LECTURE 13: THE EXTREME VALUE THEOREM
In this lecture the least upper bound axiom is applied to prove a second important
theorem about continuous function on closed, finite intervals. Here it is really vital
we work with closed intervals: by “pinning” down the function at the endpoints of
the interval, we place some significant restrictions on the behaviour of f .
Definition. Let A „ R. We say a function f : A Ñ R is bounded above (respectively, below) if there exists some M P R such that f pxq ¤ M (respectively,
M ¤ f pxq) for all x P A. If f is bounded above and below, then we say f is
bounded.
Note f : A Ñ R is bounded above, bounded below or bounded if and only if the
set
ty P R : y f pxq for some x P A u
is bounded above, bounded below or bounded, respectively.
Theorem (Boundedness theorem). If f : ra, bs
then f is bounded.
Ñ R is continuous, where a b,
Remark. Many of the hypotheses cannot be weakened.
Continuity is clearly vital.
The theorem does not hold for continuous functions on open intervals pa, bq.
The proof of the theorem requires the following “local lemma”.
Lemma. Suppose f : ra, bs Ñ R is continuous at z P ra, bs. Then there exists some
δ ¡ 0 such that f is bounded on the interval ra, bs X pz δ, z δ q.
Proof. By the definition of continuity, there exists some δ
ra, bs X pz δ, z δq, then |f pxq f pzq| 1. Consequently,
¡ 0 such that if x P
|f pxq| ¤ |f pxq f pzq| |f pzq| 1 |f pzq|,
by the triangle inequality. Hence f is bounded above by 1 |f pz q| and below by
1 |f pzq| on the interval ra, bs X pz δ, z δq.
One may now proceed to prove the theorem.
Proof (of Boundedness theorem). Let f : ra, bs Ñ R be continuous and define the
set
A : tz P ra, bs : f is bounded above on ra, z su.
Clearly a P A so A H and A is bounded above by b. Hence, by the least upper
bound axiom, it follows that α : sup A exists and a ¤ α ¤ b.
We claim α b. Suppose not, so that a ¤ α b. By the lemma there exists
some δ ¡ 0 such that f is bounded above on ra, bs X pα δ, α δ q and α δ b.
Since α is the least upper bound for A, there exists some z0 P pα δ, αs such that
z0 P A. Thus, by the definition of A, the function f is also bounded above on
ra, z0 s and so f is bounded above on ra, α δ{2s ra, z0 s Y rz0 , α δ{2s. But in
this case, α δ {2 P A, contradicting the assertion that α is an upperbound for A
and so α b, as claimed.
To conclude the proof, note that by the lemma there exists some η ¡ 0 such that
f is bounded above on pb η, bs. Since b sup A, it follows that y0 P A for some
1
2
LECTURE 13: THE EXTREME VALUE THEOREM
y0 P pb η, bs and therefore f is bounded on ra, y0 s. Hence f is bounded above on
ra, bs ra, y0 s Y pb η, bs.
To see that f is bounded below, apply the previous argument to the function
f : ra, bs Ñ R to conclude it is bounded above.
The last theorem can be strengthened.
Theorem (Extreme value theorem). Let f : ra, bs Ñ R be continuous, where a b.
There exist xmin , xmax P ra, bs such that
f pxmin q ¤ f pxq ¤ f pxmax q
for all x P ra, bs.
In other words, a continuous function f : ra, bs Ñ R attains its extrema.
Proof (of Extreme value theorem). Define
A : tf pxq : x P ra, bsu.
Clearly A is non-empty and, by the boundedness theorem above, it is bounded. By
the least upper bound axiom, α : sup A exists.
We claim there exists some xmax P ra, bs such that f pxmax q α. Indeed, if not,
then the function g : ra, bs Ñ R defined by
1
g p xq :
α f p xq
is well-defined, continuous and strictly positive. By the boundedness theorem it
follows that for some M ¡ 0 we have g pxq M for all x P ra, bs. Now since α is the
least upper bound of A, there exists some x0 P ra, bs such that f px0 q P pα1{2M, αq.
But in this case, 0 α f px0 q 1{2M and so
1
M ¡ g px0 q ¡ 2M,
α f px0 q
a contradiction.
To find the point xmin one may apply the previous analysis to the function
f : ra, bs Ñ R.
Jonathan Hickman, Department of mathematics, University of Chicago, 5734 S. University Avenue, Eckhart hall Room 414, Chicago, Illinois, 60637.
E-mail address: [email protected]