# Thread: Help with proving the example

1. ## Help with proving the example

Provide an example of a continuous function f:R-->R which satisfies all of the following properties

(a) non-decreasing
(b) not constant
(c) not increasing

Be sure to justify your example

I have an example as f( x ) = { 1, x<0 or x, x >= 0 with the graph that the function is constant on the left of the x axis, and increasing on the right of the x axis, but I dont know how to prove this

2. Originally Posted by 450081592
Provide an example of a continuous function f:R-->R which satisfies all of the following properties

(a) non-decreasing
(b) not constant
(c) not increasing

Be sure to justify your example

I have an example as f( x ) = { 1, x<0 or x, x >= 0 with the graph that the function is constant on the left of the x axis, and increasing on the right of the x axis, but I dont know how to prove this
$\displaystyle \sin x\,,\,\,\cos x\,,\,\,x^2\,,\,\,etc.$.

Your example is also called weakly increasing since $\displaystyle x_1<x_2\,\Longrightarrow\,f(x_1)\leq f(x_2)$ , so it may not qualify as a right example.

Tonio

3. Originally Posted by tonio
$\displaystyle \sin x\,,\,\,\cos x\,,\,\,x^2\,,\,\,etc.$.

Your example is also called weakly increasing since $\displaystyle x_1<x_2\,\Longrightarrow\,f(x_1)\leq f(x_2)$ , so it may not qualify as a right example.

Tonio
sinx con x and x^2 won't work cause they decrease somewhere, the function should never decrease in order to sasitify the conditions

4. Originally Posted by 450081592
sinx con x and x^2 won't work cause they decrease somewhere, the function should never decrease in order to sasitify the conditions
No, a function is non-decreasing if for ALL $\displaystyle x_1 < x_2 \in \mathbb{R} , \ f(x_1) \leq f(x_2)$ -- not for SOME $\displaystyle x_1,x_2$.. tonio's answers are perfectly fine.

5. Originally Posted by 450081592
sinx con x and x^2 won't work cause they decrease somewhere, the function should never decrease in order to sasitify the conditions

There's no such a thing! If a function is not constant then there are points $\displaystyle x < y\,\, s.t.\,\, f(x) <f(y)\,\,or\,\,f(x)>f(y)$ !
ANY non-constant function will increase/decrease "somewhere"...

Tonio

6. Originally Posted by tonio
There's no such a thing! If a function is not constant then there are points $\displaystyle x < y\,\, s.t.\,\, f(x) <f(y)\,\,or\,\,f(x)>f(y)$ !
ANY non-constant function will increase/decrease "somewhere"...

Tonio
not decrease is not the same as non-decrease, non-decrease means the function never decreases, not decrease means the function not always decrease, that what my prof said

7. Originally Posted by 450081592
not decrease is not the same as non-decrease, non-decrease means the function never decreases, not decrease means the function not always decrease, that what my prof said

Whatever: there is no such a thing like a non-constant function that never decreases and never increases: it just is an oxymoron, and the reason was given in my last post.

Tonio

8. Originally Posted by tonio
Whatever: there is no such a thing like a non-constant function that never decreases and never increases: it just is an oxymoron, and the reason was given in my last post.

Tonio
This is not obvius.
Consider the function:

$\displaystyle f(x)=1$ if $\displaystyle x\in\mathbb{Q}$, and $\displaystyle f(x)=0$ if $\displaystyle x\in\mathbb{R}\setminus\mathbb{Q}$

Now there exist points $\displaystyle x_1<x_2$ such that $\displaystyle f(x_1)<f(x_2)$,
but the function is still not increasing in $\displaystyle [x_1,x_2]$ no matter what points you pick.

Weierstrass function - Wikipedia, the free encyclopedia is a continuous function with the same attribute. There is no interval where the function is increasing or decreasing.

9. Originally Posted by tonio
Whatever: there is no such a thing like a non-constant function that never decreases and never increases: it just is an oxymoron, and the reason was given in my last post.

Tonio
nope, what the 3 conditions mean
(a) non-decreasing means never decrease, it can never decrease
(b) not constant means not always constant, so it can be contant in a subinterval
(c) not increasing means not always increase, so it can increase in a subinterval

10. Originally Posted by hjortur
This is not obvius.
Consider the function:

$\displaystyle f(x)=1$ if $\displaystyle x\in\mathbb{Q}$, and $\displaystyle f(x)=0$ if $\displaystyle x\in\mathbb{R}\setminus\mathbb{Q}$

Now there exist points $\displaystyle x_1<x_2$ such that $\displaystyle f(x_1)<f(x_2)$,
but the function is still not increasing in $\displaystyle [x_1,x_2]$ no matter what points you pick.

Weierstrass function - Wikipedia, the free encyclopedia is a continuous function with the same attribute. There is no interval where the function is increasing or decreasing.
The only one that mentioned "interval" that I'm aware of was Defunkt trying to explain my answer to the OP according to what he wrote. And in your example $\displaystyle 0<\sqrt{2}\,\,and\,\,f(0)>f(\sqrt{2})$.
Perhaps the OP meant that there is no INTERVAL where the function increases or decreases and then yours or Wierstrass' function work fine.
It never minds though, since the OP knows better because "his professor told him".

Tonio