Proving a function has no extreme points.

Question

Is there any possible way I can make this proof shorter?

Question 5

a.

Consider the function $f:\ R\ ➜\ R,\ f\left(x\right)\ =\ \frac{x}{\left|x\right|+1}$

Note that by the algebra of continuous functions f is continuous.

We show the limits of f as x tends to $∞,-∞$ exist.

$\lim _{x\to \infty }\left(\frac{x}{\left|x\right|+1}\right)=\lim _{x\to \infty }\left(\frac{x}{x+1}\right)=\lim _{x\to \infty }\left(\frac{x}{x\left(1+\frac{1}{x}\right)}\right)=\lim \:_{x\to \infty \:}\left(\frac{1}{1+\frac{1}{x}}\right)=\left(\frac{1}{1+0}\right)=1$ $\lim \:\:_{x\to \:-\infty \:\:}\left(\frac{x}{\left|x\right|+1}\right)=\lim \:_{x\to -\infty \:}\left(\frac{x}{-x+1}\right)=\lim _{x\to -\infty }\left(\frac{x}{x\left(-1-\frac{1}{x}\right)}\right)=\lim \:_{x\to -\infty \:}\left(\frac{1}{-1-\frac{1}{x}}\right)=\left(\frac{1}{-1-0}\right)=-1\:$

Now we show that for every $x ∈ R \ \ $ we have $-1 < f(x) < 1$.

Assume toward contradiction that there exist $x ∈ R$ such that $f(x) ≥ 1$ or $f(x) ≤ -1$

Case 1: $x = 0$. In this case $f$ will equal to $0$ which is obviously in $(-1,1)$, so we have a contradiction.

Case 2: $x > 0$.

$\frac{x}{\left|x\right|+1}\ge1 \ ➜ \ \frac{x}{x+1}\ge1 \ ➜ \ x\ge x+1 \ ➜ \ 0\ge1$ Contradiction!

$\frac{x}{\left|x\right|+1}\le-1 \ ➜ \ \frac{x}{x+1}\le-1 \ ➜ \ x\le-x-1 \ ➜ \ 2x\le-1$ Contradiction! Since $2x > 0$.

Case 3: $x < 0$.

$\frac{x}{\left|x\right|+1}\ge1 \ ➜ \ \frac{x}{-x+1}\ge1 \ ➜ \ x\ge-x+1 \ ➜ \ 2x\ge1$ Contradiction! since $2x < 0$.

$\frac{x}{\left|x\right|+1}\le-1 \ ➜ \ \frac{x}{-x+1}\le-1 \ ➜ \ x\le x-1 \ ➜ \ 0\le-1$ Contradiction!

Therefore -1 < f(x) < 1 for every $x ∈ R$.

Now we show f(x) is strictly increasing. Assume toward contradiction that f(a) ≥ f(b) for every a,b ∈ R such that a < b.

$\frac{a}{\left|a\right|+1}\ge\frac{b}{\left|b\right|+1}\ \ ➜\ \ a\left(\left|b\right|+1\right)\ge b\left(\left|a\right|+1\right)\ $

Case 1: b = 0, 0 > a.

$a\left(\left|0\right|+1\right)\ge0\left(\left|a\right|+1\right)\ \ ➜\ \ a\ \ge0$ Contradiction!

Case 2: b > 0, a = 0.

$0\left(\left|b\right|+1\right)\ge b\left(\left|0\right|+1\right)\ \ ➜\ \ 0\ \ge b$ Contradiction!

Case 3: b > 0, a > 0

$a\left(b+1\right)\ge b\left(a+1\right)\ \ ➜\ \ ab\ +a\ge ab+b\ \ \ ➜\ \ a\ \ge b\ $ Contradiction!

Case 4: 0 > b, 0 > a.

$a\left(-b+1\right)\ge b\left(-a+1\right)\ \ ➜\ \ -ab\ +a\ge-ab+b\ \ \ ➜\ \ a\ \ge b\ $ Contradiction!

Case 5: b > 0, 0 > a.

$a\left(b+1\right)\ge b\left(-a+1\right)\ \ ➜\ \ ab\ +a\ge-ab+b\ \ \ ➜\ \ 2ab\ \ge b-a$ Contradiction! Since $2ab < 0$ and $b - a > 0$

We've reached a contradiction for every possible case of $a,b$, therefore $f(a) < f(b)$ for every $a,b ∈ R$, such that $a < b$ which means $f$ is strictly increasing.

Now we show $f$ has no extreme points.

Assume toward contradiction that $f$ takes a maximum at a point, we will denote it $a$.

Since $f$ is strictly increasing, there exist a point $b ∈ R$ such that $a < b$ for which $f(a) < f(b)$,

in contradiction to the minimality of $a$.

Assume toward contradiction that $f$ takes a minimum at a point, we will denote it $c$.

Since $f$ is strictly increasing, there exist a point $d ∈ R$ such that $d < c$ for which $f(d) < f(c)$,

in contradiction to the minimality of $c$.

Therefore we've shown that $f$ satisfies all requirements.

Answer 1

You can have much simpler examples... For instance, $f(x)=\arctan x$. Here, $L_{1,2} = \pm \frac{\pi}{2}$ and, since $f$ is diffrenciable in an open set and $f' \ne 0$, there cannot be any max / min.

Regarding your example, you can make your proof way shorter... Your function is $$ f(x)=\begin{cases}\dfrac{x}{-x+1}, & x \leq 0 \\[1em] \dfrac{x}{x+1}, & x>0 \end{cases} $$ and it is actually differentiable. Since it has no critical points, there can be no max/min.

If you cannot use derivatives, just take $a - b < 0$ and check that $f(a)-f(b)< 0$

$$ f(a)-f(b) = \frac{a}{|a|+1}-\frac{b}{|b|+1} = \frac{a-b + (a|b|-b|a|)}{(|a|+1)(|b|+1)} $$

If $a$ and $b$ are both positive or both negative $a|b|-b|a| =0$ and it follows that $f(a)-f(b) < 0$. If $a<0$ and $b>0$, $a|b|-b|a| = 2a b < 0$ and you still get $f(a)-f(b)<0$.

Answer 2

That's some serious struggle you got there. here's a simpler example:

Consider the function $f\left(x\right)=\frac{e^{x}}{1+e^{x}}$

$f$ is continuous. Also We have:

$$\underset{x\rightarrow\infty}{\lim}\frac{e^{x}}{1+e^{x}}=1$$

$$\underset{x\rightarrow-\infty}{\lim}\frac{e^{x}}{1+e^{x}}=0$$

$f$ doesn't have a max because for every $x_1\in\mathbb{R}$ we can take $x_1+1$ such that:

$$\underbrace{f\left(x_1+1\right)}=\frac{e^{x_{1}+1}}{1+e^{x_{1}+1}}=\frac{e\cdot e^{x_{1}}}{1+e\cdot e^{x_{1}}}=\frac{e^{x_{1}}}{\frac{1}{e}+e^{x_{1}}}\underset{\frac{1}{e}<1}{\underbrace >}\frac{e^{x_{1}}}{1+e^{x_{1}}}=\underbrace{f\left(x_{1}\right)}$$

Also, $f$ doesn't have a min because for every $x_2\in\mathbb{R}$ we can take $x_2-1$ such that:

$$\underbrace{f\left(x_2-1\right)}=\frac{e^{x_{2}-1}}{1+e^{x_{2}-1}}=\frac{\frac{e^{x_{2}}}{e}}{1+\frac{e^{x_{2}}}{e}}=\frac{e^{x_{2}}}{e+e^{x_{2}}}\underset{e>1}{\underbrace <}\frac{e^{x_{2}}}{1+e^{x_{2}}}=\underbrace{f\left(x_{2}\right)}$$