What is in the distributional sense ?

[inlaid]

What is "in the distributional sense"?

What is "in the distributional sense"? How to comprehend it meanings?
What is the difference between "in the distributional sense" and "in the normal sense" ?

 

[Hurkyl]

It means you're taking the limit (or whatever) in the some space of distributions, rather than in some space of functions.

 

[HallsofIvy]

"Distributions", also called "generalized functions" are essentially functionals, that is, things that map functions to numbers.

We can think of regular functions as being "distributions" in this way: the function f(x) maps any function g(x) to the number \int_{-\infty}^\infty f(x)g(x)dx.

Another example of a distribution is the "Dirac delta function" (which is what started the investigation into distributions because it is NOT a function). The Dirac delta function, that P.A.M. Dirac used in quantum physics papers as the "exact position" operator, was defined in a very rough way as a function that was 0 for every x except x= 0 and was infinite at x= 0 in such a way that the integral \int_{-\infty}^\infty \delta(x) dx= 1.

Of course, there is no such "function" but calculations done with it worked! It could be formalized by thinking of it as the functional that, to every function f(x), assigned the value f(0). Then, in a rough sense, \int_{-\infty}^\infty \delta(x)f(x)dx= f(0).

One useful property that distributions have that functions don't is that every distribution is infinitely differentiable. Since we can think of functions as being as subset of distributions, a function may not have a function as its derivative but it can have a distribution, that is not a function, as its derivative. For example, the function f(x)= 0 if x< 0, f(x)= 1 if x\ge 0 has the Dirac delta "function" as its derivative. It is not differentiable, at x= 0, in the sense of functions, but it is differentiable "in the distributional sense".

 

[disregardthat]

Another example of a distribution is the "Dirac delta function" (which is what started the investigation into distributions because it is NOT a function). The Dirac delta function, that P.A.M. Dirac used in quantum physics papers as the "exact position" operator, was defined in a very rough way as a function that was 0 for every x except x= 0 and was infinite at x= 0 in such a way that the integral \int_{-\infty}^\infty \delta(x) dx= 1.

Can this be formalized in non-standard analysis? My suggestion would be to define \delta(x) on the non-standard reals by being 0 everywhere but on [-\frac{\sigma}{2},\frac{\sigma}{2}], where its value is \frac{1}{\sigma} for an infinitesimal \sigma. In which case \int^{\infty}_{-\infty} \delta(x) dx = \frac{1}{\sigma}(\sigma) = 1, by using the riemann definition of the integral.

If we extend the definition of f(x) on the non-standard reals as f(st(x)), then \int^{\infty}_{-\infty} \delta(x)f(x) dx = \int^{\frac{\sigma}{2}}_{-\frac{\sigma}{2}} \delta(x)f(x) dx = \int^{\frac{\sigma}{2}}_{-\frac{\sigma}{2}} \frac{f(0)}{\sigma} dx = f(0).

And so we can define \delta in terms of functions. Or is this nonsense?

 

[Hurkyl]

Sure. Every distribution is a limit (in the distributional sense) functions, so I expect the general machinery of NSA let's you represent a distribution as the standard part of integration against a non-standard function.

And the particular function you wrote is one of the many different non-standard functions that name the dirac delta distribution.

For example, the function

f(x) = \begin{cases} 1/\epsilon &amp; |x| \in \left[ \frac{\epsilon}{2}, \epsilon \right] \\<br /> 0 &amp; \text{otherwise} \end{cases}​

also names the dirac delta distribution.

 

[Hurkyl]

Incidentally, you normally extend a function to the nonstandard reals via transfer. I have no idea if the extension you used is sufficiently well-behaved. At the very least, it's a problem that it's not defined for transfinite values!

I expect you probably want to truncate the limit at a non-standard real number too.

 

[disregardthat]

At the very least, it's a problem that it's not defined for transfinite values!

Yes, I forgot that. But I guess we can leave it undefined for transfinite values if we define our improper integral to be the limit over only real values.

 

[Hurkyl]

Yes, I forgot that. But I guess we can leave it undefined for transfinite values if we define our improper integral to be the limit over only real values.

Now that one definitely doesn't make sense, since the set of finite hyperreal numbers and the set of real numbers is are both external sets.

Ah yes, that's the reason why f(st x) isn't usable too.

 

[disregardthat]

Now that one definitely doesn't make sense, since the set of finite hyperreal numbers and the set of real numbers is are both external sets.

What is an external set? Any interval in the non-standard reals (a,b) where a and b are real values is empty of transfinite values. Thus taking the limit as a and b tends to positive and negative infinity is well-defined. I don't see how f(st(x)) needs to be defined on the transfinite values.

 

[Hurkyl]

What is an external set? Any interval in the non-standard reals (a,b) where a and b are real values is empty of transfinite values. Thus taking the limit as a and b tends to positive and negative infinity is well-defined.

And that limit is the entirety of the hyperreal line. Don't forget that +\infty is bigger than all hyperreal numbers too!

I don't see how f(st(x)) needs to be defined on the transfinite values.

No, I mean f(st(x)) is an external function, so you can't do things like take integrals of it or what-not.

 

[Hurkyl]

"Internal" objects are those available to do analysis with.

e.g. every internal, non-empty subset of the hyperreal numbers has a least upper bound. External subsets generally do not -- e.g. the set of all finite hyperreal numbers.

Internal things come from the transfer principle. e.g. there is a type of object called "real-valued function of the reals", and so there is a corresponding type in non-standard analysis. This type consists of hyperreal-valued functions of the hyperreals, but it does not include everything that can be defined set-theoretically.

An internal function is a member of this type.

If you define it via transfer, it's automatically internal. Generally speaking, if you define something through arithmetic, calculus, analysis, or what-not, it's internal. Generally speaking, if you define it by comparing the standard and non-standard models (e.g. the notion of "finite" is such a comparison), then it is external.

 

[disregardthat]

Thanks for the explanation.

And that limit is the entirety of the hyperreal line. Don't forget that +\infty is bigger than all hyperreal numbers too!

Sure, but one can e.g. take the limit of f(n) as n ranges through integers to infinity, even though f(n) is defined on the reals. The definition I want to make is similar.

Suppose both \lim_{x \to \pm \infty} g(x) exists (over the reals), and g(x)=g(st(x)) for finite hyperreals. If an hyperreal number y is larger than any positive number, can we consistently define f(y) as \lim_{x \to \infty} g(x) (taken over the reals), and similarly for negative infinite hyperreal numbers?

 

[Hurkyl]

You certainly can define a function that way. But as I said, you won't be able to do things like integrate it.