Extreme value theorem

Extreme value theorem
In calculus, the extreme value theorem states that if a real-valued function is continuous on the closed interval , then must attain a maximum and a minimum, each at least once. That is, there exist numbers and in such that:
The extreme value theorem is more specific than the related boundedness theorem, which states merely that a continuous function on the closed interval is bounded on that interval; that is, there exist real numbers and such that:
This does not say that and are necessarily the maximum and minimum values of on the interval which is what the extreme value theorem stipulates must also be the case.
The extreme value theorem is used to prove Rolle's theorem. In a formulation due to Karl Weierstrass, this theorem states that a continuous function from a non-empty compact space to a subset of the real numbers attains a maximum and a minimum.
History
The extreme value theorem was originally proven by Bernard Bolzano in the 1830s in a work Function Theory but the work remained unpublished until 1930. Bolzano's proof consisted of showing that a continuous function on a closed interval was bounded, and then showing that the function attained a maximum and a minimum value. Both proofs involved what is known today as the Bolzano–Weierstrass theorem.
Functions to which the theorem does not apply
The following examples show why the function domain must be closed and bounded in order for the theorem to apply. Each fails to attain a maximum on the given interval.
defined over is not bounded from above.
defined over is bounded but does not attain its least upper bound .
defined over is not bounded from above.
defined over is bounded but never attains its least upper bound .
Defining in the last two examples shows that both theorems require continuity on .
Generalization to metric and topological spaces
When moving from the real line to metric spaces and general topological spaces, the appropriate generalization of a closed bounded interval is a compact set. A set is said to be compact if it has the following property: from every collection of open sets such that , a finite subcollection can be chosen such that . This is usually stated in short as "every open cover of has a finite subcover". The Heine–Borel theorem asserts that a subset of the real line is compact if and only if it is both closed and bounded. Correspondingly, a metric space has the Heine–Borel property if every closed and bounded set is also compact.
The concept of a continuous function can likewise be generalized. Given topological spaces , a function is said to be continuous if for every open set , is also open. Given these definitions, continuous functions can be shown to preserve compactness:
Theorem. If are topological spaces, is a continuous function, and is compact, then is also compact.
In particular, if , then this theorem implies that is closed and bounded for any compact set , which in turn implies that attains its supremum and infimum on any (nonempty) compact set . Thus, we have the following generalization of the extreme value theorem:
Theorem. If is a compact set and is a continuous function, then is bounded and there exist such that and .
Slightly more generally, this is also true for an upper semicontinuous function. (see compact space#Functions and compact spaces).
Proving the theorems
We look at the proof for the upper bound and the maximum of . By applying these results to the function , the existence of the lower bound and the result for the minimum of follows. Also note that everything in the proof is done within the context of the real numbers.
We first prove the boundedness theorem, which is a step in the proof of the extreme value theorem. The basic steps involved in the proof of the extreme value theorem are:
Prove the boundedness theorem.
Find a sequence so that its image converges to the supremum of .
Show that there exists a subsequence that converges to a point in the domain.
Use continuity to show that the image of the subsequence converges to the supremum.
Proof of the boundedness theorem
Statement If is continuous on then it is bounded on
Suppose the function is not bounded above on the interval . Then, for every natural number , there exists an such that . This defines a sequence . Because is bounded, the Bolzano–Weierstrass theorem implies that there exists a convergent subsequence of . Denote its limit by . As is closed, it contains . Because is continuous at , we know that converges to the real number (as is sequentially continuous at ). But for every , which implies that diverges to , a contradiction. Therefore, is bounded above on .
Alternative proof
Statement If is continuous on then it is bounded on
Proof Consider the set of points in such that is bounded on . We note that is one such point, for is bounded on by the value . If is another point, then all points between and also belong to . In other words is an interval closed at its left end by .
Now is continuous on the right at , hence there exists such that for all in . Thus is bounded by and on the interval so that all these points belong to .
So far, we know that is an interval of non-zero length, closed at its left end by .
Next, is bounded above by . Hence the set has a supremum in ; let us call it . From the non-zero length of we can deduce that .
Suppose . Now is continuous at , hence there exists such that for all in so that is bounded on this interval. But it follows from the supremacy of that there exists a point belonging to , say, which is greater than . Thus is bounded on which overlaps so that is bounded on . This however contradicts the supremacy of .
We must therefore have . Now is continuous on the left at , hence there exists such that for all in so that is bounded on this interval. But it follows from the supremacy of that there exists a point belonging to , say, which is greater than . Thus is bounded on which overlaps so that is bounded on .
∎
Proof of the extreme value theorem
By the boundedness theorem, f is bounded from above, hence, by the Dedekind-completeness of the real numbers, the least upper bound (supremum) M of f exists. It is necessary to find a point d in [a, b] such that M = f(d). Let n be a natural number. As M is the least upper bound, M – 1/n is not an upper bound for f. Therefore, there exists dn in [a, b] so that M – 1/n < f(dn). This defines a sequence {dn}. Since M is an upper bound for f, we have M – 1/n < f(dn) ≤ M for all n. Therefore, the sequence {f(dn)} converges to M.
The Bolzano–Weierstrass theorem tells us that there exists a subsequence {}, which converges to some d and, as [a, b] is closed, d is in [a, b]. Since f is continuous at d, the sequence {f()} converges to f(d). But {f(dnk)} is a subsequence of {f(dn)} that converges to M, so M = f(d). Therefore, f attains its supremum M at d. ∎
Alternative proof of the extreme value theorem
The set is a bounded set. Hence, its least upper bound exists by least upper bound property of the real numbers. Let on . If there is no point x on [a, b] so that f(x) = M ,then
on [a, b]. Therefore, is continuous on [a, b].
However, to every positive number ε, there is always some x in [a, b] such that because M is the least upper bound. Hence which means that is not bounded. Since every continuous function on a [a, b] is bounded, this contradicts the conclusion that was continuous on [a, b]. Therefore, there must be a point x in [a, b] such that f(x) = M. ∎
Proof using the hyperreals
In the setting of non-standard calculus, let N be an infinite hyperinteger. The interval [0, 1] has a natural hyperreal extension. Consider its partition into N subintervals of equal infinitesimal length 1/N, with partition points xi = i /N as i "runs" from 0 to N. The function ƒ is also naturally extended to a function ƒ* defined on the hyperreals between 0 and 1. Note that in the standard setting (when N is finite), a point with the maximal value of ƒ can always be chosen among the N+1 points xi, by induction. Hence, by the transfer principle, there is a hyperinteger i0 such that 0 ≤ i0 ≤ N and for all i = 0, ..., N. Consider the real point
where st is the standard part function. An arbitrary real point x lies in a suitable sub-interval of the partition, namely , so that st(xi) = x. Applying st to the inequality , we obtain . By continuity of ƒ we have
.
Hence ƒ(c) ≥ ƒ(x), for all real x, proving c to be a maximum of ƒ.
Proof from first principles
Statement If is continuous on then it attains its supremum on
Proof By the Boundedness Theorem, is bounded above on and by the completeness property of the real numbers has a supremum in . Let us call it , or . It is clear that the restriction of to the subinterval where has a supremum which is less than or equal to , and that increases from to as increases from to .
If then we are done. Suppose therefore that and let . Consider the set of points in such that .
Clearly ; moreover if is another point in then all points between and also belong to because is monotonic increasing. Hence is a non-empty interval, closed at its left end by .
Now is continuous on the right at , hence there exists such that for all in . Thus is less than on the interval so that all these points belong to .
Next, is bounded above by and has therefore a supremum in : let us call it . We see from the above that . We will show that is the point we are seeking i.e. the point where attains its supremum, or in other words .
Suppose the contrary viz. . Let and consider the following two cases:
. As is continuous at , there exists such that for all in . This means that is less than on the interval . But it follows from the supremacy of that there exists a point, say, belonging to which is greater than . By the definition of , . Let then for all in , . Taking to be the minimum of and , we have for all in . Hence so that . This however contradicts the supremacy of and completes the proof.
. As is continuous on the left at , there exists such that for all in . This means that is less than on the interval . But it follows from the supremacy of that there exists a point, say, belonging to which is greater than . By the definition of , . Let then for all in , . Taking to be the minimum of and , we have for all in . This contradicts the supremacy of and completes the proof.
Extension to semi-continuous functions
If the continuity of the function f is weakened to semi-continuity, then the corresponding half of the boundedness theorem and the extreme value theorem hold and the values –∞ or +∞, respectively, from the extended real number line can be allowed as possible values. More precisely:
Theorem: If a function is upper semi-continuous, meaning that
for all x in [a,b], then f is bounded above and attains its supremum.
Proof: If f(x) = –∞ for all x in [a,b], then the supremum is also –∞ and the theorem is true. In all other cases, the proof is a slight modification of the proofs given above. In the proof of the boundedness theorem, the upper semi-continuity of f at x only implies that the limit superior of the subsequence {f(xnk)} is bounded above by f(x) < ∞, but that is enough to obtain the contradiction. In the proof of the extreme value theorem, upper semi-continuity of f at d implies that the limit superior of the subsequence {f(dnk)} is bounded above by f(d), but this suffices to conclude that f(d) = M. ∎
Applying this result to −f proves:
Theorem: If a function is lower semi-continuous, meaning that
for all x in [a,b], then f is bounded below and attains its infimum.
A real-valued function is upper as well as lower semi-continuous, if and only if it is continuous in the usual sense. Hence these two theorems imply the boundedness theorem and the extreme value theorem.
References
Further reading
External links
A Proof for extreme value theorem at cut-the-knot
Extreme Value Theorem by Jacqueline Wandzura with additional contributions by Stephen Wandzura, the Wolfram Demonstrations Project.
Mizar system proof: http://mizar.org/version/current/html/weierstr.html#T15
Articles containing proofs
Theorems in calculus
Theorems in real analysis