Re: How do you prove Delta(2*t)=1/2*Delta(t)

Andy,

I don't want to start anther Diracian thread (this is a comp.dsp joke)
- people have gone to jail over this issue here. Anycase, I did as you
suggested (went and looked at my favourite functional analysis book),
and 'lo, it was fun! I learnt something about distributions, a topic
that we never really covered in our functional analysis lecture.

Jazz wrote:
> Andor wrote:
> > Hi Andy,
> >
> > you are right. We should do away with the handwaving. See my comments
> > below:
> >
> > Jazz wrote:
> > ...
> > > Defining distributions by limits is not really elegant to say the
> > > least. A very common definition that allows for a simple proof of the
> > > statement would be:
> > >
> > > Integral_-infty^infty f(x) delta(x) dx := f(0)
> >
> > Unfortunately, you find yourself back in handwaving domain again using
> > this defintion. It can quite easily be shown that such a function
> > delta(x) cannot exist.
> >
>
> Sorry, but that's not handwaving, that's exactly the point. You define
> a new class of objects called distributions that DO have this property.
> We're not talking about functions here. And there is no handwaving
> involved at all, that's a solid mathematical definition. If you don't
> believe me check your favourite book on functional analysis or
> distribution theory.

It is as you say. Let me recapulate what I understood from my read
yesterday evening (please correct me): First, you define the test
function space (usually, this is some well behaved space like the
smooth functions with compact support), call it D(E) (where E is the
domain of the functions you want to apply to functional to). It is
important to note that the functional depends on the test-function
space, even using the same mapping. Then you define how the functional
acts on the test-functions, for example, the functional l_g (where g is
some continuous complex-valued function on E) is defined as

l_g[\phi] = integral_E g(x)* \phi(x) dx =: < g, \phi > (*
denotes complex conjugate).

One verifies that this is indeed a linear functional. A functional that
can be written this way (using an integral over a continuous function)
is called a regular distribution.

Now, on the other hand, if you have some functional l, you wonder
whether there exists some (continuous) function g, such that l = l_g.
Specifically, you wonder whether there is some (continuous) function
delta(x), such the the Dirac delta functional can be written as an
inner product with delta(x). It turns out there is no such delta(x)
(continuous or otherwise) - such functionals are called singular
distributions. But you are stubborn, and say: "Well, what if I simply
_define_ the integral sign and the term delta(x), such that

DiracDelta[f] = integral_E delta(x)* f(x) dx holds?"

I think at this point you accept that the integral is just notation, as
I have not seen any formal definition of integrals over such
"generalized" functions (remember how Rieman and Lebesgue gave a formal
defintion and an exact recipe on how to compute an integral). The book
I read does mention that one can get such integrals to work by using
different measures than Lebesgue. For the Dirac delta, one can
integrate with the Lebesgue-Stieltjes measure induced by the Heaviside
function. But if you do that however, you throw away the idea of a
"generalized" function delta(x) (I quite like this approach).

> > I think a simple approach to define the Dirac delta functional using
> > integrals is as the following limit:
> >
> > delta[f,x_0] := lim_{n->infty} integral_{R} f(x) delta_n(x-x_0) dx,
> >
> > where R denotes the real numbers, and delta_n(x) := n, -1/(2n) < x <
> > 1/(2n), zero otherwise (or any other function sequence with similar
> > properties). The salient point is that one cannot exchange the limit
> > and the integral, as you correctly note. I think it is ok now to apply
> > your derivation below to show the required property (just insert the
> > limit in front of the integral).
>
> That definition has several problems. First of all, there is no real
> reason for chosing this sequence and not a different one with the same
> limit, say a gaussian with constant area and shrinking support.

One can state regulatory conditions on the function sequence such that
the above mentioned limit is well defined. Such sequences are known as,
surprise, delta sequences (by the way, a Gaussian delta sequence does
not have shrinking support).

> The problem with that is that you can construct functions (even lebesgue
> measureable real functions) that make the limit depend on the used
> sequence, or even worse, the limit not exist or even not have the
> integration property from above. So your definition is problematic, to
> say the least.

What is the reason that one wants an integral definition of the Dirac
functional? It would be very simple to just define DiracDelta[f] (I
use the square brackets to indicate a functional acting on its
argument, which is the function f) by way of

DiracDirac[f] := f(0), for any f continuous on, say, the real numbers.

Why bother at all with the integral definition? I think (this also
isn't mentioned anywhere in the book) the main point is that one wants
a defintion that works for continuous functions f, but that can be
extended to other function spaces (such as L^p) as well. After all, the
Dirac delta functions is the quantum mechanics favourite, and isn't L^2
quite important in quantum mechanics?

The main problem one faces in such spaces is that its elements (which
are not really functions, but function equivalence classes) are not
defined point-wise, ie. the expression f(0) has no meaning for an
element f in L^p, and therefore cannot be used for definitions of
anything.

To illustrate, consider the zero-function (call it f1) and the
characteristic function of the rational numbers (call it f2). Both
functions are defined on R (real numbers), and are integrable for any p
\geq 1. However, in the space L^p, these two functions belong into the
same equivalence class - there is no way you can differentiate the one
from the other, even though their value differs on a set that lies
dense in the real numbers. Specifically, f1(0) = 0 =/= 1 = f2(0).

So it's obvious that the functional (defined as the limit over a series
of standard integrals) might not have the property that DiracDelta[f] =
f(0), for above reasons. It is however a more general definition
(encompasses a larger class of functions), exactly for that reason.

> To make myself clear, that's how a mathematician sees things, you may
> well use sequences to visualize the action of distributions if it helps
> you understand. But as soon as you need mathematical rigor you won't be
> able to evade the integral definition.

And as soon as you need the Dirac functional in an L^p space, you
cannot evade the limit over integrals definition. My definition is
bigger than yours! (That was a joke :-).

>
> Cheers,
>
> Andy

Regards,
Andor

.

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