If $fin L^1(mathbb{R})$ and $M>0$ is it true that $mathcal{F}left( chi_{[-M,M]}mathcal{F}(f) right)in...
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If $fin L^1(mathbb{R})$ and $M>0$ is it true that $mathcal{F}left( chi_{[-M,M]}mathcal{F}(f) right)in L^1(mathbb{R})$, where $mathcal{F}$ is the Fourier transform? It is pretty clear that $fin L^p(mathbb{R})$ for every $p>1$ (e.g. via Young's convolution inequality) but I can't see if the same holds true for $p=1$.
Any suggestion?
fourier-transform
asked Jan 12 at 20:10
BobBob
1,7311725
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add a comment |
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If $fin L^1(mathbb{R})$ and $M>0$ is it true that $mathcal{F}left( chi_{[-M,M]}mathcal{F}(f) right)in L^1(mathbb{R})$, where $mathcal{F}$ is the Fourier transform? It is pretty clear that $fin L^p(mathbb{R})$ for every $p>1$ (e.g. via Young's convolution inequality) but I can't see if the same holds true for $p=1$.
Any suggestion?
fourier-transform
asked Jan 12 at 20:10
BobBob
1,7311725
$endgroup$
add a comment |
$begingroup$
If $fin L^1(mathbb{R})$ and $M>0$ is it true that $mathcal{F}left( chi_{[-M,M]}mathcal{F}(f) right)in L^1(mathbb{R})$, where $mathcal{F}$ is the Fourier transform? It is pretty clear that $fin L^p(mathbb{R})$ for every $p>1$ (e.g. via Young's convolution inequality) but I can't see if the same holds true for $p=1$.
Any suggestion?
fourier-transform
asked Jan 12 at 20:10
BobBob
1,7311725
$endgroup$
If $fin L^1(mathbb{R})$ and $M>0$ is it true that $mathcal{F}left( chi_{[-M,M]}mathcal{F}(f) right)in L^1(mathbb{R})$, where $mathcal{F}$ is the Fourier transform? It is pretty clear that $fin L^p(mathbb{R})$ for every $p>1$ (e.g. via Young's convolution inequality) but I can't see if the same holds true for $p=1$.
Any suggestion?
fourier-transform
fourier-transform
asked Jan 12 at 20:10
BobBob
1,7311725
asked Jan 12 at 20:10
BobBob
1,7311725
asked Jan 12 at 20:10
BobBob
1,7311725
asked Jan 12 at 20:10
BobBob
1,7311725
asked Jan 12 at 20:10
BobBob
1,7311725
1,7311725
add a comment |
add a comment |
1 Answer
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Fix a function $g$ from Schwarz class such that $g = 1$ on $|x| leq 1$, and define $f$ to be the inverse Fourier transform of $g$, namely $f = mathcal{F}^{-1}(g)$. Then $f$ is again from Schwarz class, in particular from $L^1(mathbb{R})$, and taking $M=1$ we get
$$
mathcal{F}(chi_{[-1,1]}mathcal{F}(f)) = mathcal{F}(chi_{[-1,1]}g) = mathcal{F}(chi_{[-1,1]})
$$
which is not $L^1$.
answered Jan 12 at 21:05
HaykHayk
2,6271214
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1 Answer
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$begingroup$
Fix a function $g$ from Schwarz class such that $g = 1$ on $|x| leq 1$, and define $f$ to be the inverse Fourier transform of $g$, namely $f = mathcal{F}^{-1}(g)$. Then $f$ is again from Schwarz class, in particular from $L^1(mathbb{R})$, and taking $M=1$ we get
$$
mathcal{F}(chi_{[-1,1]}mathcal{F}(f)) = mathcal{F}(chi_{[-1,1]}g) = mathcal{F}(chi_{[-1,1]})
$$
which is not $L^1$.
answered Jan 12 at 21:05
HaykHayk
2,6271214
$endgroup$
add a comment |
$begingroup$
Fix a function $g$ from Schwarz class such that $g = 1$ on $|x| leq 1$, and define $f$ to be the inverse Fourier transform of $g$, namely $f = mathcal{F}^{-1}(g)$. Then $f$ is again from Schwarz class, in particular from $L^1(mathbb{R})$, and taking $M=1$ we get
$$
mathcal{F}(chi_{[-1,1]}mathcal{F}(f)) = mathcal{F}(chi_{[-1,1]}g) = mathcal{F}(chi_{[-1,1]})
$$
which is not $L^1$.
answered Jan 12 at 21:05
HaykHayk
2,6271214
$endgroup$
add a comment |
$begingroup$
Fix a function $g$ from Schwarz class such that $g = 1$ on $|x| leq 1$, and define $f$ to be the inverse Fourier transform of $g$, namely $f = mathcal{F}^{-1}(g)$. Then $f$ is again from Schwarz class, in particular from $L^1(mathbb{R})$, and taking $M=1$ we get
$$
mathcal{F}(chi_{[-1,1]}mathcal{F}(f)) = mathcal{F}(chi_{[-1,1]}g) = mathcal{F}(chi_{[-1,1]})
$$
which is not $L^1$.
answered Jan 12 at 21:05
HaykHayk
2,6271214
$endgroup$
Fix a function $g$ from Schwarz class such that $g = 1$ on $|x| leq 1$, and define $f$ to be the inverse Fourier transform of $g$, namely $f = mathcal{F}^{-1}(g)$. Then $f$ is again from Schwarz class, in particular from $L^1(mathbb{R})$, and taking $M=1$ we get
$$
mathcal{F}(chi_{[-1,1]}mathcal{F}(f)) = mathcal{F}(chi_{[-1,1]}g) = mathcal{F}(chi_{[-1,1]})
$$
which is not $L^1$.
answered Jan 12 at 21:05
HaykHayk
2,6271214
answered Jan 12 at 21:05
HaykHayk
2,6271214
answered Jan 12 at 21:05
HaykHayk
2,6271214
answered Jan 12 at 21:05
HaykHayk
2,6271214
2,6271214
add a comment |
add a comment |
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