From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Functions with compact support
Date: 16 Feb 1999 04:00:20 -0500
Newsgroups: sci.math.num-analysis
Keywords: Topology of space of functions with compact support
In article <36C917AE.A79FD45F@meteo.fr>,
Valere wrote:
>Hi!
>
>I have to work with the set of smooth functions with compact support...
>and I'd like to know if it's a Banach space...
With questions like this, a norm needs to be specified before an answer
can be given. If the convergence of a sequence to zero is meant to be
"uniform convergence of all derivatives, and all supports of these
functions can be included in the same compact set", then this vector space
is not even metrizable, much less normable, and much less normable with a
complete norm (to become a Banach space). Consult any Functional Analysis
textbook which has a chapter on distributions.
Cheers, ZVK(Slavek).
==============================================================================
From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Functions with compact support
Date: 16 Feb 1999 10:11:03 -0500
Newsgroups: sci.math.num-analysis
In article <36C94803.539B1D35@meteo.fr>,
Valere wrote:
>>
>
>It's rue that I have to be more precise.
>The norm is the norm that you suggest. But I'd like to have the answer for a
>bounded open set in IRn instead of all IRn.
I did _not_ suggest any norm; if anything, I suggested a (customary)
sequence of seminorms plus an extra restriction on the supports (which
prevents the space from being metrizable). If you drop that restriction,
the space becomes metrizable but not complete (if it were complete, it
would be called Frechet space, rather than Banach space. And a Frechet
space can fail to be locally convex, and can have only a trivial dual,
making many problems much harder (such as optimization).)
My imagination does not go far enough to give me an idea what the
completion would be. And the boundedness of the open domain makes no
difference: lack of completeness persists.