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# 9.1 Extreme Values

9.1   Definition (Maximum, Minimum.) Let be a function from a set to , and let . We say that has a maximum at if for all , and we say has a minimum at if for all .

9.2   Definition (Maximizing set.) Let be a function and let be a subset of . We say is a maximizing set for on if for each there is a point such that .

9.3   Examples. If has a maximum at then is a maximizing set for on .

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 is a maximizing set for , since if we can find with ; then , so , so and . This argument shows that is also a maximizing set for .

9.4   Remark. Let be a set, and let , and let be a subset of . If is not a maximizing set for on , then there is some point such that for all .

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 be a subset of , let . We say is an upper bound for if for all , and we say is a lower bound for if for all .

9.8   Remark. If is a bounded subset of and is a bound for , then is an upper bound for and is a lower bound for , since

Conversely, if a subset of has an upper bound and a lower bound ,then is bounded, and is a bound for , 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