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Topic: test functions (Read 3263 times) |
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Jen
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test functions
« on: Apr 15th, 2010, 12:55pm » |
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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
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Obob
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Re: test functions
« Reply #1 on: Apr 15th, 2010, 3:56pm » |
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on Apr 15th, 2010, 12:55pm, 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? |
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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.
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| « Last Edit: Apr 15th, 2010, 4:13pm by Obob » |
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Jen
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Re: test functions
« Reply #2 on: Apr 15th, 2010, 9:43pm » |
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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?
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| « Last Edit: Apr 15th, 2010, 9:47pm by Jen » |
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Obob
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Re: test functions
« Reply #3 on: Apr 16th, 2010, 4:15am » |
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The bump function should be 1 on the support of f.
Of course f+g is a test function.
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Jen
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Re: test functions
« Reply #4 on: Apr 16th, 2010, 4:46am » |
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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?
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| « Last Edit: Apr 16th, 2010, 4:49am by Jen » |
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Obob
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Re: test functions
« Reply #5 on: Apr 16th, 2010, 6:15am » |
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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.
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Jen
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Re: test functions
« Reply #6 on: Apr 16th, 2010, 6:19am » |
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why the argument in 2 doesn't contradicts our example in 1?
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| « Last Edit: Apr 16th, 2010, 7:04am by Jen » |
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Obob
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Re: test functions
« Reply #7 on: Apr 16th, 2010, 9:01am » |
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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.
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| « Last Edit: Apr 16th, 2010, 9:01am by Obob » |
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