Why isn't the parallel between the Fourier transform and the Laplace transform complete?

Question

I mean the question in the following sense. For Fourier, we can do it on compact intervals and then we get a sequence of coefficients. We can do it continuum-style, and then we get a superposition of waves of continuum-varying frequencies. We can even do it abstractly on compact groups.

So why is the only Laplace transform (at least that I've ever heard about) on $[0, \infty)$?

Answer 1

Consider the space $L=L^1(\mathbb{R})$, this is a Banach space which becomes a commutative Banach algebra under the convolution $$f*g(x)=\int_\mathbb{R}f(y)g(x-y)dy$$ Similarly, the space $L^+=L^1(\mathbb{R}^+)$ is a Banach space which becomes a commutative Banach algebra under the convolution $$f*g(x)=\int_0^xf(y)g(x-y)dy$$

• The Gelfand transform on $L$ is the Fourier transform.
• The Gelfand transform on $L^+$ is the Laplace transform.

There are also discrete versions of this.

Consider the space $\ell=\ell^1(\mathbb{Z})=L^1(\mathbb{Z})$, this is a Banach space which becomes a commutative Banach algebra under the convolution $$f*g(k)=\sum_\mathbb{Z}f(n)g(k-n)$$ Similarly, the space $\ell^+=\ell^1(\mathbb{N})=L^1(\mathbb{N})$ is a Banach space which becomes a commutative Banach algebra under the convolution $$f*g(k)=\sum_{n=0}^kf(n)g(k-n)$$

• The Gelfand transform on $\ell$ is an absolute convergent Fourier series.
• The Gelfand transform on $\ell^+$ is an absolute convergent Taylor series.

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If we have a locally compact group $G$ then we may consider complex valued functions $f:G\to\mathbb{C}$ and there is a generalisation of Lebesgue measure due to Alfred Haar which is left invariant, i.e. $\int f(yx) d\mu(x)= \int f(x) d\mu(x)$. If it happens that $\mu$ is right invariant too, then $\mu$ is called unimodular (which is the case for Abelian groups).

Having a Haar measure it makes sense to talk about convolution, that is $$f*g(x)=\int_G f(y)g(y^{-1}x)d\mu(y)$$ The hope to understand $L^1(G)$ through diagonalising the convolution using a kind of Fourier transform is not easy. Loosely speaking we look for $\phi$ such that $\phi(xy)=\phi(x)\phi(y)$ (note, for the additive group $\mathbb{R}$ this reads $\phi(x+y)=\phi(x)\phi(y)$ and leads to the exponential function) and thus $$\int_G f*g(x)\phi(x)d\mu(x)=\int_G\int_Gf(y)g(y^{-1}x\phi(x)d\mu(y)d\mu(x)=\\ \int_G\int_Gf(y)g(z)\phi(yz)d\mu(y)d\mu(z)= \int_Gf(y)g(z)\phi(y)\phi(z)d\mu(y)d\mu(z)$$ For LCA-groups (locally compact Abelian groups) can more or less be written in the form $\mathbb{R}^n\times\mathbb{Z}^m\times(\mathbb{R}/\mathbb{Z})^l$. For non-Abelian groups there is a whole subject known as Representation theory, in particular you will find several books on $SL(2,\mathbb{R})$.

Answer 2

As AD. said above, for LCA semigroups one can construct the semigroup algebra just as in the group case and apply Gelfand theory.

It might be worth noting that the characters of the resulting Banach algebra change when we lose the group structure. For a LCA group $G$, the character space (the Pontryagin dual) consist of group homomorphisms $\phi: G \rightarrow S^1$. In the case $G = \mathbb{R}$, a typical character is $\phi_t(x) = e^{itx}$. The Gelfand-Fourier transform $f \mapsto \hat{f}$ is then given by

$$ \hat{f} (\phi_t) = \int f(x) \phi_t(x)dx. $$

This is why $e^{itx}$ appears in the Fourier transform. When we just have a semigroup $A$, the characters are now semigroup maps $\psi: A \rightarrow \mathbb{R}_+^{\times}$. For $A = \mathbb{R}_+$, you get maps $\{\psi_{-s}(x) = e^{-sx}\}$, which appear in the Laplace transform.

The two-sided Laplace transform can be generalized to LCA groups a la Gelfand also, by relaxing the definition of a character. The circle is replaced by $\mathbb{C}^{\times}$. These are called "generalized characters".