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Topic: Question of generalized function definition
Replies: 7   Last Post: Mar 23, 2012 9:33 AM

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 David C. Ullrich Posts: 21,553 Registered: 12/6/04
Re: Question of generalized function definition
Posted: Mar 23, 2012 9:33 AM

On Thu, 22 Mar 2012 18:25:25 -0700 (PDT), vv <vanamali@netzero.net>
wrote:

>On Mar 22, 6:56 pm, David C. Ullrich <ullr...@math.okstate.edu> wrote:
>> Anything one can do with "generalized functions"  one can do
>> without them. But one can say the same about _any_
>> mathematical construction! There's no need to introduce
>> the concept of X, with a definition; instead of defining
>> an "X" to be a so-and-so one could always talk about
>> so-and-sos.

>
>Thanks for taking the time to explain these nuances.
>

>> The best example of why we care about these things
>> is from Fourier analysis. Keep reading...

>
>Will do!

You caught me by surprise yesterday, switching the question
to why we care about generalized functions.

Here's one of my favorite examples, which has the virtue
that it can be completely explained with no need for
any high-powered background in analysis:

Consider convex functions (what calculus books tend
to call "concave up" - mathematicians say "convex"

Say f : R -> R. _If_ f is twice differentiable
_then_ f is convex if and only if f'' >= 0. Unfortunately,
we can't just leave out the "if f is twice differnentiable"
part, since a convex function need not be differentiable.

For example f(t) = |t| is convex, but not differentiable
at the origin. That function has f'(t) = 1 for t > 0 and
f'(t) = -1 for t < 0. So in a vague sort of way, the
sort of thing that people considered before Schwarz
got everything nice and systematic, a person might
say that f = |t| does have a second derivative, that
second derivative being 2 time a delta function at the
origin - saying f'' = 2 delta_0 says that f' should have
a jump equal to 2 at the origin, which it does.

Now here's the fun part. Once we've learned the theory
of generalized functions, _every_ continuous function
has a derivative, in the "generalized" sense, and that
derivative has a derivative in the same sense. And
a convex function _does_ have to be continuous; that's
not hard to prove.

And it turns out the following is true: Say f : R -> R.
Then f is convex if and only if f is continuous and
f'' >= 0 (where f'' is the generalized second derivative).

This is a much nicer thing than we get if we don't
know about generalized functions. It applies to _all_
convex functions. Given continuity (which as mentioned
above is no restriction here, as opposed to differnetiability,
which does rule out most convex functions), convexity
is equivalent to having a non-negative second derivative,
period.

>--vv

Date Subject Author
3/21/12 Vanamali
3/21/12 David C. Ullrich
3/21/12 Vanamali
3/21/12 David C. Ullrich
3/21/12 Vanamali
3/22/12 David C. Ullrich
3/22/12 Vanamali
3/23/12 David C. Ullrich