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Test functions for tempered distributions: analytic?

  1. Feb 28, 2014 #1

    jasonRF

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    When considering tempered distributions, I am only aware of the definition of test functions of a real variable. However, is it okay to use test functions of a complex variable z that are analytic in a strip that includes the real axis? (of course they still must fall off fast enough as the real part of z goes to +/- infinity). I know of at least one test function of a real variable, [itex]e^{-x^2}[/itex] for whom the analytic continuation to the strip is trivial.

    I am asking because (just for fun) I am looking at the Fourier transform of the Heaviside step function, [itex]u(t)[/itex] that is zero for t<0 and one for t>0. If we let [itex]\hat{f} [/itex] denote the Fourier transform of an arbitrary [itex]f[/itex] (test function or distribution), and let [itex]\psi[/itex] be a test function, then by definition of the Fourier transform for distributions
    [tex]
    \langle \hat{u}, \psi \rangle = \langle u, \hat{\psi} \rangle = \int_0^{\infty} dx\, \int_{-\infty}^\infty dt\, e^{-i x t} \psi(t)
    [/tex]
    I want to swap the order of integration but cannot since the reverse order integral is not defined. However, the above integral is well behaved so it should be equal to
    [tex]
    \lim_{\epsilon \rightarrow 0} \int_0^{\infty} dx\, e^{-\epsilon x} \int_{-\infty}^\infty dt\, e^{-i x t} \psi(t).
    [/tex]
    Now I can swap the order of integration and perform the first integration to obtain,
    [tex]
    \lim_{\epsilon \rightarrow 0} \frac{1}{i} \int_{-\infty}^\infty dt\, \frac{\psi(t)}{t - i \epsilon}.
    [/tex]

    This is where I would like [itex]\psi[/itex] to be analytic in a strip including the real axis and extending into the lower half plane (even by just a little bit). In that case the above integral is simple by contour integral techniques - the integral is along the real axis and as [itex]\epsilon \rightarrow 0[/itex] we indent the contour into the lower half plane and I get
    [tex]
    \langle \hat{u}, \psi \rangle = \pi \psi(0) + PV\int_{-\infty}^\infty dt\, \frac{\psi(t)}{it}
    [/tex]
    where [itex]PV[/itex] indicates the Cauchy principle value (symmetric limits about t=0). Hence,
    [tex]
    \hat{u}(x) = \pi \delta(x) + PV \frac{1}{ix}.
    [/tex]
    It works out so nice it seems like it should be fine to do this, but lots of things can be done formally that make no sense! Everything up to where I want to use contour integration is easy to find in textbooks (right now I am looking at "waves and distributions" by Jonsson and Yngvason), but I don't see authors doing the last step with contour integration and I am wondering if there is a reason - I am guessing there is but I just don't see it.

    Thanks,

    Jason
     
  2. jcsd
  3. Feb 28, 2014 #2

    jasonRF

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    I have convinced myself that what I did was legitimate. For a test functions of a real variable, [itex]f(t)[/itex], the fourier tranform
    [tex]
    \hat{f}(z) = \int_{-\infty}^{\infty} dt\, e^{-i z t} f(t)
    [/tex]
    is an analytic function of [itex]z[/itex] in a strip including the real axis. Likewise for inverse fourier transforms. Combined with the fact that fourier transforms of test functions are also test functions and it follows that test functions are analytic in a strip including the real axis.

    I think my hang-up was that I wasn't sure all test functions could be continued off of the real axis - now I am convinced that they can be, although it was in a roundabout, inelegant way!

    jason
     
  4. Feb 28, 2014 #3

    pwsnafu

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    This is called the Carleman theory of generalised functions. See Chapter 4 of Hoskins and Pinto Distributions, ultradistributions and other generalised functions.

    Not analytically. Analytic functions have only isolated zeroes, so there is no extension of a non-zero test function which happens to be zero on an interval. However
    Theorem Let ##f \in L^p(\mathbb{R})## be continuous with ##p \geq 1##. Then there exists ##f^{\circ} : \mathbb{C}\setminus\mathbb{R} \to\mathbb{C}## which is analytic and satisfies
    ##\lim_{\epsilon\to 0+}\left[ f^{\circ}(x+i\epsilon) - f^{\circ}(x-i\epsilon) \right] = f(x)##
    the convergence being locally uniform on ##\mathbb{R}##
     
  5. Mar 1, 2014 #4

    jasonRF

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    Good point. Since I was using tempered distributions the test functions are not required to have compact support so I was implicitly assuming they did not vanish on an interval - but certainly functions with compact support (or that vanish on an interval) are a subset of the test functions.

    I'll have to see if our library has that book.

    Nice. I supposedly learned some distribution theory as an undergrad (an applied analysis class that covered 2/3 complex analysis and 1/3 distribution theory) but I didn't really understand the details at the time and I am sure we never saw such a nice connection to analytic functions.

    Thanks
     
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