Kernel of linear operator with functions having Fourier transform supported in $[0,infty)$
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A bounded linear operator $ T: L^2 to L^2$ with this properties :
1. Commutes with translation
2. Commutes with dilation
3. Has in its kernel functions $ f $ such that support $hat f subseteq [0,infty) $
I already can prove that $T= f* mu$ where $mu $ is a tempered distribution.
Then $int_-infty^0hat f(s)e^2ipi s x ds =c Tf (x)$ where $ c $ is s constant. How to prove this ?
real-analysis
edited Aug 26 at 9:19
Bernard
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A bounded linear operator $ T: L^2 to L^2$ with this properties :
1. Commutes with translation
2. Commutes with dilation
3. Has in its kernel functions $ f $ such that support $hat f subseteq [0,infty) $
I already can prove that $T= f* mu$ where $mu $ is a tempered distribution.
Then $int_-infty^0hat f(s)e^2ipi s x ds =c Tf (x)$ where $ c $ is s constant. How to prove this ?
real-analysis
edited Aug 26 at 9:19
Bernard
111k635103
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
Aug 26 at 7:59
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up vote
0
down vote
favorite
up vote
0
down vote
favorite
A bounded linear operator $ T: L^2 to L^2$ with this properties :
1. Commutes with translation
2. Commutes with dilation
3. Has in its kernel functions $ f $ such that support $hat f subseteq [0,infty) $
I already can prove that $T= f* mu$ where $mu $ is a tempered distribution.
Then $int_-infty^0hat f(s)e^2ipi s x ds =c Tf (x)$ where $ c $ is s constant. How to prove this ?
real-analysis
edited Aug 26 at 9:19
Bernard
111k635103
A bounded linear operator $ T: L^2 to L^2$ with this properties :
1. Commutes with translation
2. Commutes with dilation
3. Has in its kernel functions $ f $ such that support $hat f subseteq [0,infty) $
I already can prove that $T= f* mu$ where $mu $ is a tempered distribution.
Then $int_-infty^0hat f(s)e^2ipi s x ds =c Tf (x)$ where $ c $ is s constant. How to prove this ?
real-analysis
edited Aug 26 at 9:19
Bernard
111k635103
edited Aug 26 at 9:19
Bernard
111k635103
edited Aug 26 at 9:19
Bernard
111k635103
edited Aug 26 at 9:19
Bernard
111k635103
111k635103
asked Aug 26 at 7:53
user586523
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
Aug 26 at 7:59
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Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
Aug 26 at 7:59
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
Aug 26 at 7:59
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
Aug 26 at 7:59
add a comment |Â
1 Answer
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Take the fourier transform of $Tf=f*mu$
$hatTf=m(s)hatf$ where $m(s)$ is the fourier transform of $mu$
So $m(s)$ must be zero on the positive real line and non zero on the negative real line. Combining this with the dilation property implies the result you asked.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Take the fourier transform of $Tf=f*mu$
$hatTf=m(s)hatf$ where $m(s)$ is the fourier transform of $mu$
So $m(s)$ must be zero on the positive real line and non zero on the negative real line. Combining this with the dilation property implies the result you asked.
add a comment |Â
up vote
0
down vote
Take the fourier transform of $Tf=f*mu$
$hatTf=m(s)hatf$ where $m(s)$ is the fourier transform of $mu$
So $m(s)$ must be zero on the positive real line and non zero on the negative real line. Combining this with the dilation property implies the result you asked.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Take the fourier transform of $Tf=f*mu$
$hatTf=m(s)hatf$ where $m(s)$ is the fourier transform of $mu$
So $m(s)$ must be zero on the positive real line and non zero on the negative real line. Combining this with the dilation property implies the result you asked.
Take the fourier transform of $Tf=f*mu$
$hatTf=m(s)hatf$ where $m(s)$ is the fourier transform of $mu$
So $m(s)$ must be zero on the positive real line and non zero on the negative real line. Combining this with the dilation property implies the result you asked.
answered Aug 26 at 9:50
user587275
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Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
Aug 26 at 7:59