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Is this homomorphism, actually isomorphism of groups?

  1. Dec 6, 2013 #1
    Example:
    ##\mathcal{L}[f(t)*g(t)]=F(s)G(s) ##
    Is this homomorphism, actually isomorphism of groups? ##\mathcal{L}## is Laplace transform.
     
  2. jcsd
  3. Dec 6, 2013 #2

    jbunniii

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    Is the domain actually a group? What is the identity with respect to convolution? How about inverses?
     
  4. Dec 6, 2013 #3
    This depends on what kind of functions you want ##f## and ##g## to be. If you want the group under convolution to be on ##\mathcal{C}^0(\mathbb{R})##, you won't have an identity, for example.
     
  5. Dec 6, 2013 #4

    jbunniii

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    Any identity ##e## would have to satisfy ##f = e*f##, for all ##f## in the domain, which forces
    $$F(s) = \mathcal{L}[f(t)] = \mathcal{L}[e*f(t)] = E(s)F(s)$$
    So for any given ##s##, if there is some ##f## in the domain for which ##F(s) \neq 0##, then we can divide by ##F(s)## to force ##E(s) = 1##. Moreover, for the special case ##f = e##, we have ##E(s) = E(s)E(s)##, which means that ##E(s)## must be either 0 or 1 for every ##s##. I believe that if ##E## is the Laplace transform of an integrable function ##e##, then ##E## must be continuous (I know this is true for the Fourier transform), in which case this forces ##E(s) = 1## for all ##s##.

    Thus there is only one identity candidate, and that is a "function" ##e## satisfying ##\mathcal{L}[e(t)] = 1##. Indeed there is no such function, so we would have to enlarge the domain to include the Dirac delta distribution ##\delta##.

    But the domain also needs to include inverses, i.e. for every ##f## there must be some ##g## such that ##f*g = \delta##. This will not be possible in general. (Consider ##f = 0## for one extreme case.)
     
    Last edited: Dec 6, 2013
  6. Dec 6, 2013 #5
    It's going to be an homomorphism of rings though. Even of algebras. Even of Banach algebras! (depending on domain and codomain)
     
  7. Dec 7, 2013 #6

    jbunniii

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    Yes, for the domain one has to choose a set of functions that is actually a ring. In particular it has to be closed under the convolution operation. The set of continuous functions with compact support would work as a ring (without identity).

    Then to address the rest of the question: is it an isomorphism? Well, if the domain is a ring, then the Laplace transform is a homomorphism of rings, so its image is a ring, and it's surjective onto its image. So if we choose the codomain equal to the image, we just need injectivity. This will depend on the domain. Does ##\mathcal{L}(f) = \mathcal{L}(g)## imply ##f = g##? By linearity of the Laplace transform, this is equivalent to asking whether ##\mathcal{L}(f) = 0## implies ##f = 0##. This will be true if the domain contains only continuous functions, but not in general.

    [edit] In fact, the domain must be restricted to continuous functions, because otherwise it won't be closed under convolution. (Convolutions are continuous.)
     
    Last edited: Dec 7, 2013
  8. Dec 8, 2013 #7
    So could you construct any space of function where this is isomorphism?
     
  9. Dec 8, 2013 #8

    jbunniii

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    Sure, for example the set ##C_c(\mathbb{R})## of all continuous functions with compact support is a ring (without identity) under the operations of addition and convolution. The Laplace transform ##\mathcal{L}##, restricted to ##C_c(\mathbb{R})##, is an injective ring homomorphism, so it is an isomorphism from ##C_c(\mathbb{R})## to its image ##\mathcal{L}(C_c(\mathbb{R}))##.
     
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