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,

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 $\lbrack a,b\rbrack$ 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.

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

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

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

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

Calculus: An Intuitive and Physical Approach, Second Edition, Morris Kline, Chapter 8，Section 3↩︎
Introduction to Calculus and Analysis Volume I, Reprint of the 1989 edition, Richard Courant, Fritz John, p101↩︎