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A continuous function ƒ(x) on the closed interval [a,b] showing the absolute max (red) and the absolute min (blue).

In calculus, the extreme value theorem states that if a real-valued function f is continuous on the closed interval [a,b], then f must attain a maximum and a minimum, each at least once. That is, there exist numbers c and d in [a,b] such that:

A related theorem is the boundedness theorem which states that a continuous function f in the closed interval [a,b] is bounded on that interval. That is, there exist real numbers m and M such that:

The extreme value theorem enriches the boundedness theorem by saying that not only is the function bounded, but it also attains its least upper bound as its maximum and its greatest lower bound as its minimum.

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.

Contents

HistoryEdit

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value theorem was originally proven by Bernard Bolzano in the 1830s in a work Function Theory but the work remained unpublished until 1930.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.[1] The result was also discovered later by Weierstrass in 1860.[citation needed]

Functions to which the theorem does not applyEdit

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.

  1. ƒ(x) = x defined over [0, ∞) is not bounded from above.
  2. ƒ(x) = x / (1 + x) defined over [0, ∞) is bounded but does not attain its least upper bound 1.
  3. ƒ(x) = 1 / x defined over (0, 1] is not bounded from above.
  4. ƒ(x) = 1 – x defined over (0, 1] is bounded but never attains its least upper bound 1.

Defining ƒ(0) = 0 in the last two examples shows that both theorems require continuity on [ab].

Generalization to metric and topological spacesEdit

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 called the Heine–Borel property, and it 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.

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:[2]

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:[2]

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 theoremsEdit

We look at the proof for the upper bound and the maximum of f. By applying these results to the function –f, the existence of the lower bound and the result for the minimum of f 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:

  1. Prove the boundedness theorem.
  2. Find a sequence so that its image converges to the supremum of f.
  3. Show that there exists a subsequence that converges to a point in the domain.
  4. Use continuity to show that the image of the subsequence converges to the supremum.

Proof of the boundedness theoremEdit

Statement   If   is continuous on   then it is bounded on  

Suppose the function f is not bounded above on the interval [a,b]. Then, for every natural number n, there exists an xn in [a,b] such that f(xn) > n. This defines a sequence {xn}. Because [a,b] is bounded, the Bolzano–Weierstrass theorem implies that there exists a convergent subsequence { } of {xn}. Denote its limit by x. As [a,b] is closed, it contains x. Because f is continuous at x, we know that {f( )} converges to the real number f(x) (as f is sequentially continuous at x.) But f(xnk) > nkk for every k, which implies that {f(xnk)} diverges to +∞, a contradiction. Therefore, f is bounded above on [a,b]. 

Alternative proofEdit

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 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 theoremEdit

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 + 1/n 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 theoremEdit

The set {yR : y = f(x) for some x ∈ [a,b]} is a bounded set. Hence, its least upper bound exists by least upper bound property of the real numbers. Let M = sup(f(x)) on [ab]. If there is no point x on [ab] so that f(x) = M then f(x) < M on [ab]. Therefore, 1/(M − f(x)) is continuous on [a, b].

However, to every positive number ε, there is always some x in [ab] such that M − f(x) < ε because M is the least upper bound. Hence, 1/(M − f(x)) > 1/ε, which means that 1/(M − f(x)) is not bounded. Since every continuous function on a [a, b] is bounded, this contradicts the conclusion that 1/(M − f(x)) was continuous on [ab]. Therefore, there must be a point x in [ab] such that f(x) = M.

Proof using the hyperrealsEdit

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.[3]

Proof from first principlesEdit

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 :

(1)     .   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.

(2)     .   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 functionsEdit

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 f : [a,b] → [–∞,∞) 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 f : [a,b] → (–∞,∞] 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.

ReferencesEdit

  1. ^ Rusnock, Paul; Kerr-Lawson, Angus (2005). "Bolzano and Uniform Continuity". Historia Mathematica. 32 (3): 303–311. doi:10.1016/j.hm.2004.11.003.
  2. ^ a b Rudin, Walter (1976). Principles of Mathematical Analysis. New York: McGraw Hill. pp. 89–90. ISBN 0-07-054235-X.
  3. ^ Keisler, H. Jerome (1986). Elementary Calculus : An Infinitesimal Approach (PDF). Boston: Prindle, Weber & Schmidt. p. 164. ISBN 0-87150-911-3.

Further readingEdit

External linksEdit