Is a smooth function with compact support in $mathbb{R}$ can be written as the convolution of two square...
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Suppose h is smooth compact supported function on $mathbb{R}$. How to show that there exist $f, g in L^2(mathbb{R}, m)$, where $m$ is the usual Lebsegue measure, such that $h = f * g $, where $*$ denotes the convolution?
real-analysis functional-analysis fourier-transform convolution
asked Dec 16 '18 at 21:15
chelseachelsea
61
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add a comment |
$begingroup$
Suppose h is smooth compact supported function on $mathbb{R}$. How to show that there exist $f, g in L^2(mathbb{R}, m)$, where $m$ is the usual Lebsegue measure, such that $h = f * g $, where $*$ denotes the convolution?
real-analysis functional-analysis fourier-transform convolution
asked Dec 16 '18 at 21:15
chelseachelsea
61
$endgroup$
add a comment |
$begingroup$
Suppose h is smooth compact supported function on $mathbb{R}$. How to show that there exist $f, g in L^2(mathbb{R}, m)$, where $m$ is the usual Lebsegue measure, such that $h = f * g $, where $*$ denotes the convolution?
real-analysis functional-analysis fourier-transform convolution
asked Dec 16 '18 at 21:15
chelseachelsea
61
$endgroup$
Suppose h is smooth compact supported function on $mathbb{R}$. How to show that there exist $f, g in L^2(mathbb{R}, m)$, where $m$ is the usual Lebsegue measure, such that $h = f * g $, where $*$ denotes the convolution?
real-analysis functional-analysis fourier-transform convolution
real-analysis functional-analysis fourier-transform convolution
asked Dec 16 '18 at 21:15
chelseachelsea
61
asked Dec 16 '18 at 21:15
chelseachelsea
61
asked Dec 16 '18 at 21:15
chelseachelsea
61
asked Dec 16 '18 at 21:15
chelseachelsea
61
asked Dec 16 '18 at 21:15
chelseachelsea
61
61
add a comment |
add a comment |
1 Answer
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Here is an outline:
Consider this in the Fourier domain: $widehat{h}=widehat{f}widehat{g}$. $widehat{h}$ is smooth and rapidly decaying based on the properties of $h$. Take $widehat{f}$ to be something positive and smooth with polynomial decay. Then $widehat{g} = widehat{h}/widehat{f}$.
answered Dec 16 '18 at 23:38
DunhamDunham
2,084614
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$begingroup$
Thank you very much. Just a simple trick on the Fourier Transform. I see it..
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– chelsea
Dec 17 '18 at 4:20
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is an outline:
Consider this in the Fourier domain: $widehat{h}=widehat{f}widehat{g}$. $widehat{h}$ is smooth and rapidly decaying based on the properties of $h$. Take $widehat{f}$ to be something positive and smooth with polynomial decay. Then $widehat{g} = widehat{h}/widehat{f}$.
answered Dec 16 '18 at 23:38
DunhamDunham
2,084614
$endgroup$
$begingroup$
Thank you very much. Just a simple trick on the Fourier Transform. I see it..
$endgroup$
– chelsea
Dec 17 '18 at 4:20
add a comment |
$begingroup$
Here is an outline:
Consider this in the Fourier domain: $widehat{h}=widehat{f}widehat{g}$. $widehat{h}$ is smooth and rapidly decaying based on the properties of $h$. Take $widehat{f}$ to be something positive and smooth with polynomial decay. Then $widehat{g} = widehat{h}/widehat{f}$.
answered Dec 16 '18 at 23:38
DunhamDunham
2,084614
$endgroup$
$begingroup$
Thank you very much. Just a simple trick on the Fourier Transform. I see it..
$endgroup$
– chelsea
Dec 17 '18 at 4:20
add a comment |
$begingroup$
Here is an outline:
Consider this in the Fourier domain: $widehat{h}=widehat{f}widehat{g}$. $widehat{h}$ is smooth and rapidly decaying based on the properties of $h$. Take $widehat{f}$ to be something positive and smooth with polynomial decay. Then $widehat{g} = widehat{h}/widehat{f}$.
answered Dec 16 '18 at 23:38
DunhamDunham
2,084614
$endgroup$
Here is an outline:
Consider this in the Fourier domain: $widehat{h}=widehat{f}widehat{g}$. $widehat{h}$ is smooth and rapidly decaying based on the properties of $h$. Take $widehat{f}$ to be something positive and smooth with polynomial decay. Then $widehat{g} = widehat{h}/widehat{f}$.
answered Dec 16 '18 at 23:38
DunhamDunham
2,084614
answered Dec 16 '18 at 23:38
DunhamDunham
2,084614
answered Dec 16 '18 at 23:38
DunhamDunham
2,084614
answered Dec 16 '18 at 23:38
DunhamDunham
2,084614
2,084614
$begingroup$
Thank you very much. Just a simple trick on the Fourier Transform. I see it..
$endgroup$
– chelsea
Dec 17 '18 at 4:20
add a comment |
$begingroup$
Thank you very much. Just a simple trick on the Fourier Transform. I see it..
$endgroup$
– chelsea
Dec 17 '18 at 4:20
$begingroup$
Thank you very much. Just a simple trick on the Fourier Transform. I see it..
$endgroup$
– chelsea
Dec 17 '18 at 4:20
$begingroup$
Thank you very much. Just a simple trick on the Fourier Transform. I see it..
$endgroup$
– chelsea
Dec 17 '18 at 4:20
add a comment |
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