Continuity of Mean Value Theorem

[jostpuur]

TL;DR Summary

Can the points produced by Mean Value Theorem be chosen to be continuous?

Suppose [itex]f:]a,b[\to\mathbb{R}[/itex] is some differentiable function. Then it is possible to define a new function

[tex]
]a,b[\to [a,b],\quad x\mapsto \xi_x
[/tex]

in such way that

[tex]
f(x) - f(a) = (x - a)f'(\xi_x)
[/tex]

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

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

However, for any fixed [itex]x[/itex], there can exist many different [itex]\xi_x[/itex] with the same mean value property. So Mean Value Theorem doesn't fix the function [itex]x\mapsto \xi_x[/itex] 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 [itex]\xi_x[/itex] in a such way that the function [itex]x\mapsto\xi_x[/itex] becomes continuous?

 

[pasmith]

If [itex]f[/itex] is a polynomial of degree [itex]n[/itex] in [itex]x[/itex] then [itex]f(x) - f(a) = (x- a) f'(\xi_x)[/itex] is a polynomial of degree [itex]n - 1[/itex] in [itex]\xi_x[/itex]. Roots of polynomials are generally continuous functions of the coefficients, which here are continuous functions of [itex]x[/itex].

 

[fresh_42]

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.

 

[Office_Shredder]

Does anyone have a concrete example where ##\xi## is forced to jump no matter what previous choices we make?

 

[fresh_42]

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.

 

[mathwonk]

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.

 

[Office_Shredder]

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?

 

[mathwonk]

the interval [a,b] is [-1/2, 1] so a = -1/2.

 

[Office_Shredder]

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.

 

[mathwonk]

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.

 

[jostpuur]

I realised soon after posting the question that I had made a mistake with the sets [itex]]a,b[[/itex] and [itex][a,b][/itex], 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.

 

[Office_Shredder]

I realised soon after posting the question that I had made a mistake with the sets [itex]]a,b[[/itex] and [itex][a,b][/itex], 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.

 

[fresh_42]

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.