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- Newsgroups: sci.math
- Path: sparky!uunet!snorkelwacker.mit.edu!galois!riesz!jbaez
- From: jbaez@riesz.mit.edu (John C. Baez)
- Subject: Re: Lebesgue integral (was: Couple of questions
- Message-ID: <1992Sep15.060716.8281@galois.mit.edu>
- Sender: news@galois.mit.edu
- Nntp-Posting-Host: riesz
- Organization: MIT Department of Mathematics, Cambridge, MA
- References: <1992Sep11.130033.26063@watson.ibm.com> <1992Sep15.020445.26398@unixg.ubc.ca> <1992Sep15.023801.12444@mp.cs.niu.edu>
- Date: Tue, 15 Sep 92 06:07:16 GMT
- Lines: 60
-
- rickert@mp.cs.niu.edu (Neil Rickert) writes:
- >In article <1992Sep15.020445.26398@unixg.ubc.ca> ramsay@unixg.ubc.ca (Keith Ramsay) writes:
-
- >>I've heard that much (nearly all, perhaps) of what is done with L^2,
- >>regarded as Lebesgue square-integrable functions, modulo functions
- >>supported on sets of measure zero, is just as nicely done (if not
- >>better) by regarding L^2 as the formal completion (relative to the L^2
- >>norm) of a convenient dense subset (where the Lebesgue measure theory
- >>is not needed). I'm not enough of an analyst to confirm this.
-
- I wouldn't want to say you can do everything better, since someone
- will probably come up with something you can't do better, but it is a
- very nice approach and one I often use myself. Just take, say,
- Schwartz functions, which we can integrate a la
- Riemann, and complete 'em in the L2 norm. Now say you want to show
- something. Well, what do you want to show? Say you want to do
- Fourier transforms and prove the Plancherel theorem (that the Fourier
- transform is an isometry on L2.) The easiest way
- is to note:
-
- 1) Schwartz functions are in L2
- 2) Schwartz functions are dense in L2
- 3) The Fourier transform of a Schwartz function is Schwartz
- 4) The Fourier transform is an isometry on Schwartz functions
-
- and you conclude by easy abstract nonsense that there is a unique way
- of extending the Fourier transform from Schwartz functions to an
- isometry on all of L2, thus simultaneously defining the Fourier
- transform and proving Plancherel.
-
- The nice thing is that 1-4 are all really easy and only use Riemann
- integration. Check out Reed and Simon vol. 2 - Fourier Analysis,
- Self-adjointness. They do all this in essentially 3 pages, pages 2-4!
- In those same two pages they prove the Fourier inversion theorem.
-
- In my book with Segal and Zhou the real power of this approach shows
- through when we define L2 of an infinite-dimensional Hilbert space
- with Gaussian "measure". The point is, we all know how to integrate
- polynomials in finitely many variables against a Gaussian, so using
- that formula we can define their integral in the infinite-dimensional
- case and define the L2 by completion. That's really the best way of
- making sense of these "measures" in QFT - realize that they are just
- a trace on an algebra of "nice" functions, in this case polynomials.
-
- >This leaves the physicist with a function that he can't be sure he can
- >get a Fourier series for, and a solution to a differential equation
- >which has a Fourier series but is not a function since it is in some
- >mysterious formal completion space.
-
- I don't think so. I just sketched how you can do Fourier transforms
- (Fourier series are easier). As for "mysterious formal completions",
- 1) it's not so mysterious, 2) elements of L2 as defined in the
- old-fashioned way are "mysterious equivalence classes" of functions
- mod null functions - they sure aren't functions - they don't have
- values at points - so we're not losing much.
-
- Of course a true analyst should know lots of different approaches.
-
-
- .
-