Home  ›  Continuous Function on a Bounded and Open Set With No Maximum

Continuous Function on a Bounded and Open Set With No Maximum

Written By Friday, October 14, 2022 Add Comment Edit
next up previous index
Next: 12.4 The Mean Value Up: 12. Extreme Values of Previous: 12.2 A Nowhere Differentiable Index

12.3 Maxima and Minima

12.10 Definition (Maximum, minimum, extreme points.) Let $A$ be a set, let

$f\colon A\to\mbox{{\bf R}}$ and let $a \in A$. We say that $f$ has a maximum at $a$ if

\begin{displaymath}f(a)\geq f(x) \mbox{ for all } x\in A,\end{displaymath}

and we say that $f$ has a minimum at $a$ if

\begin{displaymath}f(a)\leq f(x) \mbox{ for all } x\in A.\end{displaymath}

Points $a$ where $f$ has a maximum or a minimum are called extreme points of$f$.

\psfig{file=ch12c.eps,width=4in}

12.11 Example. Let

$f\colon [0,1]\to\mbox{{\bf R}}$ be defined by

\begin{displaymath} f(x)=\cases{ x &if $0\leq x<1$\cr 0 &if $x=1$.\cr} \end{displaymath} (12.12)

\psfig{file=ch12d.eps,width=1.5in}

Then $f$ has a minimum at $0$ and at $1$, but $f$ has no maximum. To see that $f$ has no maximum, observe that if $a\in [0,1)$ then

$\displaystyle { {{1+a}\over 2}\in [0,1)}$ and

\begin{displaymath}f( {{1+a}\over 2})={{1+a}\over 2}> {{a+a}\over 2}=a=f(a).\end{displaymath}

If $g$ is the function whose graph is shown, then $g$ has a maximum at $a$, and $g$ has minimums at $b$ and $c$.

\psfig{file=ch12e.eps,width=3in}

12.13 Assumption (Extreme value property.) If $f$ is a continuous function on the interval $[a,b]$, then $f$ has a maximum and a minimum on $[a,b]$.

The extreme value property is another assumption that is really a theorem, (although the proof requires yet another assumption, namely completeness of the real numbers.)

The following exercise shows that all of the hypotheses of the extreme value property are necessary.

12.17 Theorem (Critical point theorem I.) Let $f$ be a real valued function with $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$. Let $a\in\mbox{{\bf R}}$. If $f$ has a maximum (or a minimum) at $a$, and $f$ is differentiable at $a$, then $f^\prime (a)=0$.


Proof: We will consider only the case where $f$ has a maximum. Suppose $f$ has a maximum at $a$ and $f$ is differentiable at $a$. Then $a$ is an interior point of $\mbox{{\rm dom}}(f)$ so we can find sequences $\{p_n\}$ and $\{q_n\}$ in $\mbox{{\rm dom}}(f)\setminus\{a\}$ such that $\{p_n\}\to a$, $\{q_n\}\to a$, $p_n>a$ for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$, and $q_n<a$ for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$.

\psfig{file=ch12f.eps,width=2in}

Since $f$ has a maximum at $a$, we have $f(p_n)-f(a)\leq 0$ and $f(q_n)-f(a)\leq 0$ for all $n$. Hence

\begin{displaymath}{{f(p_n)-f(a)}\over {p_n-a}}\leq 0 \mbox{ and } {{f(q_n)-f(a)}\over {q_n-a}}\geq 0 \mbox{ for all } n.\end{displaymath}


Hence by the inequality theorem for limits,

\begin{displaymath}f^\prime (a)=\lim \Big\{ {{f(p_n)-f(a)}\over {p_n-a}}\Big\} \... ...rime (a)=\lim \Big\{ {{f(q_n)-f(a)}\over {q_n-a}}\Big\} \geq 0.\end{displaymath}


It follows that $f^\prime (a)=0$. $\diamondsuit$

12.18 Definition (Local maximum and minimum.) Let $f$ be a real valued function whose domain is a subset of

$\mbox{{\bf R}}$. Let

$a\in\mbox{{\rm dom}}(f)$. We say that $f$ has a local maximum at $a$ if there is a positive number $\delta$ such that

\begin{displaymath}f(a)\geq f(x) \mbox{ for all } x\in\mbox{{\rm dom}}(f)\cap (a-\delta ,a+\delta ),\end{displaymath}

and we say that $f$ has a local minimum at $a$ if there is a positive number $\delta$ such that

\begin{displaymath}f(a)\leq f(x) \mbox{ for all } x\in\mbox{{\rm dom}}(f)\cap (a-\delta ,a+\delta ).\end{displaymath}

Sometimes we say that $f$ has a global maximum at $a$ to mean that $f$ has a maximum at $a$, when we want to emphasize that we do not mean local maximum. If $f$ has a local maximum or a local minimum at $a$ we say $f$ has a local extreme point at $a$.

12.19 Theorem (Critical point theorem II.) Let $f$ be a real valued function with $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$. Let $a\in\mbox{{\bf R}}$. If $f$ has a local maximum or minimum at $a$, and $f$ is differentiable at $a$, then $f^\prime (a)=0$.

Proof: The proof is the same as the proof of theorem 12.17.

From the critical point theorem, it follows that to investigate the extreme points of $f$, we should look at critical points, or at points where $f$ is not differentiable (including endpoints of domain $f$).

12.21 Example. Let $f(x)=x^3-3x$ for

$-2\leq x\leq 2$. Then $f$ is differentiable everywhere on

$\mbox{{\rm dom}}(f)$ except at $2$ and $-2$. Hence, any local extreme points are critical points of $f$ or are in $\{2,-2\}$. Now

\begin{displaymath}f^\prime (x)=3x^2-3=3(x^2-1)=3(x-1)(x+1).\end{displaymath}

From this we see that the critical set for $f$ is $\{-1,1\}$. Since $f$ is a continuous function on a closed interval $[-2,2]$ we know that $f$ has a maximum and a minimum on $[-2,2]$. Now

\begin{displaymath}f(-2)=-2,\; f(-1)=2,\; f(1)=-2,\; f(2)=2.\end{displaymath}

Hence $f$ has global maxima at $-1$ and $2$, and $f$ has global minima at $-2$ and $1$. The graph of $f$ is shown.

\psfig{file=ch12h.eps,width=2.5in}

12.22 Example. Let

\begin{displaymath}f(x)={1\over {1+x^2}}.\end{displaymath}

Here

$\mbox{{\rm dom}}(f)=\mbox{{\bf R}}$ and clearly $f(x)>0$ for all $x$. I can see by inspection that $f$ has a maximum at $0$; i.e.,

\begin{displaymath}f(x)={1\over {1+x^2}}\leq {1\over {1+0}}=1=f(0) \mbox{ for all } x\in\mbox{{\bf R}}\end{displaymath}

I also see that $f(-x)=f(x)$, and that $f$ is strictly decreasing on

$\mbox{${\mbox{{\bf R}}}^{+}$}$

\begin{displaymath}0<x<t\hspace{1em}\mbox{$\Longrightarrow$}\hspace{1em} x^2<t^2... ...$\Longrightarrow$}\hspace{1em}{1\over {1+x^2}}>{1\over {1+t^2}}\end{displaymath}

thus $f$ has no local extreme points other than $0$. Also $f(x)$ is very small when $x$ is large. There is no point in calculating the critical points here because all the information about the extreme points is apparent without the calculation.

\psfig{file=ch12i.eps,width=3.5in}

next up previous index
Next: 12.4 The Mean Value Up: 12. Extreme Values of Previous: 12.2 A Nowhere Differentiable Index
Ray Mayer 2007-09-07

whiteaffathe.blogspot.com

Source: http://people.reed.edu/~mayer/math111.html/header/node64.html

0 Response to "Continuous Function on a Bounded and Open Set With No Maximum"

Post a Comment

Subscribe to: Post Comments (Atom)

Iklan Atas Artikel

Iklan Tengah Artikel 1

Iklan Tengah Artikel 2

Iklan Bawah Artikel