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value theorem was originally proven by Bernard Bolzano in the 1830s in a work Function Theory but the work remained unpublished until 1930.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 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. ${\displaystyle f(x)=x}$  defined over ${\displaystyle [0,\infty )}$  is not bounded from above.
2. ${\displaystyle f(x)={\frac {x}{1+x}}}$  defined over ${\displaystyle [0,\infty )}$  is bounded but does not attain its least upper bound ${\displaystyle 1}$ .
3. ${\displaystyle f(x)={\frac {1}{x}}}$  defined over ${\displaystyle (0,1]}$  is not bounded from above.
4. ${\displaystyle f(x)=1-x}$  defined over ${\displaystyle (0,1]}$  is bounded but never attains its least upper bound ${\displaystyle 1}$ .

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

When moving from the real line ${\displaystyle \mathbb {R} }$  to metric spaces and general topological spaces, the appropriate generalization of a closed bounded interval is a compact set. A set ${\displaystyle K}$  is said to be compact if it has the following property: from every collection of open sets ${\displaystyle U_{\alpha }}$  such that ${\textstyle \bigcup U_{\alpha }\supset K}$ , a finite subcollection ${\displaystyle U_{\alpha _{1}},\ldots ,U_{\alpha _{n}}}$ 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 ${\displaystyle K}$  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 ${\displaystyle V,\ W}$ , a function ${\displaystyle f:V\to W}$  is said to be continuous if for every open set ${\displaystyle U\subset W}$ , ${\displaystyle f^{-1}(U)\subset V}$  is also open. Given these definitions, continuous functions can be shown to preserve compactness:^[2]

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

In particular, if ${\displaystyle W=\mathbb {R} }$ , then this theorem implies that ${\displaystyle f(K)}$  is closed and bounded for any compact set ${\displaystyle K}$ , which in turn implies that ${\displaystyle f}$  attains its supremum and infimum on any (nonempty) compact set ${\displaystyle K}$ . Thus, we have the following generalization of the extreme value theorem:^[2]

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

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 ${\displaystyle f}$ . By applying these results to the function ${\displaystyle -f}$ , the existence of the lower bound and the result for the minimum of ${\displaystyle f}$  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 ${\displaystyle f}$ .
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 theorem edit

Statement   If ${\displaystyle f(x)}$  is continuous on ${\displaystyle [a,b]}$  then it is bounded on ${\displaystyle [a,b]}$ 

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

Alternative proof edit

Statement   If ${\displaystyle f(x)}$  is continuous on ${\displaystyle [a,b]}$  then it is bounded on ${\displaystyle [a,b]}$ 

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

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

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

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

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

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

Proof of the extreme value theorem edit

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 {${\displaystyle d_{n_{k}}}$ }, 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(${\displaystyle d_{n_{k}}}$ )} 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. ∎

Alternative proof of the extreme value theorem edit

The set {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 M = sup(f(x)) on [a, b]. If there is no point x on [a, b] so that f(x) = M, then f(x) < M on [a, b]. Therefore, 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 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 [a, b]. Therefore, there must be a point x in [a, b] such that f(x) = M. ∎

Proof using the hyperreals edit

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 ${\displaystyle f^{*}(x_{i_{0}})\geq f^{*}(x_{i})}$   for all i = 0, ..., N. Consider the real point

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

Hence ƒ(c) ≥ ƒ(x), for all real x, proving c to be a maximum of ƒ.^[3]

Proof from first principles edit

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

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

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

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

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

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

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

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

Theorem: If a function f : [a, b] → [–∞, ∞) is upper semi-continuous, meaning that

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(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:

Theorem: If a function f : [a, b] → (–∞, ∞] is lower semi-continuous, meaning that

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.

• Adams, Robert A. (1995). Calculus : A Complete Course. Reading: Addison-Wesley. pp. 706–707. ISBN 0-201-82823-5.
• Protter, M. H.; Morrey, C. B. (1977). "The Boundedness and Extreme–Value Theorems". A First Course in Real Analysis. New York: Springer. pp. 71–73. ISBN 0-387-90215-5.

• 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.
• Weisstein, Eric W. "Extreme Value Theorem". MathWorld.
• Mizar system proof: http://mizar.org/version/current/html/weierstr.html#T15