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



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

> 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. 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.

> One more point: the identity delta[f,x_0] = f(x_0) (for delta as
> defined above) does not hold for all classes of f and all points x_0 -
> for example, for locally integrable functions on R, you can only
> guarantee that the identity holds for almost all x_0. You can actually
> use the identity as defintion if you restrict yourself to continuous
> functions.

I don't know what your notation delta[f,x0] is supposed to mean, you
didn't introduce or explain it. However, with the integral definition
from above you always have a well defined integration behaviour,
because f(x) is defined at every point of the integration domain. And
that IS the property you want distributions to have, nothing else. All
derived properties are in then additional features. And if those don't
hold for all functions then they're not fundamental properties of
distributions. For example the properties of delta'(x) (the
'derivative' of delta(x)) are not defined for all functions that it
acts on, and thus it's only a distribution on a reduced function space.

Distribution theory may appear counter intuitive to some people, but
it's backed up very well by hard mathematics. The problem is that you
can not introduce distributions on a general (rigged) hilbert space of
measurable functions using limits of function sequences, for the
reasons I listed above. That is why in modern distribution theory the
definition of the integration properties is the fundament and
everything else derived.

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.

Cheers,

Andy

.

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