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What is the minimal requirement for a Fourier transformation?


Is this a correct way of thinking of Fourier transformsRepresenting a real sampled signal with N samples as a complex sampled signal with N/2 samplesA vector space with countable and uncountable basis at the same timeWhat's amiss in this Fourier convergence analysis?Magnitude and Angle of Discrete Fourier TransformWays to generate triangle wave function.Fourier series of piecewise-defined function and convergenceDirac delta distribution and sin(x) - what can be a test function?Decomposition of periodic spikes into frequenciesDetermine the fourier cosine series of $f(x)$













1












$begingroup$


I have a lot of comprehension questions that I can't really figure out by googling:



  1. The difference between a Fourier transformation and a Fourier series. Am I correct in thinking that the difference: transforms are for continous cases and series are for discrete cases ?


  2. What functions can be Fourier transformed? Which not? I read/learned once that piecewise continuity is enough for a function to be transformed as long as it can be extended to 2 - pi periodic continuity (so the pieces can be "stringed together to a periodic wave")


Is this correct?










share|cite|improve this question









$endgroup$











  • $begingroup$
    1. is more or less correct but 2. not so much. To be fourier transformed a function has to have some integrability properties because the fourier transform is defined as an integral over all of $mathbbR$.
    $endgroup$
    – Tony S.F.
    Mar 21 at 13:08










  • $begingroup$
    thanks for your prompt response. i'm not looking for complicated counter examples and such, just the most generally applicable case , as in, what is the "weakest" function, that is still transformable
    $endgroup$
    – shakshoka1234
    Mar 21 at 13:22










  • $begingroup$
    We can readily talk about the fourier transform of $L^1$ functions because $|e^2pi i x xi|leq 1$ and we can use a density argument to extend to $L^2$. Beyond that we can go to distributions and say something about Schwartz functions but I don't really know what is the "weakest' function for this.
    $endgroup$
    – Tony S.F.
    Mar 21 at 13:31










  • $begingroup$
    So all L1 and L2 functions are fourier transformable?
    $endgroup$
    – shakshoka1234
    Mar 21 at 13:39










  • $begingroup$
    Yes. The proof is not so bad, you first develop the theory in $L^1$, then look at functions in $L^1cap L^2$, and use a density argument to extend this to $L^2$ itself.
    $endgroup$
    – Tony S.F.
    Mar 21 at 13:55















1












$begingroup$


I have a lot of comprehension questions that I can't really figure out by googling:



  1. The difference between a Fourier transformation and a Fourier series. Am I correct in thinking that the difference: transforms are for continous cases and series are for discrete cases ?


  2. What functions can be Fourier transformed? Which not? I read/learned once that piecewise continuity is enough for a function to be transformed as long as it can be extended to 2 - pi periodic continuity (so the pieces can be "stringed together to a periodic wave")


Is this correct?










share|cite|improve this question









$endgroup$











  • $begingroup$
    1. is more or less correct but 2. not so much. To be fourier transformed a function has to have some integrability properties because the fourier transform is defined as an integral over all of $mathbbR$.
    $endgroup$
    – Tony S.F.
    Mar 21 at 13:08










  • $begingroup$
    thanks for your prompt response. i'm not looking for complicated counter examples and such, just the most generally applicable case , as in, what is the "weakest" function, that is still transformable
    $endgroup$
    – shakshoka1234
    Mar 21 at 13:22










  • $begingroup$
    We can readily talk about the fourier transform of $L^1$ functions because $|e^2pi i x xi|leq 1$ and we can use a density argument to extend to $L^2$. Beyond that we can go to distributions and say something about Schwartz functions but I don't really know what is the "weakest' function for this.
    $endgroup$
    – Tony S.F.
    Mar 21 at 13:31










  • $begingroup$
    So all L1 and L2 functions are fourier transformable?
    $endgroup$
    – shakshoka1234
    Mar 21 at 13:39










  • $begingroup$
    Yes. The proof is not so bad, you first develop the theory in $L^1$, then look at functions in $L^1cap L^2$, and use a density argument to extend this to $L^2$ itself.
    $endgroup$
    – Tony S.F.
    Mar 21 at 13:55













1












1








1





$begingroup$


I have a lot of comprehension questions that I can't really figure out by googling:



  1. The difference between a Fourier transformation and a Fourier series. Am I correct in thinking that the difference: transforms are for continous cases and series are for discrete cases ?


  2. What functions can be Fourier transformed? Which not? I read/learned once that piecewise continuity is enough for a function to be transformed as long as it can be extended to 2 - pi periodic continuity (so the pieces can be "stringed together to a periodic wave")


Is this correct?










share|cite|improve this question









$endgroup$




I have a lot of comprehension questions that I can't really figure out by googling:



  1. The difference between a Fourier transformation and a Fourier series. Am I correct in thinking that the difference: transforms are for continous cases and series are for discrete cases ?


  2. What functions can be Fourier transformed? Which not? I read/learned once that piecewise continuity is enough for a function to be transformed as long as it can be extended to 2 - pi periodic continuity (so the pieces can be "stringed together to a periodic wave")


Is this correct?







functional-analysis fourier-analysis fourier-transform






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 21 at 13:06









shakshoka1234shakshoka1234

61




61











  • $begingroup$
    1. is more or less correct but 2. not so much. To be fourier transformed a function has to have some integrability properties because the fourier transform is defined as an integral over all of $mathbbR$.
    $endgroup$
    – Tony S.F.
    Mar 21 at 13:08










  • $begingroup$
    thanks for your prompt response. i'm not looking for complicated counter examples and such, just the most generally applicable case , as in, what is the "weakest" function, that is still transformable
    $endgroup$
    – shakshoka1234
    Mar 21 at 13:22










  • $begingroup$
    We can readily talk about the fourier transform of $L^1$ functions because $|e^2pi i x xi|leq 1$ and we can use a density argument to extend to $L^2$. Beyond that we can go to distributions and say something about Schwartz functions but I don't really know what is the "weakest' function for this.
    $endgroup$
    – Tony S.F.
    Mar 21 at 13:31










  • $begingroup$
    So all L1 and L2 functions are fourier transformable?
    $endgroup$
    – shakshoka1234
    Mar 21 at 13:39










  • $begingroup$
    Yes. The proof is not so bad, you first develop the theory in $L^1$, then look at functions in $L^1cap L^2$, and use a density argument to extend this to $L^2$ itself.
    $endgroup$
    – Tony S.F.
    Mar 21 at 13:55
















  • $begingroup$
    1. is more or less correct but 2. not so much. To be fourier transformed a function has to have some integrability properties because the fourier transform is defined as an integral over all of $mathbbR$.
    $endgroup$
    – Tony S.F.
    Mar 21 at 13:08










  • $begingroup$
    thanks for your prompt response. i'm not looking for complicated counter examples and such, just the most generally applicable case , as in, what is the "weakest" function, that is still transformable
    $endgroup$
    – shakshoka1234
    Mar 21 at 13:22










  • $begingroup$
    We can readily talk about the fourier transform of $L^1$ functions because $|e^2pi i x xi|leq 1$ and we can use a density argument to extend to $L^2$. Beyond that we can go to distributions and say something about Schwartz functions but I don't really know what is the "weakest' function for this.
    $endgroup$
    – Tony S.F.
    Mar 21 at 13:31










  • $begingroup$
    So all L1 and L2 functions are fourier transformable?
    $endgroup$
    – shakshoka1234
    Mar 21 at 13:39










  • $begingroup$
    Yes. The proof is not so bad, you first develop the theory in $L^1$, then look at functions in $L^1cap L^2$, and use a density argument to extend this to $L^2$ itself.
    $endgroup$
    – Tony S.F.
    Mar 21 at 13:55















$begingroup$
1. is more or less correct but 2. not so much. To be fourier transformed a function has to have some integrability properties because the fourier transform is defined as an integral over all of $mathbbR$.
$endgroup$
– Tony S.F.
Mar 21 at 13:08




$begingroup$
1. is more or less correct but 2. not so much. To be fourier transformed a function has to have some integrability properties because the fourier transform is defined as an integral over all of $mathbbR$.
$endgroup$
– Tony S.F.
Mar 21 at 13:08












$begingroup$
thanks for your prompt response. i'm not looking for complicated counter examples and such, just the most generally applicable case , as in, what is the "weakest" function, that is still transformable
$endgroup$
– shakshoka1234
Mar 21 at 13:22




$begingroup$
thanks for your prompt response. i'm not looking for complicated counter examples and such, just the most generally applicable case , as in, what is the "weakest" function, that is still transformable
$endgroup$
– shakshoka1234
Mar 21 at 13:22












$begingroup$
We can readily talk about the fourier transform of $L^1$ functions because $|e^2pi i x xi|leq 1$ and we can use a density argument to extend to $L^2$. Beyond that we can go to distributions and say something about Schwartz functions but I don't really know what is the "weakest' function for this.
$endgroup$
– Tony S.F.
Mar 21 at 13:31




$begingroup$
We can readily talk about the fourier transform of $L^1$ functions because $|e^2pi i x xi|leq 1$ and we can use a density argument to extend to $L^2$. Beyond that we can go to distributions and say something about Schwartz functions but I don't really know what is the "weakest' function for this.
$endgroup$
– Tony S.F.
Mar 21 at 13:31












$begingroup$
So all L1 and L2 functions are fourier transformable?
$endgroup$
– shakshoka1234
Mar 21 at 13:39




$begingroup$
So all L1 and L2 functions are fourier transformable?
$endgroup$
– shakshoka1234
Mar 21 at 13:39












$begingroup$
Yes. The proof is not so bad, you first develop the theory in $L^1$, then look at functions in $L^1cap L^2$, and use a density argument to extend this to $L^2$ itself.
$endgroup$
– Tony S.F.
Mar 21 at 13:55




$begingroup$
Yes. The proof is not so bad, you first develop the theory in $L^1$, then look at functions in $L^1cap L^2$, and use a density argument to extend this to $L^2$ itself.
$endgroup$
– Tony S.F.
Mar 21 at 13:55










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