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        riddles >> medium >> support of continuous functions
        
(Message started by: hparty on Apr 24^th, 2010, 4:38pm)

Title: support of continuous functions
Post by hparty on Apr 24^th, 2010, 4:38pm

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 ??

Title: Re: support of continuous functions
Post by Obob on Apr 24^th, 2010, 5:00pm

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.

Title: Re: support of continuous functions
Post by hparty on Apr 24^th, 2010, 5:07pm

so how we can do that? ::)

Title: Re: support of continuous functions
Post by Obob on Apr 24^th, 2010, 5:10pm

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.

Title: Re: support of continuous functions
Post by hparty on Apr 24^th, 2010, 5:20pm

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)

???

Title: Re: support of continuous functions
Post by Obob on Apr 24^th, 2010, 5:27pm

I apologize, I was mistaken in my first post. For either definition of support, continuity of f isn't necessary. Your proof looks fine.

Title: Re: support of continuous functions
Post by hparty on Apr 24^th, 2010, 5:32pm

how I can explain that the statement remains true whether or nor f is continuous?

Title: Re: support of continuous functions
Post by Obob on Apr 24^th, 2010, 7:01pm

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.

Title: Re: support of continuous functions
Post by hparty on May 3^rd, 2010, 6:04pm

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

Title: Re: support of continuous functions
Post by Obob on May 3^rd, 2010, 6:19pm

I don't see any problem with it. Let me know if you figure it out.

Title: Re: support of continuous functions
Post by hparty on May 3^rd, 2010, 8:10pm

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 ?!! :-/

Title: Re: support of continuous functions
Post by Obob on May 4^th, 2010, 4:15am

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.