that $f$ has a maximum. Since $-f$ is also a continuous function on $[a,b]$, $-f$ has a maximum on $[a,b]$; i.e., there is a point $p\in [a,b]$ such that $-f(p)\geq -f(x)$ for all $x\in [a,b]$. Then $f(p)\leq f(x)$ for all $x\in [a,b]$, so $f$ has a minimum at $p$. $\mid\!\mid\!\mid$

9.7   Definition (Upper bound.) Let $S$ be a subset of $\mbox{{\bf R}}$, let $b,B\in\mbox{{\bf R}}$. We say $B$ is an upper bound for $S$ if $x\leq B$ for all $x\in S$, and we say $b$ is a lower bound for $S$ if $b\leq x$ for all $x\in S$.

9.8   Remark. If $S$ is a bounded subset of $R$ and $B$ is a bound for $S$, then $B$ is an upper bound for $S$ and $-B$ is a lower bound for $S$, since

$$\vert x\vert\leq B\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}-B\leq x\leq B.$$

Conversely, if a subset $S$ of $\mbox{{\bf R}}$ has an upper bound $B$ and a lower bound $b$,then $S$ is bounded, and $\max(\vert b\vert,\vert B\vert)$ is a bound for $S$, since

$$b\leq x\leq B\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}... ...eq x\leq B\leq \vert B\vert\leq\max(\vert b\vert,\vert B\vert).$$

9.9   Theorem (Boundedness theorem.) Let $a,b\in\mbox{{\bf R}}$ with $a<b$ and let $f\colon [a,b]\to\mbox{{\bf R}}$ be a continuous function. Then $f$ is bounded on $[a,b]$.

Proof: By the extreme value theorem, there are points $p,q\in[a,b]$ such that

$$f(p)\leq f(x)\leq f(q)\mbox{ for all } x\in [a,b].$$

Hence $f([a,b])$ has an upper bound and a lower bound, so $f([a,b])$ is bounded. $\mid\!\mid\!\mid$

9.10   Exercise. A Give examples of the functions described below, or explain why no such function exists. Describe your functions by formulas if you can, but pictures of graphs will do if a formula seems too complicated.

a)

$f\colon[0,1]\to\mbox{{\bf R}}$, $f$ is not bounded.

b)

$g\colon(0,1)\to\mbox{{\bf R}}$, $g$ is continuous, $g$ is not bounded.

c)

$h\colon[0,\infty)\to\mbox{{\bf R}}$, $h$ is continuous, $h$ is not bounded.

d)

$k\colon[0,\infty)\to\mbox{{\bf R}}$, $k$ is strictly increasing, $k$ is continuous, $k$ is bounded.

e)

$l\colon[0,1]\to\mbox{{\bf R}}$, $l$ is continuous, $l$ is not bounded.