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Extreme value theorem

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This article is about the calculus concept. For the statistical concept, seeFisher–Tippett–Gnedenko theorem.

Continuous real function on a closed interval has a maximum and a minimum

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

{\displaystyle f(x)} — A continuous function $f(x)$ on the closed interval $[a,b]$ showing the absolute max (red) and the absolute min (blue).

Inreal analysis, a branch ofmathematics, theextreme value theorem states that if a real-valuedfunction $f {\displaystyle f}$ iscontinuous on theclosed andbounded interval $[a,b]$ , then $f {\displaystyle f}$ must attain amaximum and aminimum, each at least once.^[1]^[2] That is, there exist numbers $c {\displaystyle c}$ and $d {\displaystyle d}$ in $[a,b]$ such that: $f(c)\leq f(x)\leq f(d)\quad \forall x\in [a,b].$

The extreme value theorem is more specific than the relatedboundedness theorem, which states merely that a continuous function $f {\displaystyle f}$ on the closed interval $[a,b]$ isbounded on that interval; that is, there exist real numbers $m {\displaystyle m}$ and $M {\displaystyle M}$ such that: $m\leq f(x)\leq M\quad \forall x\in [a,b].$

This does not say that $M {\displaystyle M}$ and $m {\displaystyle m}$ are necessarily the maximum and minimum values of $f {\displaystyle f}$ on the interval $[a,b],$ which is what the extreme value theorem stipulates must also be the case.

The extreme value theorem is used to proveRolle's theorem. In a formulation due toKarl Weierstrass, this theorem states that a continuous function from a non-emptycompact space to asubset of thereal numbers attains a maximum and a minimum.

History

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The extreme value theorem was originally proven byBernard Bolzano in the 1830s in a workFunction Theory but the work remained unpublished until 1930.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 theBolzano–Weierstrass theorem.^[3]

Functions to which the theorem does not apply

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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.

$f(x)=x$ defined over $[0,\infty )$ is not bounded from above.
$f(x)={\frac {x}{1+x}}$ defined over $[0,\infty )$ is bounded from below but does not attain its least upper bound $1 {\displaystyle 1}$ .
$f(x)={\frac {1}{x}}$ defined over $(0,1]$ is not bounded from above.
$f(x)=1-x$ defined over $(0,1]$ is bounded but never attains its least upper bound $1 {\displaystyle 1}$ .

Defining $f(0)=0$ in the last two examples shows that both theorems require continuity on $[a,b]$ .

Generalization to metric and topological spaces

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When moving from the real line $\mathbb {R}$ tometric spaces and generaltopological spaces, the appropriate generalization of a closed bounded interval is acompact set. A set $K {\displaystyle K}$ is said to be compact if it has the following property: from every collection ofopen sets $U_{\alpha }$ such that ${\textstyle \bigcup U_{\alpha }\supset K}$ , a finite subcollection $U_{\alpha _{1}},\ldots ,U_{\alpha _{n}}$ can be chosen such that ${\textstyle \bigcup _{i=1}^{n}U_{\alpha _{i}}\supset K}$ . This is usually stated in short as "every open cover of $K {\displaystyle K}$ has a finite subcover". TheHeine–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 theHeine–Borel property if every closed and bounded set is also compact.

The concept of a continuous function can likewise be generalized. Given topological spaces $V,\ W$ , a function $f:V\to W$ is said to be continuous if for every open set $U\subset W$ , $f^{-1}(U)\subset V$ is also open. Given these definitions, continuous functions can be shown to preserve compactness:^[4]

Theorem—If $V,\ W$ are topological spaces, $f:V\to W$ is a continuous function, and $K\subset V$ is compact, then $f(K)\subset W$ is also compact.

In particular, if $W=\mathbb {R}$ , then this theorem implies that $f(K)$ is closed and bounded for any compact set $K {\displaystyle K}$ , which in turn implies that $f {\displaystyle f}$ attains itssupremum andinfimum on any (nonempty) compact set $K {\displaystyle K}$ . Thus, we have the following generalization of the extreme value theorem:^[4]

Theorem—If $K {\displaystyle K}$ is a nonempty compact set and $f:K\to \mathbb {R}$ is a continuous function, then $f {\displaystyle f}$ is bounded and there exist $p,q\in K$ such that $f(p)=\sup _{x\in K}f(x)$ and $f(q)=\inf _{x\in K}f(x)$ .

Slightly more generally, this is also true for an upper semicontinuous function. (seecompact space#Functions and compact spaces).

Proving the theorems

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We look at the proof for theupper bound and the maximum of $f {\displaystyle f}$ . By applying these results to the function $-f$ , the existence of the lower bound and the result for the minimum of $f {\displaystyle f}$ follows. Also note that everything in the proof is done within the context of thereal 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 itsimage converges to thesupremum of $f {\displaystyle f}$ .
Show that there exists asubsequence that converges to a point in thedomain.
Use continuity to show that the image of the subsequence converges to the supremum.

Proof of the boundedness theorem

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Boundedness Theorem—If $f(x)$ is continuous on $[a,b],$ then it is bounded on $[a,b].$

Proof

Suppose the function $f {\displaystyle f}$ is not bounded above on the interval $[a,b]$ . Pick asequence $(x_{n})_{n\in \mathbb {N} }$ such that $x_{n}\in [a,b]$ and $f(x_{n})>n$ . Because $[a,b]$ is bounded, theBolzano–Weierstrass theorem implies that there exists a convergent subsequence $(x_{n_{k}})_{k\in \mathbb {N} }$ of $({x_{n}})$ . Denote its limit by $x {\displaystyle x}$ . As $[a,b]$ is closed, it contains $x {\displaystyle x}$ . Because $f {\displaystyle f}$ is continuous at $x {\displaystyle x}$ , we know that $f(x_{{n}_{k}})$ converges to the real number $f(x)$ (as $f {\displaystyle f}$ issequentially continuous at $x {\displaystyle x}$ ). But $f(x_{{n}_{k}})>n_{k}\geq k$ for every $k {\displaystyle k}$ , which implies that $f(x_{{n}_{k}})$ diverges to $+\infty$ , a contradiction. Therefore, $f {\displaystyle f}$ is bounded above on $[a,b]$ . ∎

Alternative proof

Consider the set $B {\displaystyle B}$ of points $p {\displaystyle p}$ in $[a,b]$ such that $f(x)$ is bounded on $[a,p]$ . We note that $a {\displaystyle a}$ is one such point, for $f(x)$ is bounded on $[a,a]$ by the value $f(a)$ . If $e>a$ is another point, then all points between $a {\displaystyle a}$ and $e {\displaystyle e}$ also belong to $B {\displaystyle B}$ . In other words $B {\displaystyle B}$ is an interval closed at its left end by $a {\displaystyle a}$ .

Now $f {\displaystyle f}$ is continuous on the right at $a {\displaystyle a}$ , hence there exists $\delta >0$ such that $|f(x)-f(a)|<1$ for all $x {\displaystyle x}$ in $[a,a+\delta ]$ . Thus $f {\displaystyle f}$ is bounded by $f(a)-1$ and $f(a)+1$ on the interval $[a,a+\delta ]$ so that all these points belong to $B {\displaystyle B}$ .

So far, we know that $B {\displaystyle B}$ is an interval of non-zero length, closed at its left end by $a {\displaystyle a}$ .

Next, $B {\displaystyle B}$ is bounded above by $b {\displaystyle b}$ . Hence the set $B {\displaystyle B}$ has a supremum in $[a,b]$ ; let us call it $s {\displaystyle s}$ . From the non-zero length of $B {\displaystyle B}$ we can deduce that $s>a$ .

Suppose $s<b$ . Now $f {\displaystyle f}$ is continuous at $s {\displaystyle s}$ , hence there exists $\delta >0$ such that $|f(x)-f(s)|<1$ for all $x {\displaystyle x}$ in $[s-\delta ,s+\delta ]$ so that $f {\displaystyle f}$ is bounded on this interval. But it follows from the supremacy of $s {\displaystyle s}$ that there exists a point belonging to $B {\displaystyle B}$ , $e {\displaystyle e}$ say, which is greater than $s-\delta /2$ . Thus $f {\displaystyle f}$ is bounded on $[a,e]$ which overlaps $[s-\delta ,s+\delta ]$ so that $f {\displaystyle f}$ is bounded on $[a,s+\delta ]$ . This however contradicts the supremacy of $s {\displaystyle s}$ .

We must therefore have $s=b$ . Now $f {\displaystyle f}$ is continuous on the left at $s {\displaystyle s}$ , hence there exists $\delta >0$ such that $|f(x)-f(s)|<1$ for all $x {\displaystyle x}$ in $[s-\delta ,s]$ so that $f {\displaystyle f}$ is bounded on this interval. But it follows from the supremacy of $s {\displaystyle s}$ that there exists a point belonging to $B {\displaystyle B}$ , $e {\displaystyle e}$ say, which is greater than $s-\delta /2$ . Thus $f {\displaystyle f}$ is bounded on $[a,e]$ which overlaps $[s-\delta ,s]$ so that $f {\displaystyle f}$ is bounded on $[a,s]$ . ∎

Proofs of the extreme value theorem

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Proof of the Extreme Value Theorem

By the boundedness theorem,f is bounded from above, hence, by theDedekind-completeness of the real numbers, the least upper bound (supremum)M off exists. It is necessary to find a pointd in [a,b] such thatM =f(d). Letn be a natural number. AsM is theleast upper bound,M − 1/n is not an upper bound forf. Therefore, there existsd_n in [a,b] so thatM − 1/n <f(d_n). This defines a sequence {d_n}. SinceM is an upper bound forf, we haveM − 1/n <f(d_n) ≤M for alln. Therefore, the sequence {f(d_n)} converges toM.

TheBolzano–Weierstrass theorem tells us that there exists a subsequence { $d_{n_{k}}$ }, which converges to somed and, as [a,b] is closed,d is in [a,b]. Sincef is continuous atd, the sequence {f( $d_{n_{k}}$ )} converges tof(d). But {f(d_{n_k})} is a subsequence of {f(d_n)} that converges toM, soM =f(d). Therefore,f attains its supremumM atd. ∎

Alternative Proof of the Extreme Value Theorem

The set{y ∈R :y =f(x) for somex ∈ [a,b]} is a bounded set. Hence, itsleast upper bound exists byleast upper bound property of the real numbers. LetM = sup(f(x)) on [a,b]. If there is no pointx on [a, b] so thatf(x) = M, thenf(x) <M on [a, b]. Therefore,1/(M −f(x)) is continuous on [a,b].

However, to every positive numberε, there is always somex in [a, b] such thatM −f(x) <ε becauseM is the least upper bound. Hence,1/(M −f(x)) > 1/ε, which means that1/(M −f(x)) is not bounded. Since every continuous function on [a,b] is bounded, this contradicts the conclusion that1/(M −f(x)) was continuous on [a, b]. Therefore, there must be a pointx in [a, b] such thatf(x) = M.∎

Proof using the hyperreals

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Proof

In the setting ofnon-standard calculus, letN be an infinitehyperinteger. The interval [0, 1] has a natural hyperreal extension. Consider its partition intoN subintervals of equalinfinitesimal length 1/N, with partition pointsx_i =i /N asi "runs" from 0 toN. The functionƒ is also naturally extended to a functionƒ* defined on the hyperreals between 0 and 1. Note that in the standard setting (whenN is finite), a point with the maximal value ofƒ can always be chosen among theN+1 pointsx_i, by induction. Hence, by thetransfer principle, there is a hyperintegeri₀ such that 0 ≤i₀ ≤N and $f^{*}(x_{i_{0}})\geq f^{*}(x_{i})$ for alli = 0, ..., N. Consider the real point $c=\mathbf {st} (x_{i_{0}})$ wherest is thestandard part function. An arbitrary real pointx lies in a suitable sub-interval of the partition, namely $x\in [x_{i},x_{i+1}]$ , so that st(x_i) =x. Applyingst to the inequality $f^{*}(x_{i_{0}})\geq f^{*}(x_{i})$ , we obtain $\mathbf {st} (f^{*}(x_{i_{0}}))\geq \mathbf {st} (f^{*}(x_{i}))$ . By continuity ofƒ we have

\mathbf {st} (f^{*}(x_{i_{0}}))=f(\mathbf {st} (x_{i_{0}}))=f(c)

Henceƒ(c) ≥ƒ(x), for all realx, provingc to be a maximum ofƒ.^[5]∎

Proof from first principles

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Statement If $f(x)$ is continuous on $[a,b]$ then it attains its supremum on $[a,b]$

Proof

By the Boundedness Theorem, $f(x)$ is bounded above on $[a,b]$ and by the completeness property of the real numbers has a supremum in $[a,b]$ . Let us call it $M {\displaystyle M}$ , or $M[a,b]$ . It is clear that the restriction of $f {\displaystyle f}$ to the subinterval $[a,x]$ where $x\leq b$ has a supremum $M[a,x]$ which is less than or equal to $M {\displaystyle M}$ , and that $M[a,x]$ increases from $f(a)$ to $M {\displaystyle M}$ as $x {\displaystyle x}$ increases from $a {\displaystyle a}$ to $b {\displaystyle b}$ .

If $f(a)=M$ then we are done. Suppose therefore that $f(a)<M$ and let $d=M-f(a)$ . Consider the set $L {\displaystyle L}$ of points $x {\displaystyle x}$ in $[a,b]$ such that $M[a,x]<M$ .

Clearly $a\in L$ ; moreover if $e>a$ is another point in $L {\displaystyle L}$ then all points between $a {\displaystyle a}$ and $e {\displaystyle e}$ also belong to $L {\displaystyle L}$ because $M[a,x]$ is monotonic increasing. Hence $L {\displaystyle L}$ is a non-empty interval, closed at its left end by $a {\displaystyle a}$ .

Now $f {\displaystyle f}$ is continuous on the right at $a {\displaystyle a}$ , hence there exists $\delta >0$ such that $|f(x)-f(a)|<d/2$ for all $x {\displaystyle x}$ in $[a,a+\delta ]$ . Thus $f {\displaystyle f}$ is less than $M-d/2$ on the interval $[a,a+\delta ]$ so that all these points belong to $L {\displaystyle L}$ .

Next, $L {\displaystyle L}$ is bounded above by $b {\displaystyle b}$ and has therefore a supremum in $[a,b]$ : let us call it $s {\displaystyle s}$ . We see from the above that $s>a$ . We will show that $s {\displaystyle s}$ is the point we are seeking i.e. the point where $f {\displaystyle f}$ attains its supremum, or in other words $f(s)=M$ .

Suppose the contrary viz. $f(s)<M$ . Let $d=M-f(s)$ and consider the following two cases:

$s<b$ . As $f {\displaystyle f}$ is continuous at $s {\displaystyle s}$ , there exists $\delta >0$ such that $|f(x)-f(s)|<d/2$ for all $x {\displaystyle x}$ in $[s-\delta ,s+\delta ]$ . This means that $f {\displaystyle f}$ is less than $M-d/2$ on the interval $[s-\delta ,s+\delta ]$ . But it follows from the supremacy of $s {\displaystyle s}$ that there exists a point, $e {\displaystyle e}$ say, belonging to $L {\displaystyle L}$ which is greater than $s-\delta$ . By the definition of $L {\displaystyle L}$ , $M[a,e]<M$ . Let $d_{1}=M-M[a,e]$ then for all $x {\displaystyle x}$ in $[a,e]$ , $f(x)\leq M-d_{1}$ . Taking $d_{2}$ to be the minimum of $d/2$ and $d_{1}$ , we have $f(x)\leq M-d_{2}$ for all $x {\displaystyle x}$ in $[a,s+\delta ]$ .
Hence $M[a,s+\delta ]<M$ so that $s+\delta \in L$ . This however contradicts the supremacy of $s {\displaystyle s}$ and completes the proof.
$s=b$ . As $f {\displaystyle f}$ is continuous on the left at $s {\displaystyle s}$ , there exists $\delta >0$ such that $|f(x)-f(s)|<d/2$ for all $x {\displaystyle x}$ in $[s-\delta ,s]$ . This means that $f {\displaystyle f}$ is less than $M-d/2$ on the interval $[s-\delta ,s]$ . But it follows from the supremacy of $s {\displaystyle s}$ that there exists a point, $e {\displaystyle e}$ say, belonging to $L {\displaystyle L}$ which is greater than $s-\delta$ . By the definition of $L {\displaystyle L}$ , $M[a,e]<M$ . Let $d_{1}=M-M[a,e]$ then for all $x {\displaystyle x}$ in $[a,e]$ , $f(x)\leq M-d_{1}$ . Taking $d_{2}$ to be the minimum of $d/2$ and $d_{1}$ , we have $f(x)\leq M-d_{2}$ for all $x {\displaystyle x}$ in $[a,b]$ . This contradicts the supremacy of $M {\displaystyle M}$ and completes the proof.∎

Extension to semi-continuous functions

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If the continuity of the functionf is weakened tosemi-continuity,then the corresponding half of the boundedness theorem and the extreme value theorem hold and the values −∞ or +∞, respectively, from theextended real number line can be allowed as possible values.^{[clarification needed]}

A function $f:[a,b]\to [-\infty ,\infty )$ is said to beupper semi-continuous if $\limsup _{y\to x}f(y)\leq f(x)\quad \forall x\in [a,b].$

Theorem—If a functionf : [a,b] →[–∞, ∞) is upper semi-continuous, thenf is bounded above and attains its supremum.

Proof

If $f(x)=-\infty$ for allx in [a,b], then the supremum is also $-\infty$ 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 off atx only implies that thelimit superior of the subsequence {f(x_{n_k})} is bounded above byf(x) < ∞, but that is enough to obtain the contradiction. In the proof of the extreme value theorem, upper semi-continuity off atd implies that the limit superior of the subsequence {f(d_{n_k})} is bounded above byf(d), but this suffices to conclude thatf(d) =M. ∎

Applying this result to −f proves a similar result for the infimums of lower semicontinuous functions. A function $f:[a,b]\to [-\infty ,\infty )$ is said to belower semi-continuous if $\liminf _{y\to x}f(y)\geq f(x)\quad \forall x\in [a,b].$

Theorem—If a functionf : [a,b] →(–∞, ∞] is lower semi-continuous, thenf is bounded below and attains itsinfimum.

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

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^Spivak, Michael (September 1994).Calculus. Publish or Perish publishing.ISBN 978-0-914098-89-8.
^Abbott, Stephen (2001).Understanding Analysis. Undergraduate Texts in Mathematics. New York: Springer-Verlag.ISBN 978-0387950600.
^Rusnock, Paul; Kerr-Lawson, Angus (2005). "Bolzano and Uniform Continuity".Historia Mathematica.32 (3):303–311.doi:10.1016/j.hm.2004.11.003.
^^a ^bRudin, Walter (1976).Principles of Mathematical Analysis. New York: McGraw Hill. pp. 89–90.ISBN 0-07-054235-X.
^Keisler, H. Jerome (1986).Elementary Calculus : An Infinitesimal Approach(PDF). Boston: Prindle, Weber & Schmidt. p. 164.ISBN 0-87150-911-3.

External links

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A Proof for extreme value theorem atcut-the-knot
Extreme Value Theorem by Jacqueline Wandzura with additional contributions by Stephen Wandzura, theWolfram Demonstrations Project.
Weisstein, Eric W."Extreme Value Theorem".MathWorld.
Mizar system proof:http://mizar.org/version/current/html/weierstr.html#T15

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