extreme value theorem

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value theorem was originally proven by Bernard Bolzano in the 1830s in a work Function Theory but the work remained unpublished until 1930.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.{{cite journal |first1=Paul |last1=Rusnock |first2=Angus |last2=Kerr-Lawson |title=Bolzano and Uniform Continuity |journal=Historia Mathematica |volume=32 |issue=3 |year=2005 |pages=303–311 |doi=10.1016/j.hm.2004.11.003 |doi-access= }}

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

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

When moving from the real line $\backslash mathbb\{R\}$ to metric spaces and general topological spaces, the appropriate generalization of a closed bounded interval is a compact set. A set $K$ is said to be compact if it has the following property: from every collection of open sets $U\_\backslash alpha$ such that $\backslash bigcup\; U\_\backslash alpha\; \backslash supset\; K$, a finite subcollection $U\_\{\backslash alpha\_1\},\backslash ldots,U\_\{\backslash alpha\_n\}$can be chosen such that $\backslash bigcup\_\{i=1\}^n\; U\_\{\backslash alpha\_i\}\; \backslash supset\; K$. This is usually stated in short as "every open cover of $K$ 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 $V,\backslash \; W$, a function $f:V\backslash to\; W$ is said to be continuous if for every open set $U\backslash subset\; W$, $f^\{-1\}(U)\backslash subset\; V$ is also open. Given these definitions, continuous functions can be shown to preserve compactness:{{Cite book|url=https://archive.org/details/1979RudinW|title=Principles of Mathematical Analysis|last=Rudin|first=Walter|publisher=McGraw Hill|year=1976|isbn=0-07-054235-X|location=New York|pages=89–90}}

{{Math theorem|If $V,\backslash \; W$ are topological spaces, $f:V\backslash to\; W$ is a continuous function, and $K\backslash subset\; V$ is compact, then $f(K)\backslash subset\; W$ is also compact.}}

In particular, if $W\; =\; \backslash mathbb\{R\}$, then this theorem implies that $f(K)$ is closed and bounded for any compact set $K$, which in turn implies that $f$ attains its supremum and infimum on any (nonempty) compact set $K$. Thus, we have the following generalization of the extreme value theorem:

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

}}

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

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.

{{Math theorem

|name=Boundedness Theorem

|If $f(x)$ is continuous on $[a,b],$ then it is bounded on $[a,b].$}}

{{Math proof|

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

{{Math proof

|title=Alternative proof

|Consider the set $B$ of points $p$ in $[a,b]$ such that $f(x)$ is bounded on $[a,p]$. We note that $a$ 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 $a$ and $e$ also belong to $B$. In other words $B$ is an interval closed at its left end by $a$.

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

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

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

Suppose

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

∎}}

{{Math proof

|title=Proof of the Extreme Value Theorem

|proof=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 d_n in [a, b] so that M – 1/n < f(d_n). This defines a sequence {d_n}. Since M is an upper bound for f, we have M – 1/n < f(d_n) ≤ M for all n. Therefore, the sequence {f(d_n)} converges to M.

The Bolzano–Weierstrass theorem tells us that there exists a subsequence {$d\_\{n\_k\}$}, 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($d\_\{n\_k\}$)} converges to f(d). But {f(d_nk)} is a subsequence of {f(d_n)} that converges to M, so M = f(d). Therefore, f attains its supremum M at d. 

∎}}

{{Math proof

|title=Alternative Proof of the Extreme Value Theorem

|proof=The set {{math|1= {y ∈ R : 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 {{math|1=M = sup(f(x))}} on {{closed-closed|a, b}}. If there is no point x on [a, b] so that f(x) = M, then

{{math|f(x) < M}} on [a, b]. Therefore, {{math|1/(M − f(x))}} is continuous on [a, b].

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

∎}}

{{Math proof

|name=Proof of Extreme Value Theorem

|proof=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 x_i = 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 x_i, by induction. Hence, by the transfer principle, there is a hyperinteger i₀ such that 0 ≤ i₀ ≤ N and $f^*(x\_\{i\_0\})\backslash geq\; f^*(x\_i)$  for all i = 0, ..., N. Consider the real point

$$c\; =\; \backslash mathbf\{st\}(x\_\{i\_0\})$$

where st is the standard part function. An arbitrary real point x lies in a suitable sub-interval of the partition, namely $x\backslash in\; [x\_i,x\_\{i+1\}]$, so that  st(x_i) = x. Applying st to the inequality $f^*(x\_\{i\_0\})\backslash geq\; f^*(x\_i)$, we obtain $\backslash mathbf\{st\}(f^*(x\_\{i\_0\}))\backslash geq\; \backslash mathbf\{st\}(f^*(x\_i))$. By continuity of ƒ  we have

:$\backslash mathbf\{st\}(f^*(x\_\{i\_0\}))=\; f(\backslash mathbf\{st\}(x\_\{i\_0\}))=f(c)$.

Hence ƒ(c) ≥ ƒ(x), for all real x, proving c to be a maximum of ƒ.{{cite book |last=Keisler |first=H. Jerome |title=Elementary Calculus : An Infinitesimal Approach |publisher=Prindle, Weber & Schmidt |location=Boston |year=1986 |isbn=0-87150-911-3 |url=https://www.math.wisc.edu/~keisler/chapter_3e.pdf#page=60 |page=164 }}

∎

}}

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

{{Math proof

|name=Proof of Extreme Value Theorem

|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 $M$, or $M[a,b]$. It is clear that the restriction of $f$ to the subinterval $[a,x]$ where $x\backslash le\; b$ has a supremum $M[a,\; x]$ which is less than or equal to $M$, and that $M[a,x]$ increases from $f(a)$ to $M$ as $x$ increases from $a$ to $b$.

If $f(a)=M$ then we are done. Suppose therefore that

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

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

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

Suppose the contrary viz.

1. .   As $f$ is continuous at $s$, there exists $\backslash delta>0$ such that for all $x$ in $[s-\backslash delta,s+\backslash delta]$. This means that $f$ is less than $M-d/2$ on the interval $[s-\backslash delta,s+\backslash delta]$. But it follows from the supremacy of $s$ that there exists a point, $e$ say, belonging to $L$ which is greater than $s-\backslash delta$. By the definition of $L$, . Let $d\_1=M-M[a,e]$ then for all $x$ in $[a,e]$, $f(x)\backslash le\; M-d\_1$. Taking $d\_2$ to be the minimum of $d/2$ and $d\_1$, we have $f(x)\backslash le\; M-d\_2$ for all $x$ in $[a,s+\backslash delta]$. {{pb}} Hence
2. $s=b$.   As $f$ is continuous on the left at $s$, there exists $\backslash delta>0$ such that for all $x$ in $[s-\backslash delta,s]$. This means that $f$ is less than $M-d/2$ on the interval $[s-\backslash delta,s]$. But it follows from the supremacy of $s$ that there exists a point, $e$ say, belonging to $L$ which is greater than $s-\backslash delta$. By the definition of $L$, . Let $d\_1=M-M[a,e]$ then for all $x$ in $[a,e]$, $f(x)\backslash le\; M-d\_1$. Taking $d\_2$ to be the minimum of $d/2$ and $d\_1$, we have $f(x)\backslash le\; M-d\_2$ for all $x$ in $[a,b]$. This contradicts the supremacy of $M$ and completes the proof. ∎

}}

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.{{clarify|reason=It is not clear what is meant by " –∞ or +∞, respectively...can be allowed as possible values." What is the "respectively" in respect to?|date=August 2024}}

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

{{Math theorem

|If a function {{math|f : [a, b] → {{closed-open|–∞, ∞}}}} is upper semi-continuous, then f is bounded above and attains its supremum.}}

{{Math proof

|proof=If $f(x)\; =\; -\; \backslash infty$ for all x in [a,b], then the supremum is also $-\backslash infty$ 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 of f at x only implies that the limit superior of the subsequence {f(x_nk)} 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(d_nk)} is bounded above by f(d), but this suffices to conclude that f(d) = M. ∎

}}

Applying this result to −f proves a similar result for the infimums of lower semicontinuous functions.

A function $f\; :\; [a,\; b]\; \backslash to\; [-\backslash infty,\; \backslash infty)$ is said to be lower semi-continuous if $$\backslash liminf\_\{y\backslash to\; x\}\; f(y)\; \backslash geq\; f(x)\backslash quad\; \backslash forall\; x\; \backslash in\; [a,\; b].$$

{{Math theorem

|If a function {{math|f : [a, b] → {{open-closed|–∞, ∞}}}} is lower semi-continuous, 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.

{{Reflist}}

• {{cite book |first=Robert A. |last=Adams |title=Calculus : A Complete Course |location=Reading |publisher=Addison-Wesley |year=1995 |isbn=0-201-82823-5 |pages=706–707 }}
• {{cite book |first1=M. H. |last1=Protter |author-link1=Murray H. Protter |first2=C. B. |last2=Morrey |author-link2=Charles B. Morrey Jr. |title=A First Course in Real Analysis |location=New York |publisher=Springer |year=1977 |isbn=0-387-90215-5 |pages=71–73 |chapter=The Boundedness and Extreme–Value Theorems |chapter-url=https://books.google.com/books?id=NgX3BwAAQBAJ&pg=PA71 }}

• [http://www.cut-the-knot.org/fta/fta_note.shtml A Proof for extreme value theorem] at cut-the-knot
• [http://demonstrations.wolfram.com/ExtremeValueTheorem/ Extreme Value Theorem] by Jacqueline Wandzura with additional contributions by Stephen Wandzura, the Wolfram Demonstrations Project.
• {{MathWorld |title=Extreme Value Theorem |urlname=ExtremeValueTheorem}}
• Mizar system proof: http://mizar.org/version/current/html/weierstr.html#T15

{{Calculus topics}}

Category:Articles containing proofs

Category:Theorems in calculus

Category:Theorems in real analysis