From: ullrich@math.okstate.edu (David C. Ullrich) Subject: Re: L^1 and FT Date: Fri, 27 Apr 2001 15:45:45 GMT Newsgroups: sci.math Summary: Comparing domains and ranges of Fourier Transform On Fri, 27 Apr 2001 15:54:30 +0100, "Rob Brownlee" wrote: >I'm right in saying, what has actually been proved is the following: > >If f is an L^1 function with the property that all its derivates are L^1 >then f^ is a function of rapid decent (a member of S(\R^d)). > >This is a pretty little result. It's not true. One way you can _tell_ it's not true is this: The FT is an isomorphism from S onto S, so this result would imply that f itself was in S. But just saying all the derivatives of f are integrable does not show that f is rapidly decreasing. (1/(1+x^2) is a counterexample with d=1.) The error in what I suspect your proof was is this: The hypotheses do imply that f^ decreases faster than 1/(any polynomial). But we have not proved that the _derivatives_ of f^ decrease faster than 1/(any polynomial), which is part of the definition of S. (And in fact if d=1 and f(x)==1/(1+x^2) then f^(x)=exp(-|x|), which is not even differentiable.) If you thought the definition of S was just "decreases faster than 1/(any polynomial)" and you also know that the FT is an isomorphism from S onto S that might explain some of your other dubious conclusions... >Rob > > ============================================================================== From: Wade Ramey Subject: Re: L^1 and FT Date: Fri, 27 Apr 2001 21:07:01 GMT Newsgroups: sci.math In article <9cc17g$917f$1@rook.le.ac.uk>, "Rob Brownlee" wrote: > I'm right in saying, what has actually been proved is the following: > > If f is an L^1 function with the property that all its derivates are L^1 > then f^ is a function of rapid decent >(a member of S(\R^d)). f^ is not necessarily in S(R^d), because f^ need not be smooth. Example: if f(x) = 1/(1+x^2), then f and all its derivatives are in L1, but f^(y) is essentially exp(-|y|). Wade