Is this homomorphism, actually isomorphism of groups?

[LagrangeEuler]

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

 

[jbunniii]

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

 

[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.

 

[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.)

 

[R136a1]

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

 

[jbunniii]

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

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.)

 

[LagrangeEuler]

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

 

[jbunniii]

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

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