Re: differentiable completion of C([a,b])

From: Lee Rudolph (lrudolph_at_panix.com)
Date: 02/27/05 

Date: 27 Feb 2005 11:45:50 -0500

David C. Ullrich <ullrich@math.okstate.edu> writes:

>On Sun, 27 Feb 2005 13:16:33 +0000 (UTC), anonymous@mathforum.org
>(Albert) wrote:
>>
>> I would appreciate your help with the following
>> Can we define the differentiable completion of C([a,b])?, i.e
>>How can we embed C([a,b]) densely in a metric space in which each
>>function has a derivative?. I guess this would be analogous to
>>the metric completion: We have C[a,b], and we want to find
>>X such that C[a,b] is dense in X , and every function in
>>f has a derivative.
>
>This is obviously impossible, since not every function in
>C([a,b]) has a derivative.
>
>This is _so_ obviously impossible that you must have meant
>something else. ???

I got imaginative, and decided maybe he meant the following:
is there a topological vector space V (with the topology coming
from a metric, if possible) containing C([a,b]) as a dense subspace,
and an extension D from V to V (or at least from C([a,b]) to V)
of the differentiation operator from the subspace C_1([a,b]) of
functions with continuous derivatives, such that D is a continuous
operator on V (or at least on C([a,b]))?

Now, isn't that something you distribution-heads can cobble together
out of a few pieces of old string and some bubblegum? Seems like
it oughta be. Omit hypotheses added by me as necessary until it's
possible, and the result may be what Albert wants.

Lee Rudolph

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