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 $y_0\le y$, and when $x_0>x$ then $y_0\ge y$.

The intermediate value theorem states:

If $f$ is a continuous function on a closed interval $[a,b]$, 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 $[a,b]$ 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 $[a,b]$ 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 $x^{\prime }$ such that $f(x^{\prime })<y_0$.

• for each $x$ in any right neighborhood of $c,y_0\le f(x)$.

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.

$$f(x)=\{\begin{array}{rl}{x}^2\sin \frac{1}{x},& \text{ if }x<0\\ x^2,& \text{ if }x\ge 0\end{array}$$

$$f(x)=\{\begin{array}{rl}-x^2,& \text{ if }x\le 0\\ x^2\sin \frac{1}{x},& \text{ if }x>0\end{array}$$

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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↩︎