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A list of proofs of Fourier inversion formula
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The reason for this question is to make a list of the known proofs (or proof ideas) of Fourier inversion formula for functions $fin L^1(mathbbR)$ (obviously adding appropriate hypothesis to get a meaningful result) in order to better grasp the nuances of Fourier transform since, after all, different techniques (could) shed light on different features.
Here the list I know:
- Proof: via Bochner theorem (see e.g. Rudin - Fourier analysis on groups);
- Proof: via a summability kernel whose transform is known (see e.g. Rudin - Real & complex analysis);
- Proof: via Dirichlet kernel and Riemann-Lebesgue's lemma (see e.g. Zemanian - distribution theory and transform analysis);
- Proof: periodizing $f$ with period $L$, using Fourier inversion formula for $L$-periodic functions and letting $Lrightarrowinfty$ (see e.g. the answer by David Ullrich to this question);
- Proof idea: via a Riemann series and the Fourier inversion formula for periodic functions (see e.g. this question, and feel free to answer it :) );
Now it's your turn... Let the games begin :)
soft-question fourier-analysis fourier-transform
asked yesterday
Bob
1,400422
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show 2 more comments
up vote
5
down vote
favorite
The reason for this question is to make a list of the known proofs (or proof ideas) of Fourier inversion formula for functions $fin L^1(mathbbR)$ (obviously adding appropriate hypothesis to get a meaningful result) in order to better grasp the nuances of Fourier transform since, after all, different techniques (could) shed light on different features.
Here the list I know:
- Proof: via Bochner theorem (see e.g. Rudin - Fourier analysis on groups);
- Proof: via a summability kernel whose transform is known (see e.g. Rudin - Real & complex analysis);
- Proof: via Dirichlet kernel and Riemann-Lebesgue's lemma (see e.g. Zemanian - distribution theory and transform analysis);
- Proof: periodizing $f$ with period $L$, using Fourier inversion formula for $L$-periodic functions and letting $Lrightarrowinfty$ (see e.g. the answer by David Ullrich to this question);
- Proof idea: via a Riemann series and the Fourier inversion formula for periodic functions (see e.g. this question, and feel free to answer it :) );
Now it's your turn... Let the games begin :)
soft-question fourier-analysis fourier-transform
asked yesterday
Bob
1,400422
If you permit a tempered distributions, then using $$int_-infty^infty e^iomega (t-t'),domega =2pi delta(t-t')$$makes it trivial.
– Mark Viola
23 hours ago
@mark viola I will admit it, if you manage to show that the argument is not circular: for how I understand distributional fourier transform, we first have to prove fourier inversion formula for schwartz test functions to get a sensible definition
– Bob
22 hours ago
In This Answer, that $lim_ttoinftytextPVleft(fracsin(tx)xright)=pi delta(x)$ in the sense of distributions. Then, note that in the sense of distributions $$beginalign lim_Ltoinftyint_-L^L e^iomega (t-t'),domega&=lim_Ltoinftyleft(frac2sin(L(t-t'))t-t'right)\\ &=2pi delta(t-t') endalign$$
– Mark Viola
21 hours ago
@MarkViola : the linked answer isn't the inversion formula via an integration by parts argument? :) If it is, I suggest to write an answer to provide a link to that answer
– Bob
21 hours ago
@MarkViola and basically it seems the same argument in Zemanian, i.e. to use Dirichlet kernel, but you used IBP instead of Riemann-Lebesgue lemma to conclude
– Bob
21 hours ago
 |Â
show 2 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
The reason for this question is to make a list of the known proofs (or proof ideas) of Fourier inversion formula for functions $fin L^1(mathbbR)$ (obviously adding appropriate hypothesis to get a meaningful result) in order to better grasp the nuances of Fourier transform since, after all, different techniques (could) shed light on different features.
Here the list I know:
- Proof: via Bochner theorem (see e.g. Rudin - Fourier analysis on groups);
- Proof: via a summability kernel whose transform is known (see e.g. Rudin - Real & complex analysis);
- Proof: via Dirichlet kernel and Riemann-Lebesgue's lemma (see e.g. Zemanian - distribution theory and transform analysis);
- Proof: periodizing $f$ with period $L$, using Fourier inversion formula for $L$-periodic functions and letting $Lrightarrowinfty$ (see e.g. the answer by David Ullrich to this question);
- Proof idea: via a Riemann series and the Fourier inversion formula for periodic functions (see e.g. this question, and feel free to answer it :) );
Now it's your turn... Let the games begin :)
soft-question fourier-analysis fourier-transform
asked yesterday
Bob
1,400422
The reason for this question is to make a list of the known proofs (or proof ideas) of Fourier inversion formula for functions $fin L^1(mathbbR)$ (obviously adding appropriate hypothesis to get a meaningful result) in order to better grasp the nuances of Fourier transform since, after all, different techniques (could) shed light on different features.
Here the list I know:
- Proof: via Bochner theorem (see e.g. Rudin - Fourier analysis on groups);
- Proof: via a summability kernel whose transform is known (see e.g. Rudin - Real & complex analysis);
- Proof: via Dirichlet kernel and Riemann-Lebesgue's lemma (see e.g. Zemanian - distribution theory and transform analysis);
- Proof: periodizing $f$ with period $L$, using Fourier inversion formula for $L$-periodic functions and letting $Lrightarrowinfty$ (see e.g. the answer by David Ullrich to this question);
- Proof idea: via a Riemann series and the Fourier inversion formula for periodic functions (see e.g. this question, and feel free to answer it :) );
Now it's your turn... Let the games begin :)
soft-question fourier-analysis fourier-transform
asked yesterday
Bob
1,400422
edited 8 hours ago
asked yesterday
Bob
1,400422
asked yesterday
Bob
1,400422
asked yesterday
Bob
1,400422
1,400422
If you permit a tempered distributions, then using $$int_-infty^infty e^iomega (t-t'),domega =2pi delta(t-t')$$makes it trivial.
– Mark Viola
23 hours ago
@mark viola I will admit it, if you manage to show that the argument is not circular: for how I understand distributional fourier transform, we first have to prove fourier inversion formula for schwartz test functions to get a sensible definition
– Bob
22 hours ago
In This Answer, that $lim_ttoinftytextPVleft(fracsin(tx)xright)=pi delta(x)$ in the sense of distributions. Then, note that in the sense of distributions $$beginalign lim_Ltoinftyint_-L^L e^iomega (t-t'),domega&=lim_Ltoinftyleft(frac2sin(L(t-t'))t-t'right)\\ &=2pi delta(t-t') endalign$$
– Mark Viola
21 hours ago
@MarkViola : the linked answer isn't the inversion formula via an integration by parts argument? :) If it is, I suggest to write an answer to provide a link to that answer
– Bob
21 hours ago
@MarkViola and basically it seems the same argument in Zemanian, i.e. to use Dirichlet kernel, but you used IBP instead of Riemann-Lebesgue lemma to conclude
– Bob
21 hours ago
 |Â
show 2 more comments
If you permit a tempered distributions, then using $$int_-infty^infty e^iomega (t-t'),domega =2pi delta(t-t')$$makes it trivial.
– Mark Viola
23 hours ago
@mark viola I will admit it, if you manage to show that the argument is not circular: for how I understand distributional fourier transform, we first have to prove fourier inversion formula for schwartz test functions to get a sensible definition
– Bob
22 hours ago
In This Answer, that $lim_ttoinftytextPVleft(fracsin(tx)xright)=pi delta(x)$ in the sense of distributions. Then, note that in the sense of distributions $$beginalign lim_Ltoinftyint_-L^L e^iomega (t-t'),domega&=lim_Ltoinftyleft(frac2sin(L(t-t'))t-t'right)\\ &=2pi delta(t-t') endalign$$
– Mark Viola
21 hours ago
@MarkViola : the linked answer isn't the inversion formula via an integration by parts argument? :) If it is, I suggest to write an answer to provide a link to that answer
– Bob
21 hours ago
@MarkViola and basically it seems the same argument in Zemanian, i.e. to use Dirichlet kernel, but you used IBP instead of Riemann-Lebesgue lemma to conclude
– Bob
21 hours ago
If you permit a tempered distributions, then using $$int_-infty^infty e^iomega (t-t'),domega =2pi delta(t-t')$$makes it trivial.
– Mark Viola
23 hours ago
If you permit a tempered distributions, then using $$int_-infty^infty e^iomega (t-t'),domega =2pi delta(t-t')$$makes it trivial.
– Mark Viola
23 hours ago
@mark viola I will admit it, if you manage to show that the argument is not circular: for how I understand distributional fourier transform, we first have to prove fourier inversion formula for schwartz test functions to get a sensible definition
– Bob
22 hours ago
@mark viola I will admit it, if you manage to show that the argument is not circular: for how I understand distributional fourier transform, we first have to prove fourier inversion formula for schwartz test functions to get a sensible definition
– Bob
22 hours ago
In This Answer, that $lim_ttoinftytextPVleft(fracsin(tx)xright)=pi delta(x)$ in the sense of distributions. Then, note that in the sense of distributions $$beginalign lim_Ltoinftyint_-L^L e^iomega (t-t'),domega&=lim_Ltoinftyleft(frac2sin(L(t-t'))t-t'right)\\ &=2pi delta(t-t') endalign$$
– Mark Viola
21 hours ago
In This Answer, that $lim_ttoinftytextPVleft(fracsin(tx)xright)=pi delta(x)$ in the sense of distributions. Then, note that in the sense of distributions $$beginalign lim_Ltoinftyint_-L^L e^iomega (t-t'),domega&=lim_Ltoinftyleft(frac2sin(L(t-t'))t-t'right)\\ &=2pi delta(t-t') endalign$$
– Mark Viola
21 hours ago
@MarkViola : the linked answer isn't the inversion formula via an integration by parts argument? :) If it is, I suggest to write an answer to provide a link to that answer
– Bob
21 hours ago
@MarkViola : the linked answer isn't the inversion formula via an integration by parts argument? :) If it is, I suggest to write an answer to provide a link to that answer
– Bob
21 hours ago
@MarkViola and basically it seems the same argument in Zemanian, i.e. to use Dirichlet kernel, but you used IBP instead of Riemann-Lebesgue lemma to conclude
– Bob
21 hours ago
@MarkViola and basically it seems the same argument in Zemanian, i.e. to use Dirichlet kernel, but you used IBP instead of Riemann-Lebesgue lemma to conclude
– Bob
21 hours ago
 |Â
show 2 more comments
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That answer of mine that you link to is not an actual proof of the Inversion Theorem - it only works for "suitable" $f$, where "suitable" is left undefined. Here's an actual proof.
Just to establish where we're putting the $pi$'s, we define $$hat f(xi)=int f(t)e^-itxi,dt.$$
$L^1$ Inversion Theorem. If $fin L^1(Bbb R)$ and $hat fin L^1(Bbb R)$ then $f(t)=frac12piinthat f(xi)e^ixi t,dxi$ almost everywhere.
We use that periodization argument to establish the theorem under stronger hypotheses:
Partial Inversion Theorem. If $f,f',f''in L^1(Bbb R)$ then $hat fin L^1$ and $f(t)=frac12piinthat f(xi)e^itxi,dxi$.
To be explicit, we're assuming that $f$ is differentiable, $f'$ is absolutely continuous, and $f',f''in L^1$.
Note first that $(1+xi^2)hat f(xi)$ is the Fourier transform of $f-f''$, so it's bounded: $$|hat f(xi)|lefrac c1+xi^2.tag*$$
For $L>0$ define $$f_L(t)=sum_kinBbb Zf(t+kL).$$Then $f_L$ is a function with period $L$, and as such it has Fourier coefficients $$c_L,n=frac1Lint_0^Lf_L(t)e^-2pi i n t/L,dt.$$
Inserting the definition of $f_L$ and using the periodicity of the exponential shows that in fact $$c_L,n=frac1Lhat fleft(frac2pi nLright).$$So ($*$) above shows that $sum_n|c_L,n|<infty$; hence $f_L$ is equal to its Fourier series: $$f_L(t)=frac1Lsum_nhat fleft(frac2pi nLright)e^2pi i ntL.$$That's a Riemann sum for a certain integral; we establish convergence by noting that $$frac1Lsum_nhat fleft(frac2pi nLright)e^2pi i ntL=frac12piint g_L(xi),dxi,$$where $$g_L(xi)=hat fleft(frac2pi nLright)e^2pi i ntLquad(xiin[2pi n/L,2pi(n+1)/L)).$$Since $hat f$ is continuous, DCT shows that $$lim_Ltoinftyint g_L=inthat f(xi)e^ixi t,dxi;$$note that ($*$) gives the D in DCT.
So we're done if we can show that $f_Lto f$ almost everywhere as $Ltoinfty$. In fact we don't have to worry about whether/how this follows from the hypotheses: It's clear that $f_Lto f$ in $L^1_loc$ for every $fin L^1$, hence some subsequence tends to $f$ almost everywhere.
Derving IT from PIT is very simple. Say $(phi_n)$ is an approximate identity; in particular $phi_nin C^infty_c$, the support of $phi_n$ shrinks to the origin, $||phi_n||_1=1$ and $hatphi_nto1$ pointwise. Let $f_n=f*phi_n$. Then $f_n'=f*phi_n'$, so $f'in L^1$. Similarly for $f_n''$, so PIT applies to $f_n$. But $f_nto f$ almost everywhere and DCT shows that $||hat f_n-hat f||_1to0$.
answered 14 mins ago
David C. Ullrich
53.7k33481
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1 Answer
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1 Answer
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That answer of mine that you link to is not an actual proof of the Inversion Theorem - it only works for "suitable" $f$, where "suitable" is left undefined. Here's an actual proof.
Just to establish where we're putting the $pi$'s, we define $$hat f(xi)=int f(t)e^-itxi,dt.$$
$L^1$ Inversion Theorem. If $fin L^1(Bbb R)$ and $hat fin L^1(Bbb R)$ then $f(t)=frac12piinthat f(xi)e^ixi t,dxi$ almost everywhere.
We use that periodization argument to establish the theorem under stronger hypotheses:
Partial Inversion Theorem. If $f,f',f''in L^1(Bbb R)$ then $hat fin L^1$ and $f(t)=frac12piinthat f(xi)e^itxi,dxi$.
To be explicit, we're assuming that $f$ is differentiable, $f'$ is absolutely continuous, and $f',f''in L^1$.
Note first that $(1+xi^2)hat f(xi)$ is the Fourier transform of $f-f''$, so it's bounded: $$|hat f(xi)|lefrac c1+xi^2.tag*$$
For $L>0$ define $$f_L(t)=sum_kinBbb Zf(t+kL).$$Then $f_L$ is a function with period $L$, and as such it has Fourier coefficients $$c_L,n=frac1Lint_0^Lf_L(t)e^-2pi i n t/L,dt.$$
Inserting the definition of $f_L$ and using the periodicity of the exponential shows that in fact $$c_L,n=frac1Lhat fleft(frac2pi nLright).$$So ($*$) above shows that $sum_n|c_L,n|<infty$; hence $f_L$ is equal to its Fourier series: $$f_L(t)=frac1Lsum_nhat fleft(frac2pi nLright)e^2pi i ntL.$$That's a Riemann sum for a certain integral; we establish convergence by noting that $$frac1Lsum_nhat fleft(frac2pi nLright)e^2pi i ntL=frac12piint g_L(xi),dxi,$$where $$g_L(xi)=hat fleft(frac2pi nLright)e^2pi i ntLquad(xiin[2pi n/L,2pi(n+1)/L)).$$Since $hat f$ is continuous, DCT shows that $$lim_Ltoinftyint g_L=inthat f(xi)e^ixi t,dxi;$$note that ($*$) gives the D in DCT.
So we're done if we can show that $f_Lto f$ almost everywhere as $Ltoinfty$. In fact we don't have to worry about whether/how this follows from the hypotheses: It's clear that $f_Lto f$ in $L^1_loc$ for every $fin L^1$, hence some subsequence tends to $f$ almost everywhere.
Derving IT from PIT is very simple. Say $(phi_n)$ is an approximate identity; in particular $phi_nin C^infty_c$, the support of $phi_n$ shrinks to the origin, $||phi_n||_1=1$ and $hatphi_nto1$ pointwise. Let $f_n=f*phi_n$. Then $f_n'=f*phi_n'$, so $f'in L^1$. Similarly for $f_n''$, so PIT applies to $f_n$. But $f_nto f$ almost everywhere and DCT shows that $||hat f_n-hat f||_1to0$.
answered 14 mins ago
David C. Ullrich
53.7k33481
add a comment |Â
up vote
0
down vote
That answer of mine that you link to is not an actual proof of the Inversion Theorem - it only works for "suitable" $f$, where "suitable" is left undefined. Here's an actual proof.
Just to establish where we're putting the $pi$'s, we define $$hat f(xi)=int f(t)e^-itxi,dt.$$
$L^1$ Inversion Theorem. If $fin L^1(Bbb R)$ and $hat fin L^1(Bbb R)$ then $f(t)=frac12piinthat f(xi)e^ixi t,dxi$ almost everywhere.
We use that periodization argument to establish the theorem under stronger hypotheses:
Partial Inversion Theorem. If $f,f',f''in L^1(Bbb R)$ then $hat fin L^1$ and $f(t)=frac12piinthat f(xi)e^itxi,dxi$.
To be explicit, we're assuming that $f$ is differentiable, $f'$ is absolutely continuous, and $f',f''in L^1$.
Note first that $(1+xi^2)hat f(xi)$ is the Fourier transform of $f-f''$, so it's bounded: $$|hat f(xi)|lefrac c1+xi^2.tag*$$
For $L>0$ define $$f_L(t)=sum_kinBbb Zf(t+kL).$$Then $f_L$ is a function with period $L$, and as such it has Fourier coefficients $$c_L,n=frac1Lint_0^Lf_L(t)e^-2pi i n t/L,dt.$$
Inserting the definition of $f_L$ and using the periodicity of the exponential shows that in fact $$c_L,n=frac1Lhat fleft(frac2pi nLright).$$So ($*$) above shows that $sum_n|c_L,n|<infty$; hence $f_L$ is equal to its Fourier series: $$f_L(t)=frac1Lsum_nhat fleft(frac2pi nLright)e^2pi i ntL.$$That's a Riemann sum for a certain integral; we establish convergence by noting that $$frac1Lsum_nhat fleft(frac2pi nLright)e^2pi i ntL=frac12piint g_L(xi),dxi,$$where $$g_L(xi)=hat fleft(frac2pi nLright)e^2pi i ntLquad(xiin[2pi n/L,2pi(n+1)/L)).$$Since $hat f$ is continuous, DCT shows that $$lim_Ltoinftyint g_L=inthat f(xi)e^ixi t,dxi;$$note that ($*$) gives the D in DCT.
So we're done if we can show that $f_Lto f$ almost everywhere as $Ltoinfty$. In fact we don't have to worry about whether/how this follows from the hypotheses: It's clear that $f_Lto f$ in $L^1_loc$ for every $fin L^1$, hence some subsequence tends to $f$ almost everywhere.
Derving IT from PIT is very simple. Say $(phi_n)$ is an approximate identity; in particular $phi_nin C^infty_c$, the support of $phi_n$ shrinks to the origin, $||phi_n||_1=1$ and $hatphi_nto1$ pointwise. Let $f_n=f*phi_n$. Then $f_n'=f*phi_n'$, so $f'in L^1$. Similarly for $f_n''$, so PIT applies to $f_n$. But $f_nto f$ almost everywhere and DCT shows that $||hat f_n-hat f||_1to0$.
answered 14 mins ago
David C. Ullrich
53.7k33481
add a comment |Â
up vote
0
down vote
up vote
0
down vote
That answer of mine that you link to is not an actual proof of the Inversion Theorem - it only works for "suitable" $f$, where "suitable" is left undefined. Here's an actual proof.
Just to establish where we're putting the $pi$'s, we define $$hat f(xi)=int f(t)e^-itxi,dt.$$
$L^1$ Inversion Theorem. If $fin L^1(Bbb R)$ and $hat fin L^1(Bbb R)$ then $f(t)=frac12piinthat f(xi)e^ixi t,dxi$ almost everywhere.
We use that periodization argument to establish the theorem under stronger hypotheses:
Partial Inversion Theorem. If $f,f',f''in L^1(Bbb R)$ then $hat fin L^1$ and $f(t)=frac12piinthat f(xi)e^itxi,dxi$.
To be explicit, we're assuming that $f$ is differentiable, $f'$ is absolutely continuous, and $f',f''in L^1$.
Note first that $(1+xi^2)hat f(xi)$ is the Fourier transform of $f-f''$, so it's bounded: $$|hat f(xi)|lefrac c1+xi^2.tag*$$
For $L>0$ define $$f_L(t)=sum_kinBbb Zf(t+kL).$$Then $f_L$ is a function with period $L$, and as such it has Fourier coefficients $$c_L,n=frac1Lint_0^Lf_L(t)e^-2pi i n t/L,dt.$$
Inserting the definition of $f_L$ and using the periodicity of the exponential shows that in fact $$c_L,n=frac1Lhat fleft(frac2pi nLright).$$So ($*$) above shows that $sum_n|c_L,n|<infty$; hence $f_L$ is equal to its Fourier series: $$f_L(t)=frac1Lsum_nhat fleft(frac2pi nLright)e^2pi i ntL.$$That's a Riemann sum for a certain integral; we establish convergence by noting that $$frac1Lsum_nhat fleft(frac2pi nLright)e^2pi i ntL=frac12piint g_L(xi),dxi,$$where $$g_L(xi)=hat fleft(frac2pi nLright)e^2pi i ntLquad(xiin[2pi n/L,2pi(n+1)/L)).$$Since $hat f$ is continuous, DCT shows that $$lim_Ltoinftyint g_L=inthat f(xi)e^ixi t,dxi;$$note that ($*$) gives the D in DCT.
So we're done if we can show that $f_Lto f$ almost everywhere as $Ltoinfty$. In fact we don't have to worry about whether/how this follows from the hypotheses: It's clear that $f_Lto f$ in $L^1_loc$ for every $fin L^1$, hence some subsequence tends to $f$ almost everywhere.
Derving IT from PIT is very simple. Say $(phi_n)$ is an approximate identity; in particular $phi_nin C^infty_c$, the support of $phi_n$ shrinks to the origin, $||phi_n||_1=1$ and $hatphi_nto1$ pointwise. Let $f_n=f*phi_n$. Then $f_n'=f*phi_n'$, so $f'in L^1$. Similarly for $f_n''$, so PIT applies to $f_n$. But $f_nto f$ almost everywhere and DCT shows that $||hat f_n-hat f||_1to0$.
answered 14 mins ago
David C. Ullrich
53.7k33481
That answer of mine that you link to is not an actual proof of the Inversion Theorem - it only works for "suitable" $f$, where "suitable" is left undefined. Here's an actual proof.
Just to establish where we're putting the $pi$'s, we define $$hat f(xi)=int f(t)e^-itxi,dt.$$
$L^1$ Inversion Theorem. If $fin L^1(Bbb R)$ and $hat fin L^1(Bbb R)$ then $f(t)=frac12piinthat f(xi)e^ixi t,dxi$ almost everywhere.
We use that periodization argument to establish the theorem under stronger hypotheses:
Partial Inversion Theorem. If $f,f',f''in L^1(Bbb R)$ then $hat fin L^1$ and $f(t)=frac12piinthat f(xi)e^itxi,dxi$.
To be explicit, we're assuming that $f$ is differentiable, $f'$ is absolutely continuous, and $f',f''in L^1$.
Note first that $(1+xi^2)hat f(xi)$ is the Fourier transform of $f-f''$, so it's bounded: $$|hat f(xi)|lefrac c1+xi^2.tag*$$
For $L>0$ define $$f_L(t)=sum_kinBbb Zf(t+kL).$$Then $f_L$ is a function with period $L$, and as such it has Fourier coefficients $$c_L,n=frac1Lint_0^Lf_L(t)e^-2pi i n t/L,dt.$$
Inserting the definition of $f_L$ and using the periodicity of the exponential shows that in fact $$c_L,n=frac1Lhat fleft(frac2pi nLright).$$So ($*$) above shows that $sum_n|c_L,n|<infty$; hence $f_L$ is equal to its Fourier series: $$f_L(t)=frac1Lsum_nhat fleft(frac2pi nLright)e^2pi i ntL.$$That's a Riemann sum for a certain integral; we establish convergence by noting that $$frac1Lsum_nhat fleft(frac2pi nLright)e^2pi i ntL=frac12piint g_L(xi),dxi,$$where $$g_L(xi)=hat fleft(frac2pi nLright)e^2pi i ntLquad(xiin[2pi n/L,2pi(n+1)/L)).$$Since $hat f$ is continuous, DCT shows that $$lim_Ltoinftyint g_L=inthat f(xi)e^ixi t,dxi;$$note that ($*$) gives the D in DCT.
So we're done if we can show that $f_Lto f$ almost everywhere as $Ltoinfty$. In fact we don't have to worry about whether/how this follows from the hypotheses: It's clear that $f_Lto f$ in $L^1_loc$ for every $fin L^1$, hence some subsequence tends to $f$ almost everywhere.
Derving IT from PIT is very simple. Say $(phi_n)$ is an approximate identity; in particular $phi_nin C^infty_c$, the support of $phi_n$ shrinks to the origin, $||phi_n||_1=1$ and $hatphi_nto1$ pointwise. Let $f_n=f*phi_n$. Then $f_n'=f*phi_n'$, so $f'in L^1$. Similarly for $f_n''$, so PIT applies to $f_n$. But $f_nto f$ almost everywhere and DCT shows that $||hat f_n-hat f||_1to0$.
answered 14 mins ago
David C. Ullrich
53.7k33481
answered 14 mins ago
David C. Ullrich
53.7k33481
answered 14 mins ago
David C. Ullrich
53.7k33481
answered 14 mins ago
David C. Ullrich
53.7k33481
53.7k33481
add a comment |Â
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If you permit a tempered distributions, then using $$int_-infty^infty e^iomega (t-t'),domega =2pi delta(t-t')$$makes it trivial.
– Mark Viola
23 hours ago
@mark viola I will admit it, if you manage to show that the argument is not circular: for how I understand distributional fourier transform, we first have to prove fourier inversion formula for schwartz test functions to get a sensible definition
– Bob
22 hours ago
In This Answer, that $lim_ttoinftytextPVleft(fracsin(tx)xright)=pi delta(x)$ in the sense of distributions. Then, note that in the sense of distributions $$beginalign lim_Ltoinftyint_-L^L e^iomega (t-t'),domega&=lim_Ltoinftyleft(frac2sin(L(t-t'))t-t'right)\\ &=2pi delta(t-t') endalign$$
– Mark Viola
21 hours ago
@MarkViola : the linked answer isn't the inversion formula via an integration by parts argument? :) If it is, I suggest to write an answer to provide a link to that answer
– Bob
21 hours ago
@MarkViola and basically it seems the same argument in Zemanian, i.e. to use Dirichlet kernel, but you used IBP instead of Riemann-Lebesgue lemma to conclude
– Bob
21 hours ago