# Help with proving the example

• Nov 15th 2009, 07:16 PM
450081592
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
• Nov 16th 2009, 03:07 AM
tonio
Quote:

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

$\sin x\,,\,\,\cos x\,,\,\,x^2\,,\,\,etc.$.

Your example is also called weakly increasing since $x_1 , so it may not qualify as a right example.

Tonio
• Nov 16th 2009, 04:40 AM
450081592
Quote:

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

Your example is also called weakly increasing since $x_1 , 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
• Nov 16th 2009, 05:13 AM
Defunkt
Quote:

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 $x_1 < x_2 \in \mathbb{R} , \ f(x_1) \leq f(x_2)$ -- not for SOME $x_1,x_2$.. tonio's answers are perfectly fine.
• Nov 16th 2009, 05:18 AM
tonio
Quote:

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 $x < y\,\, s.t.\,\, f(x) f(y)$ !
ANY non-constant function will increase/decrease "somewhere"...

Tonio
• Nov 16th 2009, 03:35 PM
450081592
Quote:

Originally Posted by tonio
There's no such a thing! If a function is not constant then there are points $x < y\,\, s.t.\,\, f(x) f(y)$ !
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
• Nov 16th 2009, 03:49 PM
tonio
Quote:

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
• Nov 16th 2009, 04:06 PM
hjortur
Quote:

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:

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

Now there exist points $x_1 such that $f(x_1),
but the function is still not increasing in $[x_1,x_2]$ 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.
• Nov 16th 2009, 04:11 PM
450081592
Quote:

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
• Nov 16th 2009, 06:37 PM
tonio
Quote:

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

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

Now there exist points $x_1 such that $f(x_1),
but the function is still not increasing in $[x_1,x_2]$ 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 $0<\sqrt{2}\,\,and\,\,f(0)>f(\sqrt{2})$.
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