Intermediate value theorem: Difference between revisions

[[File:Illustration for the intermediate value theorem.svg|thumb|Intermediate value theorem: Let <math>f</math> be a continuous function defined on <math>[a,b]</math> and let <math>s</math> be a number with <math>f(a) < s < f(b)</math>. Then there exists some ''<math>x</math>'' between <math>a</math> and <math>b</math> such that <math>f(x) = s</math>.]]

In [[mathematical analysis]], the '''intermediate value theorem''' states that if <math>f</math> is a [[continuous function|continuous]] [[Function (mathematics)|function]] whose [[domain of a function|domain]] contains the [[interval (mathematics)|interval]] {{closed-closed|''a'', ''b''}}, then it takes on any given value between <math>f(a)</math> and <math>f(b)</math> at some point within the interval.

This has two important [[corollary|corollaries]]:

# If a continuous function has values of opposite sign inside an interval, then it has a [[Zero of a function|root]] in that interval ('''Bolzano's theorem''').<ref>{{MathWorld |title=Bolzano's Theorem |urlname=BolzanosTheorem}}</ref><ref>{{cite book |doi=10.1007/978-3-030-11036-9|title=Cauchy's Calcul Infinitésimal |year=2019 |last1=Cates |first1=Dennis M. |isbn=978-3-030-11035-2 |s2cid=132587955|page=249 }}</ref>

# The [[image (mathematics)|image]] of a continuous function over an interval is itself an interval.

This captures an intuitive property of continuous functions over the [[real number]]s: given ''<math>f</math>'' continuous on <math>[1,2]</math> with the known values <math>f(1) = 3</math> and <math>f(2) = 5</math>, then the graph of <math>y = f(x)</math> must pass through the horizontal line <math>y = 4</math> while <math>x</math> moves from <math>1</math> to <math>2</math>. It represents the idea that the graph of a continuous function on a closed interval can be drawn without lifting a pencil from the paper.

The intermediate value theorem states the following:

*''Version II.'' the [[Image of a function|image set]] <math>f(I)</math> is also a closed interval, and it contains <math>\bigl[\min(f(a), f(b)),\max(f(a), f(b))\bigr]</math>.

 

==Relation to completeness==

The theorem depends on, and is equivalent to, the [[completeness of the real numbers]]. The intermediate value theorem does not apply to the [[rational number]]s '''Q''' because gaps exist between rational numbers; [[irrational numbers]] fill those gaps. For example, the function <math>f(x) = x^2</math> for <math>x\in\Q</math> satisfies <math>f(0) = 0</math> and <math>f(2) = 4</math>. However, there is no rational number <math>x</math> such that <math>f(x)=2</math>, because <math>\sqrt 2</math> is an irrational number.

 

==Proof==

Let <math>S</math> be the set of all <math>x \in [a,b]</math> such that <math>f(x)<u</math>. Then <math>S</math> is non-empty since <math>a</math> is an element of <math>S</math>. Since <math>S</math> is non-empty and bounded above by <math>b</math>, by completeness, the [[supremum]] <math>c=\sup S</math> exists. That is, <math>c</math> is the smallest number that is greater than or equal to every member of <math>S</math>.

Note that, due to the continuity of <math>f</math> at <math>a</math>, we can keep <math>f(x)</math> within any <math>\varepsilon>0</math> of <math>f(a)</math> by keeping <math>x</math> sufficiently close to <math>a</math>. Since <math>f(a)<u</math> is a strict inequality, consider the implication when <math>\varepsilon</math> is the distance between <math>u</math> and <math>f(a)</math>. No <math>x</math> sufficiently close to <math>a</math> can then make <math>f(x)</math> greater than or equal to <math>u</math>, which means there are values greater than <math>a</math> in <math>S</math>. A more detailed proof goes like this:

Choose <math>\varepsilon=u-f(a)>0</math>. Then <math>\exists \delta>0</math> such that <math>\forall x \in [a,b]</math>, <math display="block">|x-a|<\delta \implies |f(x)-f(a)|<u-f(a) \implies f(x)<u.</math>Consider the interval <math>[a,\min(a+\delta,b))=I_1</math>. Notice that <math>I_1 \subseteq [a,b]</math> and every <math>x \in I_1</math> satisfies the condition <math>|x-a|<\delta</math>. Therefore for every <math>x \in I_1</math> we have <math>f(x)<u</math>. Hence <math>c</math> cannot be <math>a</math>.

 

Choose <math>\varepsilon=f(b)-u>0</math>. Then <math>\exists \delta>0</math> such that <math>\forall x \in [a,b]</math>, <math display="block">|x-b|<\delta \implies |f(x)-f(b)|<f(b)-u \implies f(x)>u.</math>Consider the interval <math>(\max(a,b-\delta),b]=I_2</math>. Notice that <math>I_2 \subseteq [a,b]</math> and every <math>x \in I_2</math> satisfies the condition <math>|x-b|<\delta</math>. Therefore for every <math>x \in I_2</math> we have <math>f(x)>u</math>. Hence <math>c</math> cannot be <math>b</math>.

 

With <math>c \neq a</math> and <math>c \neq b</math>, it must be the case <math>c \in (a,b)</math>. Now we claim that <math>f(c)=u</math>.

 

Fix some <math>\varepsilon > 0</math>. Since <math>f</math> is continuous at <math>c</math>, <math>\exists \delta_1>0</math> such that <math>\forall x \in [a,b]</math>, <math>|x-c|<\delta_1 \implies |f(x) - f(c)| < \varepsilon</math>.

 

Since <math>c \in (a,b)</math> and <math>(a,b)</math> is open, <math>\exists \delta_2>0</math> such that <math>(c-\delta_2,c+\delta_2) \subseteq (a,b)</math>. Set <math>\delta= \min(\delta_1,\delta_2)</math>. Then we have

<math display="block">f(x)-\varepsilon<f(c)<f(x)+\varepsilon</math>

for all <math>x\in(c-\delta,c+\delta)</math>. By the properties of the supremum, there exists some <math>a^*\in (c-\delta,c]</math> that is contained in <math>S</math>, and so

<math display="block">f(c)<f(a^*)+\varepsilon<u+\varepsilon.</math>

Picking <math>a^{**}\in(c,c+\delta)</math>, we know that <math>a^{**}\not\in S</math> because <math>c</math> is the supremum of <math>S</math>. This means that

<math display="block">f(c)>f(a^{**})-\varepsilon \geq u-\varepsilon.</math>

Both inequalities

<math display="block">u-\varepsilon<f(c)< u+\varepsilon</math>

are valid for all <math>\varepsilon > 0</math>, from which we deduce <math>f(c) = u</math> as the only possible value, as stated.

Define <math>g(x)=f(x)-u</math> which is equivalent to <math>f(x)=g(x)+u</math> and lets us rewrite <math>f(a)<u<f(b)</math> as <math>g(a)<0<g(b)</math>, and we have to prove, that <math>g(c)=0</math>  for some <math>c\in[a,b]</math>, which is more intuitive. We further define the set <math>S=\{x\in[a,b]:g(x)\leq 0\}</math>. Because <math>g(a)<0</math> we know, that <math>a\in S</math> so, that <math>S</math> is not empty. Moreover, as <math>S\subseteq[a,b]</math>, we know that <math>S</math> is bounded and non-empty, so by Completeness, the [[supremum]] <math>c=\sup(S)</math> exists.

 

There are 3 cases for the value of <math>g(c)</math>, those being <math>g(c)<0,g(c)>0</math> and <math>g(c)=0</math>. For contradiction, let us assume, that <math>g(c)<0</math>. Then, by the definition of continuity, for <math>\epsilon=0-g(c)</math>, there exists a <math>\delta>0</math> such that <math>x\in(c-\delta,c+\delta)</math> implies, that <math>|g(x)-g(c)|<-g(c)</math>, which is equivalent to <math>g(x)<0</math>. If we just chose <math>x=c+\frac{\delta}{N}</math>, where <math>N>\frac{\delta}{b-c}+1</math>, then as <math>1 < N</math>, <math>x<c+\delta</math>, from which we get <math>g(x)<0</math> and <math>c<x<b</math>, so <math>x\in S</math>. It follows that <math>x</math> is an upper bound for <math>S</math>.  However, <math>x>c</math>, contradicting the '''upper bound''' property of the ''least upper bound'' <math>c</math>, so <math>g(c)\geq 0</math>. Assume then, that <math>g(c)>0</math>. We similarly chose <math>\epsilon=g(c)-0</math> and know, that there exists a <math>\delta>0</math> such that <math>x\in(c-\delta,c+\delta)</math> implies <math>|g(x)-g(c)|<g(c)</math>. We can rewrite this as <math>-g(c)<g(x)-g(c)<g(c)</math> which implies, that <math>g(x)>0</math>. If we now chose <math>x=c-\frac{\delta}{2}</math>, then <math>g(x)>0</math> and <math>a<x<c</math>. It follows that <math>x</math> is an upper bound for <math>S</math>. However, <math>x<c</math>, which contradict the '''least''' property of the ''least upper bound'' <math>c</math>, which means, that <math>g(c)>0</math> is impossible. If we combine both results, we get that <math>g(c)=0</math> or <math>f(c)=u</math> is the only remaining possibility.

  }}</ref> The theorem was first proved by [[Bernard Bolzano]] in 1817. Bolzano used the following formulation of the theorem:<ref>{{Cite journal| title=A translation of Bolzano's paper on the intermediate value theorem| first=S.B.| last=Russ| journal=Historia Mathematica| date=1980| volume=7| issue=2| pages=156–185| doi=10.1016/0315-0860(80)90036-1| doi-access=free}}</ref>

 

Let <math>f, \varphi</math> be continuous functions on the interval between <math>\alpha</math> and <math>\beta</math> such that <math>f(\alpha) < \varphi(\alpha)</math> and <math>f(\beta) > \varphi(\beta)</math>. Then there is an <math>x</math> between <math>\alpha</math> and <math>\beta</math> such that <math>f(x) = \varphi(x)</math>.

Earlier authors held the result to be intuitively obvious and requiring no proof. The insight of Bolzano and Cauchy was to define a general notion of continuity (in terms of [[infinitesimal]]s in Cauchy's case and using real inequalities in Bolzano's case), and to provide a proof based on such definitions.

==Converse is false==

Vrahatis<ref>{{Cite journal |last=Vrahatis |first=Michael N. |date=2016-04-01 |title=Generalization of the Bolzano theorem for simplices |url=https://www.sciencedirect.com/science/article/pii/S0166864115005994 |journal=Topology and Its Applications |language=en |volume=202 |pages=40–46 |doi=10.1016/j.topol.2015.12.066 |issn=0166-8641}}</ref> presents a similar generalization to triangles, or more generally, ''n''-dimensional [[Simplex|simplices]]. Let ''Dⁿ'' be an ''n''-dimensional simplex with ''n''+1 vertices denoted by ''v''₀,...,''v_n''. Let ''F''=(''f''₁,...,''f_n'') be a continuous function from ''Dⁿ'' to ''Rⁿ'', that never equals 0 on the boundary of ''Dⁿ''. Suppose ''F'' satisfies the following conditions:

The theorem can be proved based on the [[Knaster–Kuratowski–Mazurkiewicz lemma]]. In can be used for approximations of fixed points and zeros.<ref>{{Cite journal |last=Vrahatis |first=Michael N. |date=2020-04-15 |title=Intermediate value theorem for simplices for simplicial approximation of fixed points and zeros |journal=Topology and Its Applications |language=en |volume=275 |pages=107036 |doi=10.1016/j.topol.2019.107036 |issn=0166-8641|doi-access=free }}</ref>

* Let <math>a</math> and <math>b</math> be real numbers and <math>f:[a,b] \to R</math> be a pointwise continuous function from the [[closed interval]] <math>[a,b]</math> to the real line, and suppose that <math>f(a) < 0</math> and <math>0 < f(b)</math>. Then for every positive number <math>\varepsilon > 0</math> there exists a point <math>x</math> in the unit interval such that <math>\vert f(x) \vert < \varepsilon</math>.<ref>{{cite journal|title=Interpolating Between Choices for the Approximate Intermediate Value Theorem | author=Matthew Frank|journal=Logical Methods in Computer Science|volume=16|issue=3|date=July 14, 2020| doi=10.23638/LMCS-16(3:5)2020|arxiv=1701.02227}}</ref>