Laplace transform and frames vs Bases












3












$begingroup$


The Laplace transform



$$F(s) = int^{∞}_{0}f(t)e^{-st} dt$$ can be understood much like the fourier transform, as a change of basis of an $L^2$ function to the eigen functions of the differential operator $frac{d}{dt}$



Unlike the complex exponentials, which form an uncountable orthogonal basis, the $e^{-st}$ basis functions are not in general orthogonal to each other. Does this mean we are switching to a frame and not a minimal orthogonal basis? Does this at all imply were overcounting? Or is there just redundant information in our resulting function?






fourier-analysis laplace-transform change-of-basis







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  • $begingroup$
    I believe that "are in general orthogonal to each other" is a typo. Do you mean "are not in general orthogonal to each other?"
    $endgroup$
    – Brian Borchers
    Jan 13 at 23:06










  • $begingroup$
    @Brian Borchers fixed. Nice catch
    $endgroup$
    – Craig
    Jan 13 at 23:08






  • 1




    $begingroup$
    There is, indeed, redundant information in the Laplace transform of a function. To invert the Laplace transform, one only needs one vertical line of values in the complex plane (rather than all values in the complex plane).
    $endgroup$
    – NicNic8
    Jan 13 at 23:09










  • $begingroup$
    @NicNic8 can you expand on this, or link to relevant readings somewhere? Thank you
    $endgroup$
    – Craig
    Jan 13 at 23:11






  • 1




    $begingroup$
    @Craig Look at equation 3 of Wikipedia here: en.wikipedia.org/wiki/Laplace_transform. You only need one value of $gamma$.
    $endgroup$
    – NicNic8
    Jan 13 at 23:50
















3












$begingroup$


The Laplace transform



$$F(s) = int^{∞}_{0}f(t)e^{-st} dt$$ can be understood much like the fourier transform, as a change of basis of an $L^2$ function to the eigen functions of the differential operator $frac{d}{dt}$



Unlike the complex exponentials, which form an uncountable orthogonal basis, the $e^{-st}$ basis functions are not in general orthogonal to each other. Does this mean we are switching to a frame and not a minimal orthogonal basis? Does this at all imply were overcounting? Or is there just redundant information in our resulting function?






fourier-analysis laplace-transform change-of-basis







share|cite|improve this question











$endgroup$












  • $begingroup$
    I believe that "are in general orthogonal to each other" is a typo. Do you mean "are not in general orthogonal to each other?"
    $endgroup$
    – Brian Borchers
    Jan 13 at 23:06










  • $begingroup$
    @Brian Borchers fixed. Nice catch
    $endgroup$
    – Craig
    Jan 13 at 23:08






  • 1




    $begingroup$
    There is, indeed, redundant information in the Laplace transform of a function. To invert the Laplace transform, one only needs one vertical line of values in the complex plane (rather than all values in the complex plane).
    $endgroup$
    – NicNic8
    Jan 13 at 23:09










  • $begingroup$
    @NicNic8 can you expand on this, or link to relevant readings somewhere? Thank you
    $endgroup$
    – Craig
    Jan 13 at 23:11






  • 1




    $begingroup$
    @Craig Look at equation 3 of Wikipedia here: en.wikipedia.org/wiki/Laplace_transform. You only need one value of $gamma$.
    $endgroup$
    – NicNic8
    Jan 13 at 23:50














3












3








3


1



$begingroup$


The Laplace transform



$$F(s) = int^{∞}_{0}f(t)e^{-st} dt$$ can be understood much like the fourier transform, as a change of basis of an $L^2$ function to the eigen functions of the differential operator $frac{d}{dt}$



Unlike the complex exponentials, which form an uncountable orthogonal basis, the $e^{-st}$ basis functions are not in general orthogonal to each other. Does this mean we are switching to a frame and not a minimal orthogonal basis? Does this at all imply were overcounting? Or is there just redundant information in our resulting function?






fourier-analysis laplace-transform change-of-basis







share|cite|improve this question











$endgroup$




The Laplace transform



$$F(s) = int^{∞}_{0}f(t)e^{-st} dt$$ can be understood much like the fourier transform, as a change of basis of an $L^2$ function to the eigen functions of the differential operator $frac{d}{dt}$



Unlike the complex exponentials, which form an uncountable orthogonal basis, the $e^{-st}$ basis functions are not in general orthogonal to each other. Does this mean we are switching to a frame and not a minimal orthogonal basis? Does this at all imply were overcounting? Or is there just redundant information in our resulting function?






fourier-analysis laplace-transform change-of-basis




fourier-analysis laplace-transform change-of-basis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 23:07







Craig

















asked Jan 13 at 23:03









CraigCraig

867




867












  • $begingroup$
    I believe that "are in general orthogonal to each other" is a typo. Do you mean "are not in general orthogonal to each other?"
    $endgroup$
    – Brian Borchers
    Jan 13 at 23:06










  • $begingroup$
    @Brian Borchers fixed. Nice catch
    $endgroup$
    – Craig
    Jan 13 at 23:08






  • 1




    $begingroup$
    There is, indeed, redundant information in the Laplace transform of a function. To invert the Laplace transform, one only needs one vertical line of values in the complex plane (rather than all values in the complex plane).
    $endgroup$
    – NicNic8
    Jan 13 at 23:09










  • $begingroup$
    @NicNic8 can you expand on this, or link to relevant readings somewhere? Thank you
    $endgroup$
    – Craig
    Jan 13 at 23:11






  • 1




    $begingroup$
    @Craig Look at equation 3 of Wikipedia here: en.wikipedia.org/wiki/Laplace_transform. You only need one value of $gamma$.
    $endgroup$
    – NicNic8
    Jan 13 at 23:50


















  • $begingroup$
    I believe that "are in general orthogonal to each other" is a typo. Do you mean "are not in general orthogonal to each other?"
    $endgroup$
    – Brian Borchers
    Jan 13 at 23:06










  • $begingroup$
    @Brian Borchers fixed. Nice catch
    $endgroup$
    – Craig
    Jan 13 at 23:08






  • 1




    $begingroup$
    There is, indeed, redundant information in the Laplace transform of a function. To invert the Laplace transform, one only needs one vertical line of values in the complex plane (rather than all values in the complex plane).
    $endgroup$
    – NicNic8
    Jan 13 at 23:09










  • $begingroup$
    @NicNic8 can you expand on this, or link to relevant readings somewhere? Thank you
    $endgroup$
    – Craig
    Jan 13 at 23:11






  • 1




    $begingroup$
    @Craig Look at equation 3 of Wikipedia here: en.wikipedia.org/wiki/Laplace_transform. You only need one value of $gamma$.
    $endgroup$
    – NicNic8
    Jan 13 at 23:50
















$begingroup$
I believe that "are in general orthogonal to each other" is a typo. Do you mean "are not in general orthogonal to each other?"
$endgroup$
– Brian Borchers
Jan 13 at 23:06




$begingroup$
I believe that "are in general orthogonal to each other" is a typo. Do you mean "are not in general orthogonal to each other?"
$endgroup$
– Brian Borchers
Jan 13 at 23:06












$begingroup$
@Brian Borchers fixed. Nice catch
$endgroup$
– Craig
Jan 13 at 23:08




$begingroup$
@Brian Borchers fixed. Nice catch
$endgroup$
– Craig
Jan 13 at 23:08




1




1




$begingroup$
There is, indeed, redundant information in the Laplace transform of a function. To invert the Laplace transform, one only needs one vertical line of values in the complex plane (rather than all values in the complex plane).
$endgroup$
– NicNic8
Jan 13 at 23:09




$begingroup$
There is, indeed, redundant information in the Laplace transform of a function. To invert the Laplace transform, one only needs one vertical line of values in the complex plane (rather than all values in the complex plane).
$endgroup$
– NicNic8
Jan 13 at 23:09












$begingroup$
@NicNic8 can you expand on this, or link to relevant readings somewhere? Thank you
$endgroup$
– Craig
Jan 13 at 23:11




$begingroup$
@NicNic8 can you expand on this, or link to relevant readings somewhere? Thank you
$endgroup$
– Craig
Jan 13 at 23:11




1




1




$begingroup$
@Craig Look at equation 3 of Wikipedia here: en.wikipedia.org/wiki/Laplace_transform. You only need one value of $gamma$.
$endgroup$
– NicNic8
Jan 13 at 23:50




$begingroup$
@Craig Look at equation 3 of Wikipedia here: en.wikipedia.org/wiki/Laplace_transform. You only need one value of $gamma$.
$endgroup$
– NicNic8
Jan 13 at 23:50










1 Answer
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The operator $L_0f = -if'$ is symmetric on the domain $mathcal{D}(L_0)$ of functions $fin L^2[0,infty)$ that are absolutely continuous on $[0,infty)$, that vanish at $0$, and that satisfy $f'in L^2[0,infty)$. The operator $L_0$ is maximally symmetric, which means that it is closed and symmetric, but it is has no proper symmetric extension.



$L_0^*$ is not symmetric, but it is an extension of $L_0$. Indeed, $L_0^*f=-if'$ has the same domain except that the functions in the domain of $L_0^*$ do not necessarily vanish at $0$. The operator $L_0^*$ has spectrum equal to the upper half plane becase $e^{isx} in L^2[0,infty)$ for $Im s > 0$ and $L_0^*e^{isx}=-i(-is e^{-isx})=se^{isx}$. The real axis is also in the spectrum because the spectrum is closed.



It is possible to extend the holomorphic functiona calculus to functions $F$ that are holomorphic on a neighborhood of the upper half plane:
$$
F(L_0^*)f=frac{1}{2pi i}int_{-iepsilon-infty}^{-iepsilon+infty}F(s)(s I -L_0^*)^{-1}fds.
$$
This extends to the case where $epsilon=0$ if $F$ is a bounded holomorphic function in the upper half plane with a.e. boundary function $F(s)$ on the real axis. If you turn everything clockwise in the complex plane by looking at $L_0 f = -f'$ instead of $-if'$, then the operator calculus is implemented by a Fourier transform for the outer integral and a Laplace transform for the inner integral. The functional calculus allows for the computation of $F(L_0^*)f$ if $F$ is a bounded holomorphic function in the right half plane.






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    $begingroup$

    The operator $L_0f = -if'$ is symmetric on the domain $mathcal{D}(L_0)$ of functions $fin L^2[0,infty)$ that are absolutely continuous on $[0,infty)$, that vanish at $0$, and that satisfy $f'in L^2[0,infty)$. The operator $L_0$ is maximally symmetric, which means that it is closed and symmetric, but it is has no proper symmetric extension.



    $L_0^*$ is not symmetric, but it is an extension of $L_0$. Indeed, $L_0^*f=-if'$ has the same domain except that the functions in the domain of $L_0^*$ do not necessarily vanish at $0$. The operator $L_0^*$ has spectrum equal to the upper half plane becase $e^{isx} in L^2[0,infty)$ for $Im s > 0$ and $L_0^*e^{isx}=-i(-is e^{-isx})=se^{isx}$. The real axis is also in the spectrum because the spectrum is closed.



    It is possible to extend the holomorphic functiona calculus to functions $F$ that are holomorphic on a neighborhood of the upper half plane:
    $$
    F(L_0^*)f=frac{1}{2pi i}int_{-iepsilon-infty}^{-iepsilon+infty}F(s)(s I -L_0^*)^{-1}fds.
    $$
    This extends to the case where $epsilon=0$ if $F$ is a bounded holomorphic function in the upper half plane with a.e. boundary function $F(s)$ on the real axis. If you turn everything clockwise in the complex plane by looking at $L_0 f = -f'$ instead of $-if'$, then the operator calculus is implemented by a Fourier transform for the outer integral and a Laplace transform for the inner integral. The functional calculus allows for the computation of $F(L_0^*)f$ if $F$ is a bounded holomorphic function in the right half plane.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The operator $L_0f = -if'$ is symmetric on the domain $mathcal{D}(L_0)$ of functions $fin L^2[0,infty)$ that are absolutely continuous on $[0,infty)$, that vanish at $0$, and that satisfy $f'in L^2[0,infty)$. The operator $L_0$ is maximally symmetric, which means that it is closed and symmetric, but it is has no proper symmetric extension.



      $L_0^*$ is not symmetric, but it is an extension of $L_0$. Indeed, $L_0^*f=-if'$ has the same domain except that the functions in the domain of $L_0^*$ do not necessarily vanish at $0$. The operator $L_0^*$ has spectrum equal to the upper half plane becase $e^{isx} in L^2[0,infty)$ for $Im s > 0$ and $L_0^*e^{isx}=-i(-is e^{-isx})=se^{isx}$. The real axis is also in the spectrum because the spectrum is closed.



      It is possible to extend the holomorphic functiona calculus to functions $F$ that are holomorphic on a neighborhood of the upper half plane:
      $$
      F(L_0^*)f=frac{1}{2pi i}int_{-iepsilon-infty}^{-iepsilon+infty}F(s)(s I -L_0^*)^{-1}fds.
      $$
      This extends to the case where $epsilon=0$ if $F$ is a bounded holomorphic function in the upper half plane with a.e. boundary function $F(s)$ on the real axis. If you turn everything clockwise in the complex plane by looking at $L_0 f = -f'$ instead of $-if'$, then the operator calculus is implemented by a Fourier transform for the outer integral and a Laplace transform for the inner integral. The functional calculus allows for the computation of $F(L_0^*)f$ if $F$ is a bounded holomorphic function in the right half plane.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The operator $L_0f = -if'$ is symmetric on the domain $mathcal{D}(L_0)$ of functions $fin L^2[0,infty)$ that are absolutely continuous on $[0,infty)$, that vanish at $0$, and that satisfy $f'in L^2[0,infty)$. The operator $L_0$ is maximally symmetric, which means that it is closed and symmetric, but it is has no proper symmetric extension.



        $L_0^*$ is not symmetric, but it is an extension of $L_0$. Indeed, $L_0^*f=-if'$ has the same domain except that the functions in the domain of $L_0^*$ do not necessarily vanish at $0$. The operator $L_0^*$ has spectrum equal to the upper half plane becase $e^{isx} in L^2[0,infty)$ for $Im s > 0$ and $L_0^*e^{isx}=-i(-is e^{-isx})=se^{isx}$. The real axis is also in the spectrum because the spectrum is closed.



        It is possible to extend the holomorphic functiona calculus to functions $F$ that are holomorphic on a neighborhood of the upper half plane:
        $$
        F(L_0^*)f=frac{1}{2pi i}int_{-iepsilon-infty}^{-iepsilon+infty}F(s)(s I -L_0^*)^{-1}fds.
        $$
        This extends to the case where $epsilon=0$ if $F$ is a bounded holomorphic function in the upper half plane with a.e. boundary function $F(s)$ on the real axis. If you turn everything clockwise in the complex plane by looking at $L_0 f = -f'$ instead of $-if'$, then the operator calculus is implemented by a Fourier transform for the outer integral and a Laplace transform for the inner integral. The functional calculus allows for the computation of $F(L_0^*)f$ if $F$ is a bounded holomorphic function in the right half plane.






        share|cite|improve this answer









        $endgroup$



        The operator $L_0f = -if'$ is symmetric on the domain $mathcal{D}(L_0)$ of functions $fin L^2[0,infty)$ that are absolutely continuous on $[0,infty)$, that vanish at $0$, and that satisfy $f'in L^2[0,infty)$. The operator $L_0$ is maximally symmetric, which means that it is closed and symmetric, but it is has no proper symmetric extension.



        $L_0^*$ is not symmetric, but it is an extension of $L_0$. Indeed, $L_0^*f=-if'$ has the same domain except that the functions in the domain of $L_0^*$ do not necessarily vanish at $0$. The operator $L_0^*$ has spectrum equal to the upper half plane becase $e^{isx} in L^2[0,infty)$ for $Im s > 0$ and $L_0^*e^{isx}=-i(-is e^{-isx})=se^{isx}$. The real axis is also in the spectrum because the spectrum is closed.



        It is possible to extend the holomorphic functiona calculus to functions $F$ that are holomorphic on a neighborhood of the upper half plane:
        $$
        F(L_0^*)f=frac{1}{2pi i}int_{-iepsilon-infty}^{-iepsilon+infty}F(s)(s I -L_0^*)^{-1}fds.
        $$
        This extends to the case where $epsilon=0$ if $F$ is a bounded holomorphic function in the upper half plane with a.e. boundary function $F(s)$ on the real axis. If you turn everything clockwise in the complex plane by looking at $L_0 f = -f'$ instead of $-if'$, then the operator calculus is implemented by a Fourier transform for the outer integral and a Laplace transform for the inner integral. The functional calculus allows for the computation of $F(L_0^*)f$ if $F$ is a bounded holomorphic function in the right half plane.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 6:13









        DisintegratingByPartsDisintegratingByParts

        60.5k42681




        60.5k42681






























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