Does the existence of all the Fourier transforms imply $f in L^1$?
It is trivial that for $f in L^1$ and $y in mathbb{R}$ the integral
$$int_{mathbb{R}} f(x)e^{-2pi i xy} dx$$
converges. I was just wondering whether the converse is true:
Given a (let's say continuous) function $f:mathbb{R} to mathbb{C}$ such that the integral
$$int_{mathbb{R}} f(x)e^{-2pi i xy} dx$$
is convergent for all $y$, does it imply $f in L^1$?
I'm really not so sure what I should believe here since I neither see a counterexample nor a way to prove this.
This question must have been considered before, but I couldn't find any references online.
Edit: As mentioned in the comments, the integrals can of course not be absolutely convergent, otherwise the result would be trivial. By saying that $int_{mathbb{R}} dots$ exists I really mean that $lim_{T to infty} int_0^T dots$ and $lim_{T to infty} int_{-T}^0 dots$ both exist.
integration fourier-analysis fourier-transform
asked Nov 28 '18 at 19:36
Tintarn
2,668514
add a comment |
It is trivial that for $f in L^1$ and $y in mathbb{R}$ the integral
$$int_{mathbb{R}} f(x)e^{-2pi i xy} dx$$
converges. I was just wondering whether the converse is true:
Given a (let's say continuous) function $f:mathbb{R} to mathbb{C}$ such that the integral
$$int_{mathbb{R}} f(x)e^{-2pi i xy} dx$$
is convergent for all $y$, does it imply $f in L^1$?
I'm really not so sure what I should believe here since I neither see a counterexample nor a way to prove this.
This question must have been considered before, but I couldn't find any references online.
Edit: As mentioned in the comments, the integrals can of course not be absolutely convergent, otherwise the result would be trivial. By saying that $int_{mathbb{R}} dots$ exists I really mean that $lim_{T to infty} int_0^T dots$ and $lim_{T to infty} int_{-T}^0 dots$ both exist.
integration fourier-analysis fourier-transform
asked Nov 28 '18 at 19:36
Tintarn
2,668514
1
I don't think tat the integral is even defined (in the Lebesgue sense) if $f notin L^1$, since $|f(x)e^{-2pi i xy}| = |f(x)|$.
– MisterRiemann
Nov 28 '18 at 19:41
Okay I should have been more precise. Of course if the problem is to be interesting at all, these integrals cannot be absolutely convergent i.e. we really need to read them as $lim_{T to infty} int_0^T+lim_{T to infty} int_{-T}^0$.
– Tintarn
Nov 28 '18 at 19:56
add a comment |
It is trivial that for $f in L^1$ and $y in mathbb{R}$ the integral
$$int_{mathbb{R}} f(x)e^{-2pi i xy} dx$$
converges. I was just wondering whether the converse is true:
Given a (let's say continuous) function $f:mathbb{R} to mathbb{C}$ such that the integral
$$int_{mathbb{R}} f(x)e^{-2pi i xy} dx$$
is convergent for all $y$, does it imply $f in L^1$?
I'm really not so sure what I should believe here since I neither see a counterexample nor a way to prove this.
This question must have been considered before, but I couldn't find any references online.
Edit: As mentioned in the comments, the integrals can of course not be absolutely convergent, otherwise the result would be trivial. By saying that $int_{mathbb{R}} dots$ exists I really mean that $lim_{T to infty} int_0^T dots$ and $lim_{T to infty} int_{-T}^0 dots$ both exist.
integration fourier-analysis fourier-transform
asked Nov 28 '18 at 19:36
Tintarn
2,668514
It is trivial that for $f in L^1$ and $y in mathbb{R}$ the integral
$$int_{mathbb{R}} f(x)e^{-2pi i xy} dx$$
converges. I was just wondering whether the converse is true:
Given a (let's say continuous) function $f:mathbb{R} to mathbb{C}$ such that the integral
$$int_{mathbb{R}} f(x)e^{-2pi i xy} dx$$
is convergent for all $y$, does it imply $f in L^1$?
I'm really not so sure what I should believe here since I neither see a counterexample nor a way to prove this.
This question must have been considered before, but I couldn't find any references online.
Edit: As mentioned in the comments, the integrals can of course not be absolutely convergent, otherwise the result would be trivial. By saying that $int_{mathbb{R}} dots$ exists I really mean that $lim_{T to infty} int_0^T dots$ and $lim_{T to infty} int_{-T}^0 dots$ both exist.
integration fourier-analysis fourier-transform
integration fourier-analysis fourier-transform
asked Nov 28 '18 at 19:36
Tintarn
2,668514
asked Nov 28 '18 at 19:36
Tintarn
2,668514
edited Nov 28 '18 at 19:57
asked Nov 28 '18 at 19:36
Tintarn
2,668514
asked Nov 28 '18 at 19:36
Tintarn
2,668514
asked Nov 28 '18 at 19:36
Tintarn
2,668514
2,668514
1
I don't think tat the integral is even defined (in the Lebesgue sense) if $f notin L^1$, since $|f(x)e^{-2pi i xy}| = |f(x)|$.
– MisterRiemann
Nov 28 '18 at 19:41
Okay I should have been more precise. Of course if the problem is to be interesting at all, these integrals cannot be absolutely convergent i.e. we really need to read them as $lim_{T to infty} int_0^T+lim_{T to infty} int_{-T}^0$.
– Tintarn
Nov 28 '18 at 19:56
add a comment |
1
I don't think tat the integral is even defined (in the Lebesgue sense) if $f notin L^1$, since $|f(x)e^{-2pi i xy}| = |f(x)|$.
– MisterRiemann
Nov 28 '18 at 19:41
Okay I should have been more precise. Of course if the problem is to be interesting at all, these integrals cannot be absolutely convergent i.e. we really need to read them as $lim_{T to infty} int_0^T+lim_{T to infty} int_{-T}^0$.
– Tintarn
Nov 28 '18 at 19:56
1
1
I don't think tat the integral is even defined (in the Lebesgue sense) if $f notin L^1$, since $|f(x)e^{-2pi i xy}| = |f(x)|$.
– MisterRiemann
Nov 28 '18 at 19:41
I don't think tat the integral is even defined (in the Lebesgue sense) if $f notin L^1$, since $|f(x)e^{-2pi i xy}| = |f(x)|$.
– MisterRiemann
Nov 28 '18 at 19:41
Okay I should have been more precise. Of course if the problem is to be interesting at all, these integrals cannot be absolutely convergent i.e. we really need to read them as $lim_{T to infty} int_0^T+lim_{T to infty} int_{-T}^0$.
– Tintarn
Nov 28 '18 at 19:56
Okay I should have been more precise. Of course if the problem is to be interesting at all, these integrals cannot be absolutely convergent i.e. we really need to read them as $lim_{T to infty} int_0^T+lim_{T to infty} int_{-T}^0$.
– Tintarn
Nov 28 '18 at 19:56
add a comment |
1 Answer
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The Fourier transform can exist even if we have conditional convergence of $int_{-infty}^{infty} f(x) , dx$.
If $f in L^1$ then by the Riemann-Lebesgue lemma, the Fourier transform
$$hat{f}(y) = int_{-infty}^infty f(x) e ^{iyx} , dy$$
must decay to $0$ as $|y| to infty$. Any function with a convergent Fourier transform without this property is not in $L^1$.
An example is
$$f(x) = begin{cases}sin(x^2) & x geqslant 0 \ 0 & x < 0 end{cases}$$
where the Fourier (cosine) transform is
$$hat{f}(y) = int_0^infty sin (x^2), cos(yx) , dx = sqrt{frac{pi}{8}}left[cos frac{y^2}{4} - sin frac{y^2}{4}right]$$
answered Nov 28 '18 at 20:37
RRL
49.2k42573
Thank you for the nice answer! A natural followup question would be: Is it necessarily true that $widehat{f}(y)$ is bounded? (This would trivially be true if $f in L^1$ but remains true in your example.)
– Tintarn
Nov 28 '18 at 20:43
@Tintarn: You're welcome. Good question -- maybe worth posting -- and I will think about it.
– RRL
Nov 28 '18 at 20:45
add a comment |
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1 Answer
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active
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votes
The Fourier transform can exist even if we have conditional convergence of $int_{-infty}^{infty} f(x) , dx$.
If $f in L^1$ then by the Riemann-Lebesgue lemma, the Fourier transform
$$hat{f}(y) = int_{-infty}^infty f(x) e ^{iyx} , dy$$
must decay to $0$ as $|y| to infty$. Any function with a convergent Fourier transform without this property is not in $L^1$.
An example is
$$f(x) = begin{cases}sin(x^2) & x geqslant 0 \ 0 & x < 0 end{cases}$$
where the Fourier (cosine) transform is
$$hat{f}(y) = int_0^infty sin (x^2), cos(yx) , dx = sqrt{frac{pi}{8}}left[cos frac{y^2}{4} - sin frac{y^2}{4}right]$$
answered Nov 28 '18 at 20:37
RRL
49.2k42573
Thank you for the nice answer! A natural followup question would be: Is it necessarily true that $widehat{f}(y)$ is bounded? (This would trivially be true if $f in L^1$ but remains true in your example.)
– Tintarn
Nov 28 '18 at 20:43
@Tintarn: You're welcome. Good question -- maybe worth posting -- and I will think about it.
– RRL
Nov 28 '18 at 20:45
add a comment |
The Fourier transform can exist even if we have conditional convergence of $int_{-infty}^{infty} f(x) , dx$.
If $f in L^1$ then by the Riemann-Lebesgue lemma, the Fourier transform
$$hat{f}(y) = int_{-infty}^infty f(x) e ^{iyx} , dy$$
must decay to $0$ as $|y| to infty$. Any function with a convergent Fourier transform without this property is not in $L^1$.
An example is
$$f(x) = begin{cases}sin(x^2) & x geqslant 0 \ 0 & x < 0 end{cases}$$
where the Fourier (cosine) transform is
$$hat{f}(y) = int_0^infty sin (x^2), cos(yx) , dx = sqrt{frac{pi}{8}}left[cos frac{y^2}{4} - sin frac{y^2}{4}right]$$
answered Nov 28 '18 at 20:37
RRL
49.2k42573
Thank you for the nice answer! A natural followup question would be: Is it necessarily true that $widehat{f}(y)$ is bounded? (This would trivially be true if $f in L^1$ but remains true in your example.)
– Tintarn
Nov 28 '18 at 20:43
@Tintarn: You're welcome. Good question -- maybe worth posting -- and I will think about it.
– RRL
Nov 28 '18 at 20:45
add a comment |
The Fourier transform can exist even if we have conditional convergence of $int_{-infty}^{infty} f(x) , dx$.
If $f in L^1$ then by the Riemann-Lebesgue lemma, the Fourier transform
$$hat{f}(y) = int_{-infty}^infty f(x) e ^{iyx} , dy$$
must decay to $0$ as $|y| to infty$. Any function with a convergent Fourier transform without this property is not in $L^1$.
An example is
$$f(x) = begin{cases}sin(x^2) & x geqslant 0 \ 0 & x < 0 end{cases}$$
where the Fourier (cosine) transform is
$$hat{f}(y) = int_0^infty sin (x^2), cos(yx) , dx = sqrt{frac{pi}{8}}left[cos frac{y^2}{4} - sin frac{y^2}{4}right]$$
answered Nov 28 '18 at 20:37
RRL
49.2k42573
The Fourier transform can exist even if we have conditional convergence of $int_{-infty}^{infty} f(x) , dx$.
If $f in L^1$ then by the Riemann-Lebesgue lemma, the Fourier transform
$$hat{f}(y) = int_{-infty}^infty f(x) e ^{iyx} , dy$$
must decay to $0$ as $|y| to infty$. Any function with a convergent Fourier transform without this property is not in $L^1$.
An example is
$$f(x) = begin{cases}sin(x^2) & x geqslant 0 \ 0 & x < 0 end{cases}$$
where the Fourier (cosine) transform is
$$hat{f}(y) = int_0^infty sin (x^2), cos(yx) , dx = sqrt{frac{pi}{8}}left[cos frac{y^2}{4} - sin frac{y^2}{4}right]$$
answered Nov 28 '18 at 20:37
RRL
49.2k42573
answered Nov 28 '18 at 20:37
RRL
49.2k42573
answered Nov 28 '18 at 20:37
RRL
49.2k42573
answered Nov 28 '18 at 20:37
RRL
49.2k42573
49.2k42573
Thank you for the nice answer! A natural followup question would be: Is it necessarily true that $widehat{f}(y)$ is bounded? (This would trivially be true if $f in L^1$ but remains true in your example.)
– Tintarn
Nov 28 '18 at 20:43
@Tintarn: You're welcome. Good question -- maybe worth posting -- and I will think about it.
– RRL
Nov 28 '18 at 20:45
add a comment |
Thank you for the nice answer! A natural followup question would be: Is it necessarily true that $widehat{f}(y)$ is bounded? (This would trivially be true if $f in L^1$ but remains true in your example.)
– Tintarn
Nov 28 '18 at 20:43
@Tintarn: You're welcome. Good question -- maybe worth posting -- and I will think about it.
– RRL
Nov 28 '18 at 20:45
Thank you for the nice answer! A natural followup question would be: Is it necessarily true that $widehat{f}(y)$ is bounded? (This would trivially be true if $f in L^1$ but remains true in your example.)
– Tintarn
Nov 28 '18 at 20:43
Thank you for the nice answer! A natural followup question would be: Is it necessarily true that $widehat{f}(y)$ is bounded? (This would trivially be true if $f in L^1$ but remains true in your example.)
– Tintarn
Nov 28 '18 at 20:43
@Tintarn: You're welcome. Good question -- maybe worth posting -- and I will think about it.
– RRL
Nov 28 '18 at 20:45
@Tintarn: You're welcome. Good question -- maybe worth posting -- and I will think about it.
– RRL
Nov 28 '18 at 20:45
add a comment |
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I don't think tat the integral is even defined (in the Lebesgue sense) if $f notin L^1$, since $|f(x)e^{-2pi i xy}| = |f(x)|$.
– MisterRiemann
Nov 28 '18 at 19:41
Okay I should have been more precise. Of course if the problem is to be interesting at all, these integrals cannot be absolutely convergent i.e. we really need to read them as $lim_{T to infty} int_0^T+lim_{T to infty} int_{-T}^0$.
– Tintarn
Nov 28 '18 at 19:56