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9.1
Definition (Maximum, Minimum.)
Let
![$f\colon S\to\mbox{{\bf R}}$](img616.gif)
be a function from a set
![$S$](img427.gif)
to
![$\mbox{{\bf R}}$](img4.gif)
, and let
![$a\in S$](img897.gif)
. We
say that
has a maximum at ![$a$](img590.gif)
if
![$f(a)\geq f(x)$](img898.gif)
for all
![$x\in S$](img899.gif)
, and we
say
has a minimum at ![$a$](img590.gif)
if
![$f(a)\leq f(x)$](img900.gif)
for all
![$x\in S$](img899.gif)
.
9.2
Definition (Maximizing set.)
Let
![$f\colon S\to\mbox{{\bf R}}$](img616.gif)
be a function and let
![$M$](img481.gif)
be a subset of
![$S$](img427.gif)
. We say
![$M$](img481.gif)
is a
maximizing set for
![$f$](img13.gif)
on
![$S$](img427.gif)
if for each
![$x\in S$](img899.gif)
there is a point
![$m\in M$](img901.gif)
such that
![$f(m)\geq f(x)$](img902.gif)
.
9.3
Examples.
If
![$f$](img13.gif)
has a maximum at
![$a$](img590.gif)
then
![$\{a\}$](img903.gif)
is a maximizing set for
![$f$](img13.gif)
on
![$S$](img427.gif)
.
If
is a maximizing set for
on
, and
, then
is
also a maximizing set for
on
.
If
is any function (with
), then
is a
maximizing set for
on
, so every function with non-empty domain has a
maximizing set.
Let
Then every disc
![$D(0,\varepsilon)$](img907.gif)
is a maximizing set for
![$f$](img13.gif)
, since if
![$z\in\mbox{{\bf C}}\backslash\{0\}$](img890.gif)
we can find
![$n\in\mbox{{\bf N}}$](img9.gif)
with
![$\displaystyle {n>\max \left( {1\over
\varepsilon},{1\over {\vert z\vert}}\right)}$](img908.gif)
; then
![$\displaystyle {n>{1\over \varepsilon}}$](img909.gif)
, so
![$\displaystyle {{1\over n}<\varepsilon}$](img910.gif)
, so
![$\displaystyle { {1\over n}\in D(0,\varepsilon)}$](img911.gif)
and
![$\displaystyle {f\left({1\over n}\right)=n>{1\over {\vert z\vert}}=f(z)}$](img912.gif)
. This argument shows that
![$\displaystyle {\left\{ {1\over {n+1}}\colon n\in\mbox{{\bf N}}\right\}}$](img913.gif)
is also a maximizing set for
![$f$](img13.gif)
.
9.4
Remark.
Let
![$S$](img427.gif)
be a set, and let
![$f\colon S\to\mbox{{\bf R}}$](img616.gif)
, and let
![$M$](img481.gif)
be a subset of
![$S$](img427.gif)
. If
![$M$](img481.gif)
is
not a maximizing set for
![$f$](img13.gif)
on
![$S$](img427.gif)
, then there is some point
![$x\in S$](img899.gif)
such
that
![$f(x)>f(m)$](img914.gif)
for all
![$m\in M$](img901.gif)
.
9.5
Lemma.
Let
be a set, let
be a function, and let
be a maximizing
set for
on
. If
, then at least one of
is a maximizing set
for
on
.
Proof: Suppose
is a maximizing set for
on
, but
is not a
maximizing set for
on
. Then there is some
such that for all
,
. Since
is a maximizing set for
on
,
there is an element
in
such that
, so
for all
, so
, so
. Now, for every
there is an element
in
with
. If
, then the element
satisfies
so
there is some element
with
(if
, take
;
if
, take
.) Hence
is a
maximizing set for
on
.
9.6
Theorem (Extreme value theorem.)
Let
with
and let
be a continuous function.
Then
has a maximum and a minimum on
.
Proof: We will construct a binary search sequence
with
such that each interval
is a maximizing set for
on
. We put
By the preceding lemma (and induction), we see that each interval
is a
maximizing set for
on
. Let
be the number such that
and let
. Since
is a maximizing set for
on
,
there is a number
with
. Since
we have
, so
. By
continuity of
,
. Since
, it follows by the
inequality theorem for limits that
Hence
is a maximum point for
on
. This shows that
has a
maximum. Since
is also a continuous function on
,
has a maximum on
; i.e., there is a point
such that
for all
. Then
for all
, so
has a minimum at
.
9.7
Definition (Upper bound.)
Let
![$S$](img427.gif)
be a subset of
![$\mbox{{\bf R}}$](img4.gif)
, let
![$b,B\in\mbox{{\bf R}}$](img952.gif)
.
We say
![$B$](img258.gif)
is an
upper bound
for
![$S$](img427.gif)
if
![$x\leq B$](img953.gif)
for all
![$x\in S$](img899.gif)
, and we say
![$b$](img954.gif)
is a
lower bound for
if
![$b\leq x$](img955.gif)
for all
![$x\in S$](img899.gif)
.
9.8
Remark.
If
![$S$](img427.gif)
is a bounded subset of
![$R$](img956.gif)
and
![$B$](img258.gif)
is a bound for
![$S$](img427.gif)
, then
![$B$](img258.gif)
is an upper
bound for
![$S$](img427.gif)
and
![$-B$](img957.gif)
is a lower bound for
![$S$](img427.gif)
, since
Conversely, if a subset
![$S$](img427.gif)
of
![$\mbox{{\bf R}}$](img4.gif)
has an upper bound
![$B$](img258.gif)
and a lower bound
![$b$](img954.gif)
,then
![$S$](img427.gif)
is bounded, and
![$\max(\vert b\vert,\vert B\vert)$](img959.gif)
is a bound for
![$S$](img427.gif)
, since
9.9
Theorem (Boundedness theorem.)
Let
with
and let
be a continuous function.
Then
is bounded on
.
Proof: By the extreme value theorem, there are points
such that
Hence
has an upper bound and a lower bound, so
is bounded.
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)
-
,
is not bounded.
- b)
-
,
is continuous,
is not bounded.
- c)
-
,
is continuous,
is not bounded.
- d)
-
,
is strictly increasing,
is continuous,
is bounded.
- e)
-
,
is continuous,
is not bounded.
Next: 9.2 Intermediate Value Theorem
Up: 9. Properties of Continuous
Previous: 9. Properties of Continuous
  Index