Dirac Delta Function

There are several equivalent definitions, one common one is lim[n->oo] sin(n*x)/pi*x. Another one is F[1] where F is the FourierTransform.

Doesn't seem to be a real function, though.

The Dirac Delta Function of x has the following properties:

• Infinite at x = 0,
• Zero at all other points,
• Integrates to a HeavisideStepFunction?.

Indeed, it isn't a real function. There's a big body of theory about these things, which are strictly called "distributions". The basic point is that you never really need to use it as a function; all you need to be able to do is multiply it by something that really is a function and integrate. So, er, really it is a function, but it isn't a function from {numbers} to {numbers}; it's a function from {functions} to {numbers}. The function is conventionally written as "integral from -oo to +oo of delta(x) * f(x) dx" instead of, say, "D(f)", but that's just to avoid upsetting the physicists. :-)

Well, the physicists thought of distributions first. A nice approach is to consider distributions being to functions (in this case, f:Real->Real) as 1-forms are to vectors. Then integrating the product of a distribution and a function is the equivalent of the inner product.

It turns out that 1-forms and vectors are both tensors (as are scalars, and tensors are then just linear functions of tensors), I don't know what distributions and functions from R to R both are (higher-order functions, I guess).

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Actually, the integral statement is not a function on the Reals either. This is often misunderstood. The theory of distributions uses sequences of functions that approach the 'delta function', and defines the resulting limit to be 'the delta function' even though it isn't a function. In fact, most of the useful applications of distributions result in things that are the limit of a sequence of integrable functions, but are not, themselves, in the space. In fact this is the whole point of the delta-calculus... From the outside (not being a physicist) it always seemed a bit of a cheap hack, to me --- if you work with Lesbegue integrals instead of R-S integrals, these sequences converge in the space so you don't have all these difficulties. I am sure I am just ignorant of some of the advantages; Any physicists here who can explain why they still like distributions?

Not having to learn another breed of integral springs to mind...

It is not so bad, really. L. integrable functions are a superset of R-S. integrable functions. The integral behaves the same way on the subset. Distributions give you a way to deal with certain convergence problems, under certain conditions. Measure theory gives you a way to work in a larger space, which doesn't *have* these problems in the first place. There are other benifits of measures, of course, that aren't addressed at all by distributions. What I really want to know is if the use of distributions is just momentum (i.e., people not wanting to learn a new, albiet cleaner and more powerful, theory), or if there are problems for which they are really advantageous.

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For theoretical physicists, Dirac delta functions are very useful in calculations. Once you're used to them, they're easy to use and think about: just a sharp, narrow spike with area 1. You treat them as functions (though they aren't) the same way you treat infinity as a number (though it isn't).

It's nice that a rigorous mathematical theory of generalized functions exists to explain distributions. Still, physicists generally ignore it, in the same fashion that they use calculus while ignoring RealAnalysis?.

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CategoryMath