Is this homomorphism, actually isomorphism of groups?

1. Dec 6, 2013

LagrangeEuler

Example:
$\mathcal{L}[f(t)*g(t)]=F(s)G(s)$
Is this homomorphism, actually isomorphism of groups? $\mathcal{L}$ is Laplace transform.

2. Dec 6, 2013

jbunniii

Is the domain actually a group? What is the identity with respect to convolution? How about inverses?

3. Dec 6, 2013

Mandelbroth

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.

4. Dec 6, 2013

jbunniii

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
5. Dec 6, 2013

R136a1

It's going to be an homomorphism of rings though. Even of algebras. Even of Banach algebras! (depending on domain and codomain)

6. Dec 7, 2013

jbunniii

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.

 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
7. Dec 8, 2013

LagrangeEuler

So could you construct any space of function where this is isomorphism?

8. Dec 8, 2013

jbunniii

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 $\mathcal{L}(C_c(\mathbb{R}))$.