Minimum of Continuous Functions is Continuous

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

A continuous function f ( x ) {\displaystyle f(x)} on the closed interval [ a , b ] {\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 f {\displaystyle f} is continuous on the closed interval [ a , b ] {\displaystyle [a,b]} , then f {\displaystyle f} must attain a maximum and a minimum, each at least once. That is, there exist numbers c {\displaystyle c} and d {\displaystyle d} in [ a , b ] {\displaystyle [a,b]} such that:

f ( c ) f ( x ) f ( d ) x [ a , b ] {\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 f {\displaystyle f} on the closed interval [ a , b ] {\displaystyle [a,b]} is bounded on that interval; that is, there exist real numbers m {\displaystyle m} and M {\displaystyle M} such that:

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value theorem was originally proven by Bernard Bolzano in the 1830s in a work Function Theory but the work remained unpublished until 1930.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

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

Generalization to metric and topological spaces [edit]

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

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

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

Theorem. If K {\displaystyle K} is a compact set and f : K R {\displaystyle f:K\to \mathbb {R} } is a continuous function, then f {\displaystyle f} is bounded and there exist p , q K {\displaystyle p,q\in K} such that f ( p ) = sup x K f ( x ) {\textstyle f(p)=\sup _{x\in K}f(x)} and f ( q ) = inf x K f ( x ) {\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 f {\displaystyle f} . By applying these results to the function f {\displaystyle -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 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 {\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 f ( x ) {\displaystyle f(x)} is continuous on [ a , b ] {\displaystyle [a,b]} then it is bounded on [ a , b ] {\displaystyle [a,b]}

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

Alternative proof [edit]

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

Proof Consider the set B {\displaystyle B} of points p {\displaystyle p} in [ a , b ] {\displaystyle [a,b]} such that f ( x ) {\displaystyle f(x)} is bounded on [ a , p ] {\displaystyle [a,p]} . We note that a {\displaystyle a} is one such point, for f ( x ) {\displaystyle f(x)} is bounded on [ a , a ] {\displaystyle [a,a]} by the value f ( a ) {\displaystyle f(a)} . If e > a {\displaystyle 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 δ > 0 {\displaystyle \delta >0} such that | f ( x ) f ( a ) | < 1 {\displaystyle |f(x)-f(a)|<1} for all x {\displaystyle x} in [ a , a + δ ] {\displaystyle [a,a+\delta ]} . Thus f {\displaystyle f} is bounded by f ( a ) 1 {\displaystyle f(a)-1} and f ( a ) + 1 {\displaystyle f(a)+1} on the interval [ a , a + δ ] {\displaystyle [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 ] {\displaystyle [a,b]}  ; let us call it s {\displaystyle s} . From the non-zero length of B {\displaystyle B} we can deduce that s > a {\displaystyle s>a} .

Suppose s < b {\displaystyle s<b} . Now f {\displaystyle f} is continuous at s {\displaystyle s} , hence there exists δ > 0 {\displaystyle \delta >0} such that | f ( x ) f ( s ) | < 1 {\displaystyle |f(x)-f(s)|<1} for all x {\displaystyle x} in [ s δ , s + δ ] {\displaystyle [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 δ / 2 {\displaystyle s-\delta /2} . Thus f {\displaystyle f} is bounded on [ a , e ] {\displaystyle [a,e]} which overlaps [ s δ , s + δ ] {\displaystyle [s-\delta ,s+\delta ]} so that f {\displaystyle f} is bounded on [ a , s + δ ] {\displaystyle [a,s+\delta ]} . This however contradicts the supremacy of s {\displaystyle s} .

We must therefore have s = b {\displaystyle s=b} . Now f {\displaystyle f} is continuous on the left at s {\displaystyle s} , hence there exists δ > 0 {\displaystyle \delta >0} such that | f ( x ) f ( s ) | < 1 {\displaystyle |f(x)-f(s)|<1} for all x {\displaystyle x} in [ s δ , s ] {\displaystyle [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 δ / 2 {\displaystyle s-\delta /2} . Thus f {\displaystyle f} is bounded on [ a , e ] {\displaystyle [a,e]} which overlaps [ s δ , s ] {\displaystyle [s-\delta ,s]} so that f {\displaystyle f} is bounded on [ a , s ] {\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 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 for all n. Therefore, the sequence {f(dn )} converges to M.

The Bolzano–Weierstrass theorem tells us that there exists a subsequence { d n k {\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( d n k {\displaystyle d_{n_{k}}} )} 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 theorem [edit]

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[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/(Mf(x)) is continuous on [a, b].

However, to every positive number ε, there is always some x in [a,b] such that Mf(x) < ε because M is the least upper bound. Hence, 1/(Mf(x)) > 1/ε , which means that 1/(Mf(x)) is not bounded. Since every continuous function on a [a, b] is bounded, this contradicts the conclusion that 1/(Mf(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 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 i 0 such that 0 ≤ i 0 ≤ N and f ( x i 0 ) f ( x i ) {\displaystyle f^{*}(x_{i_{0}})\geq f^{*}(x_{i})}   for all i = 0, ...,N. Consider the real point

c = s t ( x i 0 ) {\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 x [ x i , x i + 1 ] {\displaystyle x\in [x_{i},x_{i+1}]} , so that st(xi ) = x. Applying st to the inequality f ( x i 0 ) f ( x i ) {\displaystyle f^{*}(x_{i_{0}})\geq f^{*}(x_{i})} , we obtain s t ( f ( x i 0 ) ) s t ( f ( x i ) ) {\displaystyle \mathbf {st} (f^{*}(x_{i_{0}}))\geq \mathbf {st} (f^{*}(x_{i}))} . By continuity of ƒ  we have

s t ( f ( x i 0 ) ) = f ( s t ( x i 0 ) ) = f ( c ) {\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 f ( x ) {\displaystyle f(x)} is continuous on [ a , b ] {\displaystyle [a,b]} then it attains its supremum on [ a , b ] {\displaystyle [a,b]}

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

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

Clearly a L {\displaystyle a\in L}  ; moreover if e > a {\displaystyle 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 ] {\displaystyle 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 δ > 0 {\displaystyle \delta >0} such that | f ( x ) f ( a ) | < d / 2 {\displaystyle |f(x)-f(a)|<d/2} for all x {\displaystyle x} in [ a , a + δ ] {\displaystyle [a,a+\delta ]} . Thus f {\displaystyle f} is less than M d / 2 {\displaystyle M-d/2} on the interval [ a , a + δ ] {\displaystyle [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 ] {\displaystyle [a,b]} : let us call it s {\displaystyle s} . We see from the above that s > a {\displaystyle 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 {\displaystyle f(s)=M} .

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

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

lim sup y x f ( y ) f ( x ) {\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(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

lim inf y x f ( y ) f ( x ) {\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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