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?
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2$\begingroup$ this is answered at en.wikipedia.org/wiki/Gelfand_representation#Examples $\endgroup$– Carlo BeenakkerCommented Aug 29, 2020 at 12:14
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1$\begingroup$ 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. $\endgroup$– JulesCommented Aug 29, 2020 at 16:28
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1 Answer
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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.
answered Aug 29, 2020 at 12:14
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$\begingroup$ Cool, it is concerning with one-sided Laplace transform. And for two-sided Laplace transform? $\endgroup$– ABBCommented Aug 29, 2020 at 12:22
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3$\begingroup$ This is more or less what Carlo was referring to with his wikipedia link. (But I guess it may as well be posted as an answer.) $\endgroup$ Commented Aug 30, 2020 at 5:47
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$\begingroup$ @AliBagheri One way to do it is to take the convolution algebra of the multiplicative group of positive real numbers, equipped with its Haar measure. Here the Gelfand transform is what is more classically known as the Mellin transform, and the Mellin transform is a change of variables away from the two sided Laplace transform. Maybe it's possible to get straight to the two sided Laplace transform via another choice of measure or something - I'm not sure. $\endgroup$ Commented Aug 30, 2020 at 15:09
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4$\begingroup$ @YemonChoi Yeah, I posted my answer within a minute of Carlo posting his comment. I thought for a moment about deleting it, but I noted that I got reasonably far in a PhD with an emphasis on operator algebras before I learned this fact. The question is borderline for MO, but I guess it's still open and I think this is the right answer, so :shrug: $\endgroup$ Commented Aug 30, 2020 at 15:17
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$\begingroup$ 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 :) ) $\endgroup$ Commented Aug 30, 2020 at 15:27