Fourier transform of $f=e^inx$ on $L^2[-pi,pi]$ for $1leq pleq2$
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If $fin L^2[-pi,pi]$, let $hat f$ be the Fourier transform of $f$:
$$hat f=frac12pi int_-pi^pi f(x) e^-ixi xdx , , quad xiinmathbb R, . $$
If $f=e^inx$, then
$$hat f=frac12pi int_-pi^pi e^i(n-xi)xdx=frace^i(n-xi)pi-e^-i(n-xi)pi2pi i (n-xi)=fracsin [(n-xi)pi] pi (n-xi), . quad quad (1)$$
Moreover, $fin L^p[-pi,pi]$ for $1leq pleq 2$. Does result (1) also apply to the $L^p$ case ($1leq pleq 2$)?
fourier-analysis fourier-transform
asked Sep 7 at 10:45
Mark
3,51651846
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up vote
3
down vote
favorite
If $fin L^2[-pi,pi]$, let $hat f$ be the Fourier transform of $f$:
$$hat f=frac12pi int_-pi^pi f(x) e^-ixi xdx , , quad xiinmathbb R, . $$
If $f=e^inx$, then
$$hat f=frac12pi int_-pi^pi e^i(n-xi)xdx=frace^i(n-xi)pi-e^-i(n-xi)pi2pi i (n-xi)=fracsin [(n-xi)pi] pi (n-xi), . quad quad (1)$$
Moreover, $fin L^p[-pi,pi]$ for $1leq pleq 2$. Does result (1) also apply to the $L^p$ case ($1leq pleq 2$)?
fourier-analysis fourier-transform
asked Sep 7 at 10:45
Mark
3,51651846
2
Well, not sure I undserstand what you are asking...
– nicomezi
Sep 7 at 11:06
I mean, (1) is the Fourier transform of $e^int$ on $L^2$. Is it the Fourier transform of $e^int$ also on $L^p$ ($1leq pleq 2$)?
– Mark
Sep 7 at 11:22
If you take $xiinmathbb R$ then you are probably referring to the Fourier transform on the line $mathbb R$. In that case the integral should be $int_-infty^infty$. If, instead, you want the integral $int_-pi^pi$, then you are talking about Fourier series, in which case $xiinmathbb Z$. Notice that $xin [-pi, pi]mapsto e^ixxi$ is $2pi$ periodic if and only if $xiinmathbb Z$.
– Giuseppe Negro
Sep 7 at 11:52
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If $fin L^2[-pi,pi]$, let $hat f$ be the Fourier transform of $f$:
$$hat f=frac12pi int_-pi^pi f(x) e^-ixi xdx , , quad xiinmathbb R, . $$
If $f=e^inx$, then
$$hat f=frac12pi int_-pi^pi e^i(n-xi)xdx=frace^i(n-xi)pi-e^-i(n-xi)pi2pi i (n-xi)=fracsin [(n-xi)pi] pi (n-xi), . quad quad (1)$$
Moreover, $fin L^p[-pi,pi]$ for $1leq pleq 2$. Does result (1) also apply to the $L^p$ case ($1leq pleq 2$)?
fourier-analysis fourier-transform
asked Sep 7 at 10:45
Mark
3,51651846
If $fin L^2[-pi,pi]$, let $hat f$ be the Fourier transform of $f$:
$$hat f=frac12pi int_-pi^pi f(x) e^-ixi xdx , , quad xiinmathbb R, . $$
If $f=e^inx$, then
$$hat f=frac12pi int_-pi^pi e^i(n-xi)xdx=frace^i(n-xi)pi-e^-i(n-xi)pi2pi i (n-xi)=fracsin [(n-xi)pi] pi (n-xi), . quad quad (1)$$
Moreover, $fin L^p[-pi,pi]$ for $1leq pleq 2$. Does result (1) also apply to the $L^p$ case ($1leq pleq 2$)?
fourier-analysis fourier-transform
fourier-analysis fourier-transform
asked Sep 7 at 10:45
Mark
3,51651846
asked Sep 7 at 10:45
Mark
3,51651846
asked Sep 7 at 10:45
Mark
3,51651846
asked Sep 7 at 10:45
Mark
3,51651846
asked Sep 7 at 10:45
Mark
3,51651846
3,51651846
2
Well, not sure I undserstand what you are asking...
– nicomezi
Sep 7 at 11:06
I mean, (1) is the Fourier transform of $e^int$ on $L^2$. Is it the Fourier transform of $e^int$ also on $L^p$ ($1leq pleq 2$)?
– Mark
Sep 7 at 11:22
If you take $xiinmathbb R$ then you are probably referring to the Fourier transform on the line $mathbb R$. In that case the integral should be $int_-infty^infty$. If, instead, you want the integral $int_-pi^pi$, then you are talking about Fourier series, in which case $xiinmathbb Z$. Notice that $xin [-pi, pi]mapsto e^ixxi$ is $2pi$ periodic if and only if $xiinmathbb Z$.
– Giuseppe Negro
Sep 7 at 11:52
add a comment |Â
2
Well, not sure I undserstand what you are asking...
– nicomezi
Sep 7 at 11:06
I mean, (1) is the Fourier transform of $e^int$ on $L^2$. Is it the Fourier transform of $e^int$ also on $L^p$ ($1leq pleq 2$)?
– Mark
Sep 7 at 11:22
If you take $xiinmathbb R$ then you are probably referring to the Fourier transform on the line $mathbb R$. In that case the integral should be $int_-infty^infty$. If, instead, you want the integral $int_-pi^pi$, then you are talking about Fourier series, in which case $xiinmathbb Z$. Notice that $xin [-pi, pi]mapsto e^ixxi$ is $2pi$ periodic if and only if $xiinmathbb Z$.
– Giuseppe Negro
Sep 7 at 11:52
2
2
Well, not sure I undserstand what you are asking...
– nicomezi
Sep 7 at 11:06
Well, not sure I undserstand what you are asking...
– nicomezi
Sep 7 at 11:06
I mean, (1) is the Fourier transform of $e^int$ on $L^2$. Is it the Fourier transform of $e^int$ also on $L^p$ ($1leq pleq 2$)?
– Mark
Sep 7 at 11:22
I mean, (1) is the Fourier transform of $e^int$ on $L^2$. Is it the Fourier transform of $e^int$ also on $L^p$ ($1leq pleq 2$)?
– Mark
Sep 7 at 11:22
If you take $xiinmathbb R$ then you are probably referring to the Fourier transform on the line $mathbb R$. In that case the integral should be $int_-infty^infty$. If, instead, you want the integral $int_-pi^pi$, then you are talking about Fourier series, in which case $xiinmathbb Z$. Notice that $xin [-pi, pi]mapsto e^ixxi$ is $2pi$ periodic if and only if $xiinmathbb Z$.
– Giuseppe Negro
Sep 7 at 11:52
If you take $xiinmathbb R$ then you are probably referring to the Fourier transform on the line $mathbb R$. In that case the integral should be $int_-infty^infty$. If, instead, you want the integral $int_-pi^pi$, then you are talking about Fourier series, in which case $xiinmathbb Z$. Notice that $xin [-pi, pi]mapsto e^ixxi$ is $2pi$ periodic if and only if $xiinmathbb Z$.
– Giuseppe Negro
Sep 7 at 11:52
add a comment |Â
1 Answer
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2
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Since in the compact setting the $L^p$ spaces are nested
$$ L^infty subset L^q subset L^2 subset L^p subset L^1 ;;;;(textfor; 1 le p le 2 le q le infty)$$
the Fourier-Transform of any $L^p$-function corresponds to its Fourier-Transform of it interpreted as an $L^1$-function. So the answer to your question is yes.
More generally, "the Fourier-Transforms" of a function that lives in at least two $L^p$-spaces coincide, since they are both usually defined as the limit of Fourier-Transforms of functions also present in $L^1$.
answered Sep 7 at 11:48
Joseph Adams
1167
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Since in the compact setting the $L^p$ spaces are nested
$$ L^infty subset L^q subset L^2 subset L^p subset L^1 ;;;;(textfor; 1 le p le 2 le q le infty)$$
the Fourier-Transform of any $L^p$-function corresponds to its Fourier-Transform of it interpreted as an $L^1$-function. So the answer to your question is yes.
More generally, "the Fourier-Transforms" of a function that lives in at least two $L^p$-spaces coincide, since they are both usually defined as the limit of Fourier-Transforms of functions also present in $L^1$.
answered Sep 7 at 11:48
Joseph Adams
1167
add a comment |Â
up vote
2
down vote
accepted
Since in the compact setting the $L^p$ spaces are nested
$$ L^infty subset L^q subset L^2 subset L^p subset L^1 ;;;;(textfor; 1 le p le 2 le q le infty)$$
the Fourier-Transform of any $L^p$-function corresponds to its Fourier-Transform of it interpreted as an $L^1$-function. So the answer to your question is yes.
More generally, "the Fourier-Transforms" of a function that lives in at least two $L^p$-spaces coincide, since they are both usually defined as the limit of Fourier-Transforms of functions also present in $L^1$.
answered Sep 7 at 11:48
Joseph Adams
1167
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since in the compact setting the $L^p$ spaces are nested
$$ L^infty subset L^q subset L^2 subset L^p subset L^1 ;;;;(textfor; 1 le p le 2 le q le infty)$$
the Fourier-Transform of any $L^p$-function corresponds to its Fourier-Transform of it interpreted as an $L^1$-function. So the answer to your question is yes.
More generally, "the Fourier-Transforms" of a function that lives in at least two $L^p$-spaces coincide, since they are both usually defined as the limit of Fourier-Transforms of functions also present in $L^1$.
answered Sep 7 at 11:48
Joseph Adams
1167
Since in the compact setting the $L^p$ spaces are nested
$$ L^infty subset L^q subset L^2 subset L^p subset L^1 ;;;;(textfor; 1 le p le 2 le q le infty)$$
the Fourier-Transform of any $L^p$-function corresponds to its Fourier-Transform of it interpreted as an $L^1$-function. So the answer to your question is yes.
More generally, "the Fourier-Transforms" of a function that lives in at least two $L^p$-spaces coincide, since they are both usually defined as the limit of Fourier-Transforms of functions also present in $L^1$.
answered Sep 7 at 11:48
Joseph Adams
1167
answered Sep 7 at 11:48
Joseph Adams
1167
answered Sep 7 at 11:48
Joseph Adams
1167
answered Sep 7 at 11:48
Joseph Adams
1167
1167
add a comment |Â
add a comment |Â
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2
Well, not sure I undserstand what you are asking...
– nicomezi
Sep 7 at 11:06
I mean, (1) is the Fourier transform of $e^int$ on $L^2$. Is it the Fourier transform of $e^int$ also on $L^p$ ($1leq pleq 2$)?
– Mark
Sep 7 at 11:22
If you take $xiinmathbb R$ then you are probably referring to the Fourier transform on the line $mathbb R$. In that case the integral should be $int_-infty^infty$. If, instead, you want the integral $int_-pi^pi$, then you are talking about Fourier series, in which case $xiinmathbb Z$. Notice that $xin [-pi, pi]mapsto e^ixxi$ is $2pi$ periodic if and only if $xiinmathbb Z$.
– Giuseppe Negro
Sep 7 at 11:52