6/1     Prove that gaussian is the only function family closed under
        convolution. ie, forall f,g in H, exists d in H st d=conv(f,g) ===> H
        is gaussian family.
        \_ The difficulty of the problem hinges on the definition of
           'function family.'  I could, for instance, come up with many
           'function families' of one element each for which the above would
           hold.
           \_ The difficulty does not hinge on the definition of the
              function family in the trivial way you suggest; it is deeper.
              There exist uncountable sets of functions H
              such that for all f,g in H, f * g is always in H.  One
              example of such a family is the set of sinc functions
              A sinc (w x) with w in [1 2] and A being any real number.
              This was what I was trying to point out below.  -emin
              \_ Right.  The definition could be made too narrow (to make
                 closure easy), or too wide (for instance by making H the
                 set of all continuous functions on R), which would also
                 make closure easy.  The point, in both cases, is that
                 we are missing the definition of what a function family is.
                 My guess is that a function family is a set of functions
                 with the same finitely expressible closed form (finitely
                 expressible to avoid the fact that continuous functions on
                 R all have Taylor expansions).
           \_ Utter nonsense about complete set of real functions deleted.
              Take a math course, for crying outloud. -- motd math censor.
              \_ fuck you.
        \_ This isn't true.  Let f(x) = g(x) = sinc(x), then f * g = f.
           Also, if you let f and g be generalized functions
           then f(x) = delta(x) = g(x) results in f * g = f = g.
           I belive what you are thinking of is that if you restrict
           f and g to be probability density functions then Gaussians
           are the only pdfs closed under convolution.  Not sure how
           to prove this off the top of my head.  Please let me know,
           when you find out.  -emin
           \_ Depending on the minutia of the formal definition of PDF,
              a delta function may be considered a PDF as well (since
              the integral is defined to be 1....). Then, of course,
              delta can also be considered a Gaussian with 0 st dev.
           \_ yeah, sorry i meant pdf. i'm aware of the delta function case,
              but i guess i'd like to define "in the set of" as
              "approaches an element in the set H, in some sense". also,
              i think the proof i have in mind requires L2, which deltas
              aren't. so my proof was going to look something like:
              since it's a PDF, it must be the square of some other function.
              look at the fourier transform of the square of something, and
              realize that it must be closed under multiplication in the
              fourier domain. assuming that the function is in L2, that
              constrains it to a large class of functions, which includes
              boxes and exponential functions. then i was going to use
              the fact that this is the fourier of a SQUARE of a something
              to show that within the exponential function, it must be
              gaussian. i don't know yet how to get rid of the other families
              that are closed under multiplication.  -ali
           \_ man, do you always have a comment on everything, mr. fricking
              i know everything about anything?
                 -- alum who dislikes you intensely
                 \_ Disliking ability in others is the only form of evil
                    I know. -- Ayn Rand #1 fan
                     \_ having ability is one thing, discussing
                        topics like this here is another...go to some
                        math newsgroup, don't clog the motd with this
                        extraneous garbage.
                        \_ Oh bull-fucking-shit.  Any kind of math discussion is
                           vastly better than most motd discussions, and you know
                           it.  It's no crime to be stupid, but it's a shame to
                           be bitter too.
              \_ The motd has helped me out a lot.  Consequently I feel
                 a responsibility to give back and try to help answer
                 questions which I have a clue about.  I'm not trying to
                 show off.  If you have a suggestion how I can try and
                 contribute to the motd without offending you, please
                 let me know. -emin
                 \_ i don't think "alum" was referring to you. you typically
                    provide the kind of insight the rest of the CSUA is
                    completely incapable of providing. -ali.
                     \_  i'm talking about ali, not you emin.
                       \_ you stupid shithead. i'm the one who asked the
                          initial question. if you have a problem, feel
                          free to post your own moronic "how do i make
                          text bold using html" questions. -ali
                          \_ 'Take your math elsewhere' stupidity deleted, due to
                             its inherent idiocy.  -- motd math censor.