I Understanding a proof of inexistence of max nor min

[mcastillo356]

TL;DR Summary

I've got a proof of the inexistence of a local maximum value nor a local minimum value at the origin of coordinates for a certain function, and need advice to understand it

Although a function cannot have extreme values anywhere other than at endpoints, critical points, and singular points, it need not have extreme values at such points. It is more difficult to draw the graph of a function whose domain has an endpoint at which the function fails to have an extreme value
A function with no max or min at an endpoint. Let
$f(x)=\begin{cases}{x\sin{\left(\dfrac{1}{x}\right)}}&\text{if}& x>0\\0 & \text{if}& x=0\end{cases}$
Show that $f$ is continuous on $[0,\infty)$ and differentiable on $(0,\infty)$ but it has neither a local maximum nor a local minimum value at the endpoint $x=0$
I've already proven it is continuous on $[0,\infty)$ and differentiable on $(0,\infty)$, and here is the proof that it has not loc min or loc max:
Background
A function $f$ has not a maximum local value $f(x_0)$ at $x_0$ in its domain if for all $h>0$ can always be found $x\in{\mathfrak{D}(f)}$, such that $|x-x_0|<h$ and $f(x)>f(x_0)$
A function $f$ has not a minimum local value $f(x_1)$ at $x_1$ in its domain if for all $h>0$ can always be found $x\in{\mathfrak{D}(f)}$, such that $|x-x_1|<h$ and $f(x)<f(x_1)$
$x_n=\dfrac{1}{\dfrac{\pi}{2}+2\pi\;n}$
$y_n=\dfrac{1}{-\dfrac{\pi}{2}+2\pi\;n}$
Proof I don't understand:

For all $n \in \mathbb{N}$ we have $x_n > 0$ and $y_n > 0$ , therefore $x_n , y_n \in [0,+\infty)$.
Let $h > 0$ .Exists one $n_h \in \mathbb{N}$ such that $n_h > \dfrac{1}{h}$ . Then for all $n \geq n_h$ we have:
$n \geq n_h > h$. Then $x_n < \dfrac{1}{n} \leq \dfrac{1}{n_h} < h$ and $y_n < \dfrac{1}{n} \leq \dfrac{1}{n_h} < h$
No matter how small is $h$ we have $f(y_n) < f(0) < f(x_n)$ . And is proven that $f(0)$ cannot be maximum nor minimum.

 

[FactChecker]

There could be more specific detail in the end of the proof. Like:
$f(y_n) = -y_n \lt 0 =f(0) \lt x_n = f(x_n)$
What part do you not understand?

 

[mcastillo356]

I have actually proven it, based on the same basis. I know eventually $f(y_n)<f(0)<f(x_n)$, but in a Spanish math forum I stumbled across this algebra. This proof is smarter than mine, shorter, direct, but incomprehensible

For all $n \in \mathbb{N}$ we have $x_n > 0$ and $y_n > 0$ , therefore $x_n , y_n \in [0,+\infty)$.

Right

Let $h > 0$ .Exists one $n_h \in \mathbb{N}$ such that $n_h > \dfrac{1}{h}$ . Then for all $n \geq n_h$ we have:
$n \geq n_h > h$. Then $x_n < \dfrac{1}{n} \leq \dfrac{1}{n_h} < h$ and $y_n < \dfrac{1}{n} \leq \dfrac{1}{n_h} < h$

This quote is imposible for me to understand. How does it fit the conditions of

A function $f$ has not a maximum local value $f(x_0)$ at $x_0$ in its domain if for all $h>0$ can always be found $x\in{\mathfrak{D}(f)}$, such that $|x-x_0|<h$ and $f(x)>f(x_0)$
A function $f$ has not a minimum local value $f(x_1)$ at $x_1$ in its domain if for all $h>0$ can always be found $x\in{\mathfrak{D}(f)}$, such that $|x-x_1|<h$ and $f(x)<f(x_1)$

No matter how small is $h$ we have $f(y_n) < f(0) < f(x_n)$ . And is proven that $f(0)$ cannot be maximum nor minimum.

Right

 

[Mark44]

From post #1:

A function $f$ has not a maximum local value $f(x_0)$ at $x_0$ in its domain if for all $h>0$ can always be found $x\in{\mathfrak{D}(f)}$, such that $|x-x_0|<h$ and $f(x)>f(x_0)$

Right
Let $h > 0$ .Exists one $n_h \in \mathbb{N}$ such that $n_h > \dfrac{1}{h}$ . Then for all $n \geq n_h$ we have:
$n \geq n_h > h$. Then $x_n < \dfrac{1}{n} \leq \dfrac{1}{n_h} < h$ and $y_n < \dfrac{1}{n} \leq \dfrac{1}{n_h} < h$

This inequality seems to have a typo in it: $n \geq n_h > h$, and likely should be $n \geq n_h > \frac 1 h$. The inequality as stated is certainly true, but makes the subsequent work harder to understand.

 

[mcastillo356]

Thanks, PF!