Thanks a lot for the clarification, that was very
helpful.
Does the anwer suggest that, for certain class of
functions (with more restrictions than bounded) the
"complicate convolution" get somewhat simplified? (I'm
not sure if the question has much sense, in which case
I apologize for that!)

G



--- Terry Tao <tao@...> wrote:

> --- In harmonicanalysis@yahoogroups.com, "giangy72"
> <giangy72@y...>
> wrote:
> > We know that the convolution product *of the
> fourier transform F
> of two given functions f
> > and g behaves well respect to the ordinary product
> . between
> funtions:
> > F^{-1}\hat{f}*\hat{g}=f.g.
> > Can we say something about F[min(f,g)]?
> > In the case of "crisp" functions- function which
> can only assume
> the value 0 or 1 the
> > convolution in the dual correspond to the product
> in the primal,
> which in turns correspond
> > to taking the mininum: can these be of any help
> for that?
> > Thanks!
>
> In principle there is a formula, but it is not all
> that pretty.
> If f and g are bounded, then one can approximate
> min(f,g) using
> the Weierstrass approximation theorem by some
> polynomial of f and g.
> This Fourier transforms into a complicated
> convolution between
> hat{f} and hat{g}. One can then take limits to
> obtain the exact
> Fourier transform of min(f,g).
>
> The above trick is occasionally useful. For
> instance it offers
> a Fourier-analytic proof that the minimum of two
> almost periodic
> functions is again almost periodic (though this can
> be done much more simply by a direct method), and
> gives a little
> information about the distribution of the spectrum.
>
> Terry
>
>
>




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