Continuity of Mean Value Theorem

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Discussion Overview

The discussion revolves around the continuity of the function defined by the Mean Value Theorem (MVT), specifically the mapping of points \(\xi_x\) that satisfy the theorem's conditions. Participants explore whether it is possible to choose these points in a way that maintains continuity, examining various examples and counterexamples.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the function \(x \mapsto \xi_x\) is not necessarily continuous, as the MVT only guarantees the existence of such points without ensuring their continuity.
  • Others argue that for polynomial functions, the roots of the resulting polynomial are continuous functions of the coefficients, suggesting a potential for continuity under certain conditions.
  • A participant highlights that the MVT does not guarantee that the intermediate values depend continuously on the endpoints \(a\) and \(b\), noting that the points \(\xi\) can "jump" as the interval changes.
  • One participant requests a concrete example where \(\xi\) must jump, indicating a need for a specific case to illustrate this behavior.
  • Another participant recalls a proof from functional analysis that addresses this issue, using the example of \(y = \cos(x)\) to demonstrate non-uniqueness in the derivative's zero points.
  • Several participants discuss the implications of choosing points \(c(x)\) in the context of the MVT, particularly in relation to the function \(y = x^3\) and how it behaves across different intervals.
  • There is a correction regarding the original post's domain and range definitions, with participants noting that the function is not defined at \(a\) and discussing the importance of these details in the context of the question.
  • Some participants express uncertainty about the implications of allowing endpoints to move and how that might affect the continuity of \(\xi_x\).
  • One participant suggests that extending the interval could preserve certain examples, indicating a potential for flexibility in the application of the MVT.
  • Another participant emphasizes the role of Rolle's theorem in establishing conditions for continuity and the challenges posed by the requirement \(f(a) = f(b)\).

Areas of Agreement / Disagreement

Participants generally do not reach consensus on the continuity of the function \(x \mapsto \xi_x\). Multiple competing views remain regarding the conditions under which continuity may or may not hold, and the discussion includes various examples and counterexamples that illustrate differing perspectives.

Contextual Notes

Participants note that the definitions of the intervals and the behavior of the function at the endpoints are crucial to the discussion. There are unresolved mathematical steps regarding the continuity of \(\xi_x\) and how it relates to the endpoints of the interval.

jostpuur
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TL;DR
Can the points produced by Mean Value Theorem be chosen to be continuous?
Suppose f:]a,b[\to\mathbb{R} is some differentiable function. Then it is possible to define a new function

<br /> ]a,b[\to [a,b],\quad x\mapsto \xi_x<br />

in such way that

<br /> f(x) - f(a) = (x - a)f&#039;(\xi_x)<br />

for all x\in ]a,b[. Mean Value Theorem says that these \xi_x exist.

One question that sometimes interests us is that is the function x\mapsto \xi_x continuous. The simple answer is that no, this function is not necessarily continuous.

However, for any fixed x, there can exist many different \xi_x with the same mean value property. So Mean Value Theorem doesn't fix the function x\mapsto \xi_x uniquely; it only says that at least one this kind of function exists.

So the big question that arises is that is it possible to choose the values \xi_x in a such way that the function x\mapsto\xi_x becomes continuous?
 
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If f is a polynomial of degree n in x then f(x) - f(a) = (x- a) f&#039;(\xi_x) is a polynomial of degree n - 1 in \xi_x. Roots of polynomials are generally continuous functions of the coefficients, which here are continuous functions of x.
 
The mean value theorem only provides existence. The intermediate values do not automatically depend continuously on the boundaries. This is actually a problem in many proofs because we have no control over ##\xi## if we narrow down the distance between ##a## and ##b##, e.g. The locations of the ##\xi## can jump, and if it does not, it has to be proven separately by properties the function brings with it.
 
Does anyone have a concrete example where ##\xi## is forced to jump no matter what previous choices we make?
 
I only remember that I have read a proof, IIRC somewhere in functional analysis, where this was a problem.

MVT in its differential version is proven with Rolle. ##y=\cos(x)## is an example where ##y'=0## isn't unique. The difficulty is that little word "forced" that you used. But if we only want to prove that ##\xi## doesn't depend continuously on ##a## and ##b## then ##y## might be an example.
 
In the MVT, the point c(x) is always chosen in the open interval (a,x). so that f'(c(x)) = [f(x)-f(a)]/(x-a). With this condition, c(x) cannot always depend continuously on x.

E.g. let y = x^3, on the interval [-1/2, 1]. Then when x = 1, the only admissible value of c(x) is x=1/2. As x moves to the left, once it passes 0, i.e. once x is negative, one is required to choose c(x) negative also, i.e. so it belongs to (-1/2,x).

Thus c(x) takes both positive and negative values, but c(x) never takes the value 0, since f'(0) = 0, but f is injective so [f(x)-f(a)]/(x-a) is never zero for any x.

using this principle (choosing an interval asymmetric about a flex) one could apparently make examples out of restricted ranges of y = cos(x) also.
 
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mathwonk said:
In the MVT, the point c(x) is always chosen in the open interval (a,x). so that f'(c(x)) = [f(x)-f(a)]/(x-a). With this condition, c(x) cannot always depend continuously on x.

E.g. let y = x^3, on the interval [-1/2, 1]. Then when x = 1, the only admissible value of c(x) is x=1/2. As x moves to the left, once it passes 0, i.e. once x is negative, one is required to choose c(x) negative also, i.e. so it belongs to (-1/2,x).

Thus c(x) takes both positive and negative values, but c(x) never takes the value 0, since f'(0) = 0, but f is injective so [f(x)-f(a)]/(x-a) is never zero for any x.

using this principle (choosing an interval asymmetric about a flex) one could apparently make examples out of restricted ranges of y = cos(x) also.

What is a in this example?
 
the interval [a,b] is [-1/2, 1] so a = -1/2.
 
I see. This is a nice example.

Though the original post didn't actually restrict ##\xi## to be between ##a## and ##x## but just to be between ##a## and ##b##. I didn't really think about this but maybe it's important (e.g. it breaks this counter example, which is where my confusion started at). I think the spirit of the question is probably resolved by your answer though.
 
  • #10
the original post is misstated. the function Is not defined at a, so the domain is (a,b], and the range is (a,b), more or less the opposite of what is stated.

even with these corrections, I do not have a more inclusive counterexample.
 
  • #11
I realised soon after posting the question that I had made a mistake with the sets ]a,b[ and [a,b], but I thought it wouldn't matter much, and didn't start quickly editing it. Now it looks like that these details do matter after all. I got a feeling that the question could probably be made interesting again by restating it somehow, but I'll have to think more about this.

It also looks like that it can make a difference whether only one end of the interval is allowed to move, or the both ends.
 
  • #12
jostpuur said:
I realised soon after posting the question that I had made a mistake with the sets ]a,b[ and [a,b], but I thought it wouldn't matter much, and didn't start quickly editing it. Now it looks like that these details do matter after all. I got a feeling that the question could probably be made interesting again by restating it somehow, but I'll have to think more about this.

It also looks like that it can make a difference whether only one end of the interval is allowed to move, or the both ends.
The endpoints don't matter that much. Technically the endpoints breaks mathwonk's example but if you just extend the interval a little (e.g. [-1/2, 2]) then I think it's saved.

I think the interesting extension is that the mean value theorem returns ##a<\xi_x< x ## but if you permit ##x< \xi_x<b## (so not located in the interval whose slope you are matching) then you have a lot more flexibility.
 
  • #13
MVT is proven by Rolle and Rolle by the existence of a zero. So all that is actually done results from the completeness of the real numbers. Rolle's condition ##f(a)=f(b)## is what makes considerations of continuity difficult. It establishes a constant distance where a narrowing one is asked for.
 

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