From: Dan Kotlow<dank@micrologic.com>
Subject: Re: delta function (distribution)
Date: Sun, 03 Sep 2000 07:16:54 GMT
Newsgroups: sci.math
Summary: [missing]

On Sat, 02 Sep 2000 23:07:54 GMT, prussing@aae.uiuc.edu (John
Prussing) wrote:

>I don't see any difference. The derivative of a delta function (I forget
>its name, maybe doublet?) is a positive and negative delta
>function at the same point. 

That isn't right.  In distribution theory, the derivative of the delta
function (at zero) is the distribution which assigns to each smooth
function f with compact support the value -f'(0).  In integral
notation,

  Integral[ f(x) delta'(x) dx] = -f'(0)  
      [ = -Integral[f'(x)delta(x)dx]

In general, the derivative of a distribution is defined so as to make
"integration by parts" work, as above.

>The derivative of the rectangular pulse is
>a positive delta function, followed by a zero interval of length e, followed
>by a negative delta function. As e --> 0 one gets the same result.
>--

Yes, but not relevant.  If the rectangular pulse is to approximate the
delta function it must get tall and thin as e->0, keeping area = 1.

With respect to your original question, if delta_n -> delta, then 
f * delta_n approximates f for some pretty bad functions f (* denotes
convolution).  If the delta_n are smooth, then so are the
approximatants.  This has theoretical importance, and practical value
as well.  An example is the use of windowing functions in FIR filter
design.  When you multiply the coefficients of an ideal response in
the time domain, you are convolving the ideal response in the
frequency domain with a smooth function.  Smoothness in one domain
corresponds to fast decrease at infinity in the other domain.

-Dan
==============================================================================

From: "Fabrice P. Laussy" <laussy@nat.fr>
Subject: Re: delta function
Date: Sun, 03 Sep 2000 14:15:56 +0200
Newsgroups: sci.math

John Prussing wrote:

> Not only is delta not a continuous function, it is not a function (but
> is called one for convenience).

It isn't named a function in the theory of distributions. Some
functions are called distributions, and vice versa, for convenience,
they are the so-called regular distributions. Delta, which is a
singular distribution, is not termed a function, not even for
convenience. Only it is named a function (or generalised function)
*outside* of the mathematical theory.

> But it's not a functional.

It is. It is a functional defined on the space of infinitly derivable
functions which are non-zero only on a set of finite measure.

> A functional
> maps a function into a number.

Well, doesn't delta do just that? It maps f to f(0).

> Maybe you're thinkng of the integral of
> the product of a function and a delta function.

It means nothing in the mathematical theory. Integrals are a kind of
functionnals associated to locally sommable functions (i.e., which
that are integrable on every compact subsets). They are many other
ways to define such functionnals that have nothing to do with
integrals (delta is one instance, there is also the very important
finite parts of divergents integrals, related to Cauchy integrals, or
Hadamard finite values, and those are not directly related to
integrals).

> >Correct definition is that delta maps functions to the real numbers.
> >delta: f -> R  such that delta(f) = f(0)
>
> Not true. The integral over x of f(x)*delta(x) = f(0).

No, he's right. Correct definition is that delta maps functions f
(only those which are infinitly derivable and are non-zero only on a
compact subset, however) to f(0). In the theory, we write:

<delta, f> = f(0)

So as to your first question, one *can* define delta as the limit of
non-continuous functions. Nothing is lost indeed. One can always
regulate a non-continuous function. However, they have to be
continuous at 0 (and satisfy other obvious properties: their integrals
must be defined and equal to 1, etc...). But one should pay attention
that it isn't a convergence of functions (they don't converge, in the
functionnal way). It's a convergence of distributions. In fact, the
regular distributions associated to those functions, the very one we
do call in the theory "functions", for convenience.

F.P.L.

[reformatted --djr]
==============================================================================

From: "Fabrice P. Laussy" <laussy@nat.fr>
Subject: Re: delta function
Date: Mon, 04 Sep 2000 05:19:06 +0200
Newsgroups: sci.math

John Prussing wrote:

> In <39B240FC.F764970C@nat.fr> "Fabrice P. Laussy" <laussy@nat.fr> writes:
[deletia --djr]
> > Only it is named a function (or generalised function) *outside* of
> >the mathematical theory.
>
> That's what I was talking about.

Okay, but then, delta *is* continuous (as a distribution ;)

> >Well, doesn't delta do just that? It maps f to f(0).
>
> Yes, but only after you multiply f(x) by delta(x) and integrate over
> an interval that includes x = 0.  It seems to me that I could also
> claim x^2 is a functional, because if you multiply it by a given f(x)
> and integrate between x = 0 and x = 1, the result is a real number.

Yes you're right. x^2 *is* a functional, and is a distribution too
(that is, a linear continuous functional over Schwartz's space D). It
is the distribution which associate to any phi in D the number:

  integral from minus infinity to plus infinity of x^2 times phi(x) dx

You can wonder if this is a number (integration goes from minus
infinity to plus infinity). But this is, because any phi in D is by
definition different from zero only on a compact set (so the
integration is, in fact, over a finite interval. Your between x=0 and
x=1 is not a property of the distribution, but of its argument). The
distribution x^2 is what we call (for convenience) a function, in the
theory. Now if you take, from your first post, your function in R
(real functions) defined as taking the value 1/e on a centered set
around 0 of length e, then the distributions associated to these
functions (i.e., the functionals which to any function phi in D
associate the integrals over [-e/2,e/2] of phi times 1/e), they do
converge (as distributions) towards delta. They don't converge as real
functions, but they do as distributions (named functions).

> >> Maybe you're thinkng of the integral of
> >> the product of a function and a delta function.
>
> >It means nothing in the mathematical theory. Integrals are a kind of functionnals
>
> Means nothing in mathematical theory? Perhaps we are talking about different
> things here, but this conversation is becoming tiresome.

Oh, I'm sorry if it does.

Now we're not talking about different things, just from different
point of view. It's true people sometime use delta under the integral
sign. But mathematically speaking, it means nothing. Functionals can
be defined by integration (as the distribution associated to x^2). But
some other distributions like delta can't be defined like this. You
can try, you can define some real functions which are zero everywhere
and which are taking greater and greater values at zero. They will
have a meaning as distributions, defined by the usual, integral way
(taking the integral of it with a function). But the trouble is, they
don't converge towards delta (they are, in fact, all the null
distributions, so they converge to 0). Thus the naive picture of delta
as something zero everywhere but at zero is--worst than
innacurate--wrong. It's not delta, it's zero (if something).

> >associated to locally sommable functions (i.e., which that are
> >integrable on every compact subsets). They are many other ways to
> >define such functionnals that have nothing
>
> Did I say that integrals are the only functionals? I don't think so.

No, but you seem to insist that delta should come under the integral
sign. As I said, one could attempt it this way, just that doesn't
work. One must try something else than integration.

> >No, he's right. Correct definition is that delta maps functions f
> >(only those which are infinitly derivable and are non-zero only on a
> >compact subset, however) to f(0). In the theory, we write:
>
> ><delta, f> = f(0)
>
> I think we are talking about different things. You seem to treat
> delta as an operator or transformation or mapping.
> But what is the point? I can define
> an operator (functional) Z(f) (Z for zero) such that Z(f) = f(0).
> I can also define O(f) = f(1), N(f) = f(9), etc. So what?

So you are doing it right. You have correctly defined three singular distributions (they are
all dirac distributions, however).

F.P.L.

[reformatted --djr]
==============================================================================

From: george.ivey@gallaudet.edu (G.E. Ivey)
Subject: Re:delta function (distribution)
Date: 2 Sep 2000 14:19:10 -0400
Newsgroups: sci.math

John Prussing wrote:
<There are definitions of a delta function (distribution) using smooth
<trigonometric functions that approach it in a limit.

<What is lost by the very simple definition of the delta function d(t)
<as the limit of a (non-smooth) rectangular pulse D(t) of duration e
<and height 1/e? d(t) can be defined as lim e --> 0 of D(t).  The 
<time integral of D(t) is 1 independent of e and therefore remains 1
<in the limit.

<Loosely speaking, since things are going to 0 or infinity in the
<limit, does smoothness matter? Perhaps I'm missing something.
<--

   The most common definition of distributions (also called 
"generalized functions") is that of Schwartz as functionals: that is,
functions which to each function in a given set.  The "physics"
definition of the delta function as a functions such that
integral(from -inf to inf) of delta(x)f(x) dx= f(0) is such a
definition. 
   The Polish mathematician Jan Mikusinski developed a definition
(which can be shown to be equivalent) in terms of sequences of
functions.  A sequence {f_n} of functions is called "fundamental" if
there exist some convergent sequence {g_n} and a positive integer k
such that the kth derivative of each g_n equals f_n (Notice that a
fundamental sequence is not necessarily convergent itself in the usual
sense).  We then say that two such sequences {e_n} and {f_n} are
"equivalent" if the intertwined sequence e_1, f_1, e_2, f_2, etc. is
also fundamental.  Finally, distributions are defined as equivalence
classes of fundamental sequences.  Basically, this is similar to
defining real numbers as equivalence classes of Cauchy sequences of
rational numbers.
   The quick answer to your question is that there is nothing lost in
using non-smooth functions such as the rectangular pulses to define
the delta function although some types of calculations might be easier
with smooth functions.