Understanding a proof of inexistence of max nor min

• I
• mcastillo356
In summary: All clear, just a typo. In summary, a function with no maximum or minimum at an endpoint can still be continuous and differentiable, as proven by examples using the conditions of a function not having a local maximum or minimum value. The proof may vary, but the concept remains the same.
mcastillo356
Gold Member
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.

PeroK
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
mcastillo356 said:
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

From post #1:
mcastillo356 said:
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)##
mcastillo356 said:
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!

1. What is a proof of inexistence of max nor min?

A proof of inexistence of max nor min is a mathematical demonstration that a set of numbers or values does not have a maximum or minimum value. This means that there is no single number or value that is the largest or smallest in the set.

2. How is a proof of inexistence of max nor min different from a proof of non-existence?

A proof of inexistence of max nor min specifically deals with the concept of maximum and minimum values within a set, while a proof of non-existence can refer to the absence of any type of object or entity.

3. What are some common methods used in a proof of inexistence of max nor min?

Some common methods used in a proof of inexistence of max nor min include contradiction, induction, and counterexample. These methods involve analyzing the properties and characteristics of the set in question to show that no maximum or minimum value exists.

4. Can a set have a maximum or minimum value but still have no proof of existence?

Yes, it is possible for a set to have a maximum or minimum value without a proof of existence. This may occur if the existence of the maximum or minimum value is assumed but has not been formally proven.

5. How does understanding a proof of inexistence of max nor min benefit mathematicians and scientists?

Understanding a proof of inexistence of max nor min can help mathematicians and scientists to better understand the limitations and boundaries of a set of numbers or values. It can also aid in the development of new theories and concepts in mathematics and other fields.

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