# Where does the Laplace transform come from?

The Gelfand transform on the commutative Banach *-algebra $$L^1(\mathbb{R})$$ is just the Fourier transform.

Q. What can we say concerning the Laplace transform?

• this is answered at en.wikipedia.org/wiki/Gelfand_representation#Examples – Carlo Beenakker Aug 29 '20 at 12:14
• We can evaluate a power series on the real line, or we can evaluate it on the unit circle in the complex plane. The latter gives us Fourier series. We can also evaluate it on the whole complex plane. These three options also show up in the continuous transforms, giving the Laplace transform on the real line, the Fourier transform, and the Laplace transform on the entire complex plane. The function being transformed is the analogue of the coefficients of the power series, and the function you get after transforming is the analogue of the function you get when you evaluate the power series. – Jules Aug 29 '20 at 16:28

Set $$A = L^1([0, \infty))$$, equipped with the structure of a Banach $$*$$-algebra via convolution. The spectrum of this algebra is the half plane $$\text{Re}(z) \geq 0$$, and the Gelfand transform is the Laplace transform.
• Sure, TBH I miss the days of MO when we would all share more of the things we happened to learn early on and other people didn't (the difference between Banach and Cstar algebras is one reason why examples such as $L^1({\bf R}_+)$ are baked into my "training" while e.g. I can never remember the proof of Kaplansky density :) ) – Yemon Choi Aug 30 '20 at 15:27