Rigrous justification of this eqality.

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Could anyone help me justify this equality?



$$frac1(2pi)^d/2int_mathbbT^d sum_xi in mathbbZ^d mathcalF^-1u(x+2pixi)e^ -ilangle x+2pi xi, y rangle dx = frac1(2pi)^d/2 int_mathbbR^d mathcalF^-1u(x)e^-i langle x,y rangledx$$



where $ langle . , . rangle$ is the dot product and $mathcalF^-1$ is the inverse Fourier transform from $ mathcalS(mathbbR^d) $ to $ mathcalS(mathbbR^d) $. More fundamentally I am not familiar with integration over the torus. In particular, I am not sure what is the appropriate measure in integrate over, in the sense that that if $ f: mathbbT^d rightarrow mathbbR$ is defined on the d - torus, then whether we have
$$ int_mathbbT^d f(t) dt = int_[0,1)^df(x)dx,$$
if so, where can I find a precise argument justifying this equality?



Thanks!




measure-theory fourier-analysis




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  • What is the link between $d$ and $n$ ?
    – Delta-u
    Aug 3 at 10:08










  • This is a typo. The dimensions are meant to be the same and the Fourier transform is just on $ mathbbR^d $. I apologise.
    – Meagain
    Aug 3 at 10:25














up vote
0
down vote

favorite












Could anyone help me justify this equality?



$$frac1(2pi)^d/2int_mathbbT^d sum_xi in mathbbZ^d mathcalF^-1u(x+2pixi)e^ -ilangle x+2pi xi, y rangle dx = frac1(2pi)^d/2 int_mathbbR^d mathcalF^-1u(x)e^-i langle x,y rangledx$$



where $ langle . , . rangle$ is the dot product and $mathcalF^-1$ is the inverse Fourier transform from $ mathcalS(mathbbR^d) $ to $ mathcalS(mathbbR^d) $. More fundamentally I am not familiar with integration over the torus. In particular, I am not sure what is the appropriate measure in integrate over, in the sense that that if $ f: mathbbT^d rightarrow mathbbR$ is defined on the d - torus, then whether we have
$$ int_mathbbT^d f(t) dt = int_[0,1)^df(x)dx,$$
if so, where can I find a precise argument justifying this equality?



Thanks!




measure-theory fourier-analysis




share|cite|improve this question





















  • What is the link between $d$ and $n$ ?
    – Delta-u
    Aug 3 at 10:08










  • This is a typo. The dimensions are meant to be the same and the Fourier transform is just on $ mathbbR^d $. I apologise.
    – Meagain
    Aug 3 at 10:25












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Could anyone help me justify this equality?



$$frac1(2pi)^d/2int_mathbbT^d sum_xi in mathbbZ^d mathcalF^-1u(x+2pixi)e^ -ilangle x+2pi xi, y rangle dx = frac1(2pi)^d/2 int_mathbbR^d mathcalF^-1u(x)e^-i langle x,y rangledx$$



where $ langle . , . rangle$ is the dot product and $mathcalF^-1$ is the inverse Fourier transform from $ mathcalS(mathbbR^d) $ to $ mathcalS(mathbbR^d) $. More fundamentally I am not familiar with integration over the torus. In particular, I am not sure what is the appropriate measure in integrate over, in the sense that that if $ f: mathbbT^d rightarrow mathbbR$ is defined on the d - torus, then whether we have
$$ int_mathbbT^d f(t) dt = int_[0,1)^df(x)dx,$$
if so, where can I find a precise argument justifying this equality?



Thanks!




measure-theory fourier-analysis




share|cite|improve this question













Could anyone help me justify this equality?



$$frac1(2pi)^d/2int_mathbbT^d sum_xi in mathbbZ^d mathcalF^-1u(x+2pixi)e^ -ilangle x+2pi xi, y rangle dx = frac1(2pi)^d/2 int_mathbbR^d mathcalF^-1u(x)e^-i langle x,y rangledx$$



where $ langle . , . rangle$ is the dot product and $mathcalF^-1$ is the inverse Fourier transform from $ mathcalS(mathbbR^d) $ to $ mathcalS(mathbbR^d) $. More fundamentally I am not familiar with integration over the torus. In particular, I am not sure what is the appropriate measure in integrate over, in the sense that that if $ f: mathbbT^d rightarrow mathbbR$ is defined on the d - torus, then whether we have
$$ int_mathbbT^d f(t) dt = int_[0,1)^df(x)dx,$$
if so, where can I find a precise argument justifying this equality?



Thanks!





measure-theory fourier-analysis





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 3 at 10:26
























asked Aug 3 at 6:29









Meagain

1279




1279











  • What is the link between $d$ and $n$ ?
    – Delta-u
    Aug 3 at 10:08










  • This is a typo. The dimensions are meant to be the same and the Fourier transform is just on $ mathbbR^d $. I apologise.
    – Meagain
    Aug 3 at 10:25
















  • What is the link between $d$ and $n$ ?
    – Delta-u
    Aug 3 at 10:08










  • This is a typo. The dimensions are meant to be the same and the Fourier transform is just on $ mathbbR^d $. I apologise.
    – Meagain
    Aug 3 at 10:25















What is the link between $d$ and $n$ ?
– Delta-u
Aug 3 at 10:08




What is the link between $d$ and $n$ ?
– Delta-u
Aug 3 at 10:08












This is a typo. The dimensions are meant to be the same and the Fourier transform is just on $ mathbbR^d $. I apologise.
– Meagain
Aug 3 at 10:25




This is a typo. The dimensions are meant to be the same and the Fourier transform is just on $ mathbbR^d $. I apologise.
– Meagain
Aug 3 at 10:25















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