Author |
Topic: support of continuous functions (Read 1064 times) |
|
hparty
Newbie
Gender: 
Posts: 15
|
 |
support of continuous functions
« on: Apr 24th, 2010, 4:38pm » |
 Quote  Modify
|
for any continuous function f, f can be written as f=f+ - f_; f+: positive part, f_:negative part
Prove that support(f)= supp(f+) U supp(f_)
Here is my proof:
In the proof of the second inclusion (right to left): given a point z with f+(z) nonzero
then f(z)=f+(z)
(since x=z will not be in the domain of f_(x)),
also if f_(z) is nonzero then f(z)=f_(z)
(since x=z will not be in the domain of f+(x)), and in both cases
f(z) will be nonzero, i.e. in the L.H.S.
for the first inclusion: given a with f(z) nonzero then f(z)=f+(z) - f_(z):nonzero,
then f+(z) =/= f_(z), so at least one of them nonzero, hence a in the R.H.S.
Is this right? where I supposed to use the continuity of f? and what
would happen if f is not continuous ??
|
| « Last Edit: Apr 24th, 2010, 4:39pm by hparty » |
 IP Logged |
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: support of continuous functions
« Reply #1 on: Apr 24th, 2010, 5:00pm » |
Quote Modify
|
You're not using the correct definition for support. The support should be the closure of the set where f is nonzero. With this definition of support, you will have to use the continuity hypothesis.
|
| « Last Edit: Apr 24th, 2010, 5:02pm by Obob » |
IP Logged |
|
|
|
hparty
Newbie
Gender:
Posts: 15
|
|
Re: support of continuous functions
« Reply #2 on: Apr 24th, 2010, 5:07pm » |
Quote Modify
|
so how we can do that? 
|
|
IP Logged |
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: support of continuous functions
« Reply #3 on: Apr 24th, 2010, 5:10pm » |
Quote Modify
|
I'm not here to do the problem for you. If you didn't know the correct definition until just now, you clearly haven't tried to do it yourself.
The solution you wrote would be correct if the definition of the support of f was just {x:f(x) != 0}. But this isn't the definition of support.
|
| « Last Edit: Apr 24th, 2010, 5:11pm by Obob » |
IP Logged |
|
|
|
hparty
Newbie
Gender:
Posts: 15
|
|
Re: support of continuous functions
« Reply #4 on: Apr 24th, 2010, 5:20pm » |
Quote Modify
|
supp(f)= closure{x: f(x)!= 0},
supp(f+)= closure{x: f+(x)!= 0},
supp(f_)= closure{x: f_(x)!= 0},
so supp(f+) U supp(f_)= closure{x: f+(x)!= 0}U closure{x: f_(x)!= 0}
= closure ( {x: f+(x)!= 0}U{x: f_(x)!= 0} )
... I think here we use that f is continuous:
{x: f+(x)!= 0}U{x: f_(x)!= 0}={x: f(x)!= 0} !
= closure ( {x: f(x)!= 0} )
=supp(f)
|
| « Last Edit: Apr 24th, 2010, 5:23pm by hparty » |
IP Logged |
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: support of continuous functions
« Reply #5 on: Apr 24th, 2010, 5:27pm » |
Quote Modify
|
I apologize, I was mistaken in my first post. For either definition of support, continuity of f isn't necessary. Your proof looks fine.
|
|
IP Logged |
|
|
|
hparty
Newbie
Gender:
Posts: 15
|
|
Re: support of continuous functions
« Reply #6 on: Apr 24th, 2010, 5:32pm » |
Quote Modify
|
how I can explain that the statement remains true whether or nor f is continuous?
|
|
IP Logged |
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: support of continuous functions
« Reply #7 on: Apr 24th, 2010, 7:01pm » |
Quote Modify
|
You just need to prove it without using the assumption that f is continuous, which you've already done.
The statement {x: f+(x)!= 0}U{x: f_(x)!= 0}={x: f(x)!= 0} doesn't use continuity of f: this statement can be proved as in your first post.
|
|
IP Logged |
|
|
|
hparty
Newbie
Gender:
Posts: 15
|
|
Re: support of continuous functions
« Reply #8 on: May 3rd, 2010, 6:04pm » |
Quote Modify
|
I just want to say that unfortunately this proof was not enough, this question was in my exam two days ago and I got 2 points out of 5 !!
I'm still don't know why! although I believe that it's correct.
thanks anyway
|
|
IP Logged |
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: support of continuous functions
« Reply #9 on: May 3rd, 2010, 6:19pm » |
Quote Modify
|
I don't see any problem with it. Let me know if you figure it out.
|
|
IP Logged |
|
|
|
hparty
Newbie
Gender:
Posts: 15
|
|
Re: support of continuous functions
« Reply #10 on: May 3rd, 2010, 8:10pm » |
Quote Modify
|
It says that for any function f: the support is just the complement of the set of points where f(x) is zero in a neighborhood of x, but if f is continuous then the support of f is the closure of the set {x: f(x) !=0}, and these two definitions are not equivalent !
aren't they ?!! 
|
|
IP Logged |
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: support of continuous functions
« Reply #11 on: May 4th, 2010, 4:15am » |
Quote Modify
|
Yes, those definitions are clearly equivalent for any function f (not requiring continuity).
A point is in the complement of the closure of {x: f(x)!=0} iff f is zero in a neighborhood of x.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print