From: ullrich@math.okstate.edu (David C. Ullrich) Subject: Re: Distributions...am I missing the point? Date: Fri, 20 Oct 2000 13:59:14 GMT Newsgroups: sci.math Summary: [missing] On Fri, 20 Oct 2000 14:08:49 +0100, "Rob Brownlee" wrote: >I let S denote the space of infinitely diffrentiable continuous functions of >rapid decent - endowed with the usual topology. A continuous linear >functional on S is called a (tempered) distribution. I let [f,\phi] denote >the the action of the distribution f on the test function \phi. > >I'm getting confused with how the author of the material I am reading >interchanges between refering to f say as both a function and a >distribution. >For example.... >I define T_x as the translation operator (T_x \phi)(y)=\phi(y-x). I define >(e_x)(y) via exp(ixy)). Then I can see that for all f in S we have (T_x >f)^=e_-x f^ but the author claims this is true for all distribution aswell. > >Do I define [f^,\phi]=[f,\phi^]? and [T_x f,\phi]=[f,T_-x \phi]? Yes, those are the definitions of f^ and T_x f. (The reason they count as sensible definitions is that they're consistent with the usual definitions if f is a function. Or to be awesomely precise, if f is a function and L_f is the corresponding distribution [L_f, phi] = integral(f * phi) (where * is multiplication) then these definitions give (L_f)^ = L_(f^) and T_x(L_f) = L_(T_x f).) > I still >can't see why the above is true. But you haven't given the definition of e_x f yet! You're going to need that definition first... You say you define e_x(y) "via" exp(ixy). I take it this means that e_x(y) = exp(ixy). Now what is the definition of the product of a function and a distribution? It's another distribution: If f is a distribution and g is a function then the distribution gf is defined by [gf, phi] = [f, g*phi] (* is multiplication). (Makes sense because then g L_f = L_(gf) if f is also a function.) Now the fact that (T_x f)^=e_-x (f^) should just fall out if we unravel and reravel the definitions. Let's see: [e_(-x) (f^), phi] = [f^, e_(-x) phi] = [f, (e_(-x) phi)^] = [f, T_x (phi^)] = [T_(-x) f, phi^] = [(T_(-x) f)^, phi] and since this is true for all phi we have (T_x f)^=e_(-x) (f^). >Thanks, Rob > > ============================================================================== From: ullrich@math.okstate.edu (David C. Ullrich) Subject: Re: Distributions...am I missing the point? Date: Fri, 20 Oct 2000 22:44:08 GMT Newsgroups: sci.math On Fri, 20 Oct 2000 18:52:28 GMT, popunkio@math.wisc.edu wrote: >In article <39f04bdd.177617584@nntp.sprynet.com>, > ullrich@math.okstate.edu wrote: >> On Fri, 20 Oct 2000 14:08:49 +0100, "Rob Brownlee" >> wrote: >> >> >I let S denote the space of infinitely diffrentiable continuous >functions of >> >rapid decent - endowed with the usual topology. A continuous linear >> >functional on S is called a (tempered) distribution. I let [f,\phi] >denote >> >the the action of the distribution f on the test function \phi. >> > >> >I'm getting confused with how the author of the material I am reading >> >interchanges between refering to f say as both a function and a >> >distribution. >> >For example.... >> >I define T_x as the translation operator (T_x \phi)(y)=\phi(y-x). I >define >> >(e_x)(y) via exp(ixy)). Then I can see that for all f in S we have >(T_x >> >f)^=e_-x f^ but the author claims this is true for all distribution >aswell. >> > >> >Do I define [f^,\phi]=[f,\phi^]? and [T_x f,\phi]=[f,T_-x \phi]? >> >> Yes, those are the definitions of f^ and T_x f. (The >> reason they count as sensible definitions is that they're consistent >> with the usual definitions if f is a function. Or to be awesomely >> precise, if f is a function and L_f is the corresponding distribution >> [L_f, phi] = integral(f * phi) (where * is multiplication) then >> these definitions give (L_f)^ = L_(f^) and T_x(L_f) = L_(T_x f).) >> >> > I still >> >can't see why the above is true. >> >> But you haven't given the definition of e_x f yet! You're >> going to need that definition first... >> >> You say you define e_x(y) "via" exp(ixy). I take it >> this means that e_x(y) = exp(ixy). Now what is the definition >> of the product of a function and a distribution? It's another >> distribution: If f is a distribution and g is a function then >> the distribution gf is defined by >> >> [gf, phi] = [f, g*phi] (* is multiplication). > >Actually, isn't g supposed to be C^\infty, Of course it is. In fact C^infinity is not enough, it also has to have at most polynomial growth, and all its derivatives must have only polynomial growth as well (we're talking about tempered distributions here). This was supposed to be just an informal exposition, where "function" meant "function, satisfying the appropriate technical conditions". > so that the product of g and >\phi is infinitely differentiable. Then the formula that you give makes >sense. Otherwise, I don't think it makes much sense. The point is e^x is >analytic, so it's fine to multiply distributions by it. If you insist on getting technical no that's not the point, or at least not the only one. In fact you _can't_ multiply (tempered) distributions by e^x and get (tempered) distributions back. (And yes we're talking about tempered distributions; note the "rapid decent" at the start and also that we're taking Fourier transforms - arbitrary distributions on R don't quite have Fourier transforms.) The point is that e^(ix) is infinitely differentiable, and it and all its derivatives are bounded. ============================================================================== From: Wilbert Dijkhof Subject: Re: Distributions...am I missing the point? Date: Fri, 20 Oct 2000 22:28:30 +0200 Newsgroups: sci.math Rob Brownlee wrote: > > I let S denote the space of infinitely diffrentiable continuous functions of > rapid decent - endowed with the usual topology. A continuous linear > functional on S is called a (tempered) distribution. I let [f,\phi] denote > the the action of the distribution f on the test function \phi. > > I'm getting confused with how the author of the material I am reading > interchanges between refering to f say as both a function and a > distribution. > For example.... > I define T_x as the translation operator (T_x \phi)(y)=\phi(y-x). I define > (e_x)(y) via exp(ixy)). Then I can see that for all f in S we have (T_x > f)^=e_-x f^ but the author claims this is true for all distribution aswell. > > Do I define [f^,\phi]=[f,\phi^]? and [T_x f,\phi]=[f,T_-x \phi]? I still > can't see why the above is true. Yes. Let me explain why: Suppose g is an element of L^1(R^n) then you can prove that [g^,phi] = [g,phi^]. So to be consistent we define the Fourier-transform of a tempered distribution f as [f^,phi] = [f,phi^]. So if [f,phi] = [g,phi] for some g in L^1(R^n) (with f a tempered distribution) we have [f^,phi] = [g^,phi] = [g,phi^] = [g^,phi]. It's the same with the translation operator T_x. If g is an element of L^1(R^n) then you can prove that [T_x g,phi] = [g,T_-x phi]. So to be consistent we define it for distributions in the same way. One thing remains, can this definition be extended to the Fourier-transform of an arbitrary distribution? I don't know. Wilbert ============================================================================== From: Robin Chapman Subject: Re: Distributions...am I missing the point? Date: Mon, 23 Oct 2000 17:50:22 GMT Newsgroups: sci.math In article <8t1dr1$n732$1@rook.le.ac.uk>, "Rob Brownlee" wrote: > I am still a little uneasy with the apparent disregard the author of the > paper I am reading swops between distributions and functions. For example he > lets Y denote the set of all distributions whose Fourier transform is L^2. > How can we regrad a distribution as an L^2 function? You can't in general, but you can regard L^2 functions as distributions. Each L^2 function gives rise to a distribution, and two L^2 functions give rise to the same distribution iff they are equal almost everywhere. We can thus regard L^2 as a subspace of the sace of distributions. > Do you believe he means > for Y to consist of all those distrubtions arising from funcitons whose > Fourier transforms are L^2? By arising I mean that we define [f,phi]=int > f(x) phi(x) dx. Not quite. Y consists of distributions f whose Fourier transforms f^hat lie in the subspace L^2 of the space of distributions. That is f^hat = g for some L^2 function g. (Anyway isn't this Y the same as L^2 .... ?) Anyway, you should be lucky you aren't reading the physics literature, where they do regularly fail to distinguish between functions and distributions, talking about delta *functions* and worse (shudder :-() -- Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html "`The twenty-first century didn't begin until a minute past midnight January first 2001.'" John Brunner, _Stand on Zanzibar_ (1968) Sent via Deja.com http://www.deja.com/ Before you buy.