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        general >> wanted >> test functions
        
(Message started by: Jen on Apr 15^th, 2010, 12:55pm)

Title: test functions
Post by Jen on Apr 15^th, 2010, 12:55pm

we know that any real valued function f can be written as f= (f+) _ (f_) , where f+ and f_ are two non-negative real valued functions.

Now I wanna prove that we have the same result for test functions in D (the space of test functions), that is:
any real valued function f in D can be written in the form: f= (f+) _ (f_) , but here we have f+ and
f_ are both non-negative real valued functions in D.

Moreover: Is there an example for a real valued function f in D for which f= (f+) _ (f_) but f+ and f_ are not test functions?

this is my first time I use this forum and I hope someone will help me

Thank you

Title: Re: test functions
Post by Obob on Apr 15^th, 2010, 3:56pm

on 04/15/10 at 12:55:27, Jen wrote:

Is there an example for a real valued function f in D for which f= (f+) _ (f_) but f+ and f_ are not test functions?

Yes. Just let f(x) be a bump function around the origin (so f is smooth, positive, compactly supported, and f(x) = 1 for x close to 0), and let h = xf(x). The functions h+ and h- are clearly not smooth at x=0.

For the other question, here's a hint. It's enough to find a positive test function g such that f+g is positive everywhere.

Title: Re: test functions
Post by Jen on Apr 15^th, 2010, 9:43pm

ok, I'm thinking of a function g to be
g=( the above bump function).((max value of f)+1) - f

then f+g=
( the above bump function).((max value of f) +1)

which is a test function and positive :
clearly g is positive, right !!
(of course we extend the bump function so it has the same support as f has)

but I'm not sure if it is a test function?

Title: Re: test functions
Post by Obob on Apr 16^th, 2010, 4:15am

The bump function should be 1 on the support of f.

Of course f+g is a test function.

Title: Re: test functions
Post by Jen on Apr 16^th, 2010, 4:46am

so:
our g would be g(x)= (maxvalue of f+1).b(x) - f(x)

where b(x)=1 : if x in support of f, and 0 otherwise

and this function will do the jop?!

why we cannot use exp(- x^-2) supported over [-1,1] instead of 1 in part 1 and 2 above?

also why the argument in 2 doesn't contradicts our example in 1?

Title: Re: test functions
Post by Obob on Apr 16^th, 2010, 6:15am

No, b(x) isn't 1 if x is in the support of f and 0 otherwise. b has to be smooth. So b is 1 if x is in the support of f, it is smooth, compactly supported, and positive. This doesn't uniquely specify b.

Actually constructing such a function b is a whole other exercise.

Title: Re: test functions
Post by Jen on Apr 16^th, 2010, 6:19am

why the argument in 2 doesn't contradicts our example in 1?

Title: Re: test functions
Post by Obob on Apr 16^th, 2010, 9:01am

What argument?

What you asked is if we can write any test function f as g-h, where g and h are positive test functions, and indeed we can.

This doesn't mean you can always take g and h to be the function g = f+ = max(f,0) and h = f- = -min(f,0): f+ and f- typically aren't smooth.