Help with proving the example

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November 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
November 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 $x_1<x_2\,\Longrightarrow\,f(x_1)\leq f(x_2)$ , so it may not qualify as a right example.

Tonio
November 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 $x_1<x_2\,\Longrightarrow\,f(x_1)\leq f(x_2)$ , 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
November 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.
November 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 $x < y\,\, s.t.\,\, f(x) <f(y)\,\,or\,\,f(x)>f(y)$ !
ANY non-constant function will increase/decrease "somewhere"...

Tonio
November 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 $x < y\,\, s.t.\,\, f(x) <f(y)\,\,or\,\,f(x)>f(y)$ !
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
November 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
November 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 $x_1<x_2$ such that $f(x_1)<f(x_2)$ ,
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.
November 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
November 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 $x_1<x_2$ such that $f(x_1)<f(x_2)$ ,
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