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 \(x_{0}\) if for all \(x\)-values in some interval about \(x_{0}\) it is true that when \(x_{0} < x\) then \(y_{0} < y\), and when \(x_{0} > x\) then \(y_{0} > y\).

• The function is said to be decreasing at \(x_{0}\) if for all \(x\)-val ues in some interval about \(x_{0}\) it is true that when \(x_{0} < x\) then \(y_{0} > y\), and when \(x_{0} > x\) then \(y_{0} < y\).

Then the definition of “a function is non-decreasing at \(x_{0}\)” is introduced by extend the definition of increasing above.

• The function is said to be non-decreasing at \(x_{0}\) if for all \(x\)-values in some interval about \(x_{0}\) it is true that when \(x_{0} < x\) then \(y_{0} \leq y\), and when \(x_{0} > x\) then \(y_{0} \geq y\).

The intermediate value theorem states:

If \(f\) is a continuous function on a closed interval \(\lbrack a,b\rbrack\), and if \(y_{0}\) is any value strictly between \(f(a)\) and \(f(b)\), that is \(min\{ f(a),f(b)\} < y_{0} < max\{ f(a),f(b)\}\), then \(y_{0} = f(c)\) for some \(c\) in \((a,b)\).

My original conjecture was to extend it to:

If \(f\) is a continuous function on a closed interval \(\lbrack a,b\rbrack\) with \(f(a) < f(b)\), and if \(y_{0}\) is any value strictly between \(f(a)\) and \(f(b)\), that is \(f(a) < y_{0} < f(b)\), then \(y_{0} = f(c)\) for some \(c\) in \((a,b)\) and \(\mathbf{f}\) is non-decreasing at the \(\mathbf{c}\).

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,

I realized I could save the conjecture to:

If \(f\) is a continuous function on a closed interval \(\lbrack a,b\rbrack\) with \(f(a) < f(b)\), and if \(y_{0}\) is any value strictly between \(f(a)\) and \(f(b)\), that is \(f(a) < y_{0} < f(b)\), then there is a \(c = sup(\{ x \mid f(x) < y_{0}\) in \((a,b)\}\)) in \((a,b)\ \)cause \(y_{0} = f(c)\), and such that

• in every left neighborhood of \(c\), there is always a \(x'\) such that \(f(x') < y_{0}\).

• for each \(x\) in any right neighborhood of \(c,\ y_{0} \leq f(x)\).

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.

\[ f(x)=\left\{\begin{aligned} x^{2} \sin \frac{1}{x}, & \text { if } x<0 \\ x^{2}, & \text { if } x \geq 0 \end{aligned}\right. \]

\[ f(x)=\left\{\begin{aligned} -x^{2}, & \text { if } x \leq 0 \\ x^{2} \sin \frac{1}{x}, & \text { if } x>0 \end{aligned}\right. \]

---

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↩︎