Prove That a Continuous Function on a Compact Set Has a Max and a Min

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

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

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

$${\displaystyle f(c)\geq f(x)\geq f(d)\quad \forall x\in [a,b]}$$

The extreme value theorem is more specific than the related boundedness theorem, which states merely that a continuous function ${\displaystyle f}$ on the closed interval ${\displaystyle [a,b]}$ is bounded on that interval; that is, there exist real numbers ${\displaystyle m}$ and ${\displaystyle M}$ such that:

$${\displaystyle m\leq f(x)\leq M\quad \forall x\in [a,b].}$$

This does not say that ${\displaystyle M}$ and ${\displaystyle m}$ are necessarily the maximum and minimum values of ${\displaystyle f}$ on the interval ${\displaystyle [a,b],}$ 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 [edit]

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value theorem was originally proven by Bernard Bolzano in the 1830s in a work Function Theory but the work remained unpublished until 1930.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 apply [edit]

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 ${\displaystyle [a,b]}$ .

Generalization to metric and topological spaces [edit]

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).

Proving the theorems [edit]

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

$${\displaystyle c=\mathbf {st} (x_{i_{0}})}$$

where st is the standard part function. An arbitrary real point x lies in a suitable sub-interval of the partition, namely ${\displaystyle x\in [x_{i},x_{i+1}]}$ , so that st(x_i ) = x. Applying st to the inequality ${\displaystyle f^{*}(x_{i_{0}})\geq f^{*}(x_{i})}$ , we obtain ${\displaystyle \mathbf {st} (f^{*}(x_{i_{0}}))\geq \mathbf {st} (f^{*}(x_{i}))}$ . By continuity of ƒ  we have

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

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. ∎

Extension to semi-continuous functions [edit]

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

$${\displaystyle \limsup _{y\to x}f(y)\leq f(x)}$$

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(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

$${\displaystyle \liminf _{y\to x}f(y)\geq f(x)}$$

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 [edit]

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. ISBN0-07-054235-X.
3. ^ Keisler, H. Jerome (1986). Elementary Calculus : An Infinitesimal Approach (PDF). Boston: Prindle, Weber & Schmidt. p. 164. ISBN0-87150-911-3.

Further reading [edit]

• Adams, Robert A. (1995). Calculus : A Complete Course. Reading: Addison-Wesley. pp. 706–707. ISBN0-201-82823-5.
• 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. ISBN0-387-90215-5.

External links [edit]

• 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

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Source: https://en.wikipedia.org/wiki/Extreme_value_theorem

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