# Dominate Convergence Theorem for the Dirac delta function

1. Jan 1, 2010

### friend

I'm trying to understand the multiple limit processes involved with the Dirac delta function. Does it matter which process you do first the integral or the delta parameter that approaches zero?

The closest theorem I found that addresses the order of taking limits is the Dominate Convergence Theorem, seen here at:

http://en.wikipedia.org/wiki/Dominated_convergence_theorem

But I'm not sure it applies to the Dirac delta function since there doesn't seem to be a g(x) for which |D(x)|<g(x) for all x, since D(0)=>00.

This doesn't mean, however, that there is not some other proof that integration commutes with the parameter's limiting process for the Dirac delta function, right?

2. Jan 1, 2010

### mathman

The Dirac delta function is a classic example of why you need the dominant convergence theorem. If you put the limit under the integral sign of an approximate delta function, you can easily get the wrong answer!

Illustration: Let f_n(x) be defined as follows:

f_n(0)=0, f_n(1/n)=n, f_n(2/n)=0 f_n(x)=0 outside interval (0,2/n). Define the function in the interval (0,2/n) using straight lines connecting the points. The net result is a triangle with integral=1.
Now let n->∞, f_n(x)->0 for all x, but integral limit is 1.

3. Jan 1, 2010

### friend

So is the Dirac delta an example where the order in which to take the limits is required, that you have to do the integration first before you take the limit of the delta parameter? I'm not sure I've ever seen that requirement mentioned explicitly, but it does seem like the only necessary option.

And I thought that if the answer depends on the order in which you take different limits, then that means the process is "undefined". So is the Dirac delta "undefined"?

4. Jan 2, 2010

### HallsofIvy

First the "Dirac Delta function" is NOT a function, it is a "distribution" or "generalized function". But it is not at all uncommon for different kinds of limits to be "non-commutative", that is to depend upon the order in which they are taken. That is, in fact, one of the major problems of Analysis. You may be thinking about the fact that if you take the limit "as x goes to a and then as y goes to b" of a function f(x,y) and find that you get a different result if you "take the limit sa y goes to b and then as x goes to a", then "the limit" does not exist. But that is not just a matter of changing the order. In order for the limit to exist, you must get the same thing approaching (a,b) along any path. Not the same thing as just changing the order of the limits at all.

5. Jun 20, 2010

### Klockan3

That series of functions do not converge strongly so its limit don't exist. Weak limits don't behave properly so that don't really say much.

6. Jun 20, 2010

### DrRocket

To amplify a bit on what HallsofIvy told you:

1. The "Dirac delta function" is not a function in the usual sense. You can make sense of it as 1) an atomic measure, which is a function defined on certain sets 2) a Schwartz distribution which is a function defined on a topological vector space of test funtctions -- the sace of infinitely differentiable functions having compact support with a somewhat complicated topology , 3) a tempered distribution, which is a function defined on the infinitely differentiable functions that go to zero faster than polynomially, again with a somewhat complicated topology, or 4) an identity adjoined to the Banach algebra of $$L^1$$ functions on a locally compact abelian group.

2. The limiting operation typically used in physics and engineering texts to "derive" the Dirac delta function is simply invalid. That "derivation" is pure hand waving, and is more than a bit misleading. A sequence of functions of integral one, that approaches zero almost everywhere in the limit is sometimes used in rigorous analysis in the form of what is called an "approximate identity". This the correct way to do what is done improperly in the heuristic derivation of the Dirac delta function. It turns out to be a useful concept in the study of convergence of Fourier series, -- see for instance the discussion of the the Dirichlet kernel and the Fejer kernel in Hoffman's book Banach Spaces of Analytic Functions.

3. The hand waving to which you allude is a good example of why one needs something along the lines of either the monotone convergence theoryem or the dominated convergence theorem before iinterchanging limit and integral operations -- one runs the distinct danger of making fallacious conclusions otherwise.

7. Feb 18, 2011

### newbee

None of the above comments addressed the issue which is: When can one interchange a limit operation with an integration when the integrand involves a Dirac delta? Google Parseval's Theorem using Google Books and you will find many texts which do just this sort of thing and they do it without any justification. The original author of this thread mentioned that none of the convergence theorems seems to justify such an interchange. So the questions are: Can this interchange be justified? If yes, then how?

If the answer is that the interchange is not justified then the aforementioned proofs of Parseval's Theorem are wrong. This is not to say that Parseval's Theorem is wrong (It isn't) but that a frequently given derivation is wrong and a correct derivation is to be found elsewhere.

If you are interested in this question you may want to look here for enlightenment:

http://math.nyu.edu/student_resources/wwiki/index.php/Category:Dirac_Delta_Type_Integrals

Last edited: Feb 18, 2011
8. Feb 19, 2011

### DrRocket

Viewed as an atomic measure the dominated and monotone convergence theorems of measure theory (and other related theorems) apply to the Dirac delta.

9. Feb 19, 2011

### newbee

Dr. Rocket

What do you mean by "atomic measure"? Is this an accepted mathematical term?

10. Feb 20, 2011

### DrRocket

http://en.wikipedia.org/wiki/Atom_(measure_theory [Broken])

Last edited by a moderator: May 5, 2017