A note on trying to extend the intermediate value theorem

First, it is necessary to introduce the following definitions1,

  • The function is said to be increasing at x0 if for all x-values in some interval about x0 it is true that when x0<x then y0<y, and when x0>x then y0>y.

  • The function is said to be decreasing at x0 if for all x-val ues in some interval about x0 it is true that when x0<x then y0>y, and when x0>x then y0<y.

Then the definition of “a function is non-decreasing at x0” is introduced by extend the definition of increasing above.

  • The function is said to be non-decreasing at x0 if for all x-values in some interval about x0 it is true that when x0<x then y0y, and when x0>x then y0y.

The intermediate value theorem states:

If f is a continuous function on a closed interval [a,b], and if y0 is any value strictly between f(a) and f(b), that is min{f(a),f(b)}<y0<max{f(a),f(b)}, then y0=f(c) for some c in (a,b).

My original conjecture was to extend it to:

If f is a continuous function on a closed interval [a,b] with f(a)<f(b), and if y0 is any value strictly between f(a) and f(b), that is f(a)<y0<f(b), then y0=f(c) for some c in (a,b) and f is non-decreasing at the c.

But it was said that the Weierstrass function, as a counterexample, is indeed nowhere monotonic, while I still couldn’t understand that. After reading the following proof for the intermediate value theorem2,

文本, 信件 描述已自动生成

I realized I could save the conjecture to:

If f is a continuous function on a closed interval [a,b] with f(a)<f(b), and if y0 is any value strictly between f(a) and f(b), that is f(a)<y0<f(b), then there is a c=sup({xf(x)<y0 in (a,b)}) in (a,b) cause y0=f(c), and such that

  • in every left neighborhood of c, there is always a x such that f(x)<y0.

  • for each x in any right neighborhood of c, y0f(x).

A similar argument could be drawn for f(a)>f(b).

The following two functions, which are both differentiable and continuous in neighborhoods around 0, are helpful for my consideration about the question.

f(x)={x2sin1x, if x<0x2, if x0

Plot[Evaluate[Piecewise[{{x^2 Sin[1/x], -0.5 < x < 0}, {x^2, 0 <= x < 0.5}}]], {x, -0.5, 0.5}]

f(x)={x2, if x0x2sin1x, if x>0

Plot[Evaluate[Piecewise[{{-x^2, -0.5 < x <= 0}, {x^2 Sin[1/x], 0 < x < 0.5}}]], {x, -0.5, 0.5}]


  1. Calculus: An Intuitive and Physical Approach, Second Edition, Morris Kline, Chapter 8,Section 3↩︎

  2. Introduction to Calculus and Analysis Volume I, Reprint of the 1989 edition, Richard Courant, Fritz John, p101↩︎

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