Rigorousness of this Dirac delta formula

1. Jun 7, 2015

ShayanJ

Is the following equation mathematically rigorous? How can you tell?

$\int_{-\infty}^\infty \delta(x-a) \delta(x-b) dx=\delta(a-b)$

Thanks

2. Jun 8, 2015

Svein

That's way outside my specialty, but:
1. $\delta (x-u) = 0$ for x ≠ u
2. $\int_{u-\epsilon}^{u+\epsilon}\delta(x-u)dx=1$ for any ε>0
3. Thus, if a≠b, $\int_{a-\epsilon}^{a+\epsilon}\delta(x-a)\delta(x-b)dx=0\cdot \int_{a-\epsilon}^{a+\epsilon}\delta(x-a)dx$ for ε<|b-a|
4. Symmetrical argument for (x-b)
5. Thus, if a≠b, the postulate is correct.
6. Assume a=b. Now the expression becomes $\int_{-\infty}^{\infty}(\delta(x-a))^{2}dx$.
7. Since $\delta (x-a) = 0$ for x ≠ a, obviously $(\delta (x-a))^{2} = 0$ for x ≠ a.
8. Now for the hard part: What about $\int_{a-\epsilon}^{a+\epsilon}(\delta(x-a))^{2}dx$? Is it equal to $\int_{a-\epsilon}^{a+\epsilon}\delta(x-a)dx$?
Anybody want to complete this?

3. Jun 8, 2015

ShayanJ

Oh...I fount a proof myself. Its actually very straightforward. Just substitute the equation $\delta(x-a)=\frac{1}{2\pi} \int_{-\infty}^\infty dk \ e^{ik(x-a)}$ in the integral and rearrange the resulting triple integral. Seems rigorous to me.
Actually thinking about the Svein's step 8 led me to this. I don't know why I didn't think about it before that!

EDIT:
It seems to me step 8 should be proving that $\int_{a-\epsilon}^{a+\epsilon} [\delta(x-a)]^2 dx=\delta(0)=\infty$!

Last edited: Jun 8, 2015
4. Jun 8, 2015

micromass

Staff Emeritus
It is not rigorous. You cannot multiply distributions like that. Things like $\int_{-\infty}^{+\infty} \delta(x)^2 dx$ have no meaning in the classical sense of distributions. While it is possible to extend distribution theory to include this case, this is by no means trivial (and these extensions are typically not very well-behaved).

Furthermore, something like $\delta(a-b)$ makes no sense outside an integral. And $\delta(0) = +\infty$ is most certainly incorrect.

5. Jun 8, 2015

ShayanJ

$\int_{-\infty}^\infty f(a) \left[ \int_{-\infty}^\infty \ \left( \int_{-\infty}^\infty e^{ix(k-a)} \ dx \right) \left( \int_{-\infty}^\infty e^{iy(k-b)} \ dy \right) \ dk \right] \ da$

A rearrangement of the inner integrals gives $\int_{-\infty}^\infty e^{ik(b-a)} \ dk$. I'm sure you're right about the product of distributions so it seems to me that either such a rearrangement or the equation $\delta(x-a)=\int_{-\infty}^\infty e^{ik(x-a)} \ dk$ shouldn't be applicable here for some reason.

6. Jun 8, 2015

Svein

I have thought about my last step, and I have come up with this:

Since $\int_{a-\epsilon}^{a+\epsilon}\delta(x-a) \delta(x-b)dx=0$ for |a-b|>ε, ε>0, we certainly have $\int_{a-\frac{1}{n}}^{a+\frac{1}{n}}\delta(x-a) \delta(x-a-\frac{1}{n})dx=0$ for all n (just choose ε<1/2n). Thus $\lim_{n\rightarrow\infty} \int_{a-\frac{1}{n}}^{a+\frac{1}{n}}\delta(x-a) \delta(x-a-\frac{1}{n})dx=0$ (just a tiny nagging doubt: Riemann integrals and limits are not necessary compatible).

7. Jun 8, 2015

micromass

Staff Emeritus
The more nagging doubt should be that $\delta(x-a)\delta(x-b)$ does not exist in the usual theory of distributions.

8. Jun 8, 2015

Svein

But - I know that it is possible to construct a function ƒ∈C, ƒ(x)∈[0, 1] with the following specification: Given real numbers a and b with a<b, ƒ(x)=0 for x≤a and ƒ(x)=1 for x≥b. And, of course, ƒ'(x)=0 for x<a and for x>b. Now, obviously ƒ'(x) will approach δ(x-a) as b→a.
Since ƒ is a function, multiplying two instances of ƒ' is allowable and my argument in post #6 holds with ƒa (=0 for x≤a, =1 for x≥a+(b-a)/3) and ƒb (=0 for x≤b-(b-a)/3, =1 for x≥b).

9. Jun 8, 2015

jasonRF

Of course I agree that the product of distributions is not defined int he usual theory of distributions, but the convolution of two distributions is (with some requirements on the supports of the distributions-compact support makes things work out nice I think). However, I thinkn that the intent of what the OP wrote is correct - if you convolve a delta function with a delta function you get another delta function:
$\delta_{x_1} \ast \delta_{x_2} = \delta_{x_1+x_2)}$ (or in usual engineering notation, $\delta(x-x_1)\ast\delta(x-x_2)=\delta(x-x_1-x_2)$).

The integral representation of the convolution of two deltas is of course inspired by the definition for nice functions: $\left( f \ast g \right) (x) = \int_{-\infty}^{\infty} dy \, f(y) g(x-y)$. Engineers will usually denote it $f(x) \ast g(x)$.

So while the original equation in the post may not be rigorous, if interpreted as a symbolic representation of the convolution it will work out fine in applications.

jason

10. Jun 14, 2015

δ(x) = limε→0 1/√ε exp(-x2/ε)

∫δ(x-a)δ(x-b)dx = limε→0 1/ε ∫exp(-(x-a)2/ε-(x-b)2/ε) dx

= limε→0 1/ε ∫exp(-2x2/ε+2(a+b)x/ε-a2/ε-b2/ε)dx

= limε→0 1/√ε exp(-(a-b)2/ε + O(ε) )
= δ(a-b)

11. Jun 14, 2015

pwsnafu

Dirac is an equivalence class of said functions (specifically delta sequences), so you need to demonstrate your result is independent of the choice of the function. So no that's not rigorous either.

Last edited: Jun 14, 2015
12. Jun 15, 2015

Why does that matter? You might say sin(x) is the equivalent class of all series definitions of sin(x). As long as one series converges you don't need to use the rest of them.

13. Jun 15, 2015

pwsnafu

There are results that hold for only specific choices of delta sequences. The most important example is $\delta^2$. Certain choices of delta sequences converge, other choices do not.

A more general theory is the sequential intrinstic product of Schwartz distributions. The idea is simple, if $\phi$ and $\psi$ are distributions then define $\phi \cdot \psi := \lim_{n \to \infty} \phi (\psi * \delta_n)$. It turns out that shrinking the set of allowable delta sequences allows you to multiply a larger set of distributions.

Last edited: Jun 15, 2015
14. Jun 29, 2015

Ben Niehoff

Remember that an integral is a limit, and you can't go interchanging the order of limits willy-nilly. Especially when, as in your case, they do not converge.

15. Jun 29, 2015

Lebesgue

Hello! I'm interested in learning something about distributions and how to give rigorous meaning to the Dirac function. Does anyone know about any good textbook?

16. Jun 29, 2015

micromass

Staff Emeritus
What math do you know?

17. Jun 29, 2015

Lebesgue

I have some knowledge in single and multivariable analysis, and a little bit of Lebesgue measure theory in [itex]\mathbb{R}^n[\itex]. For next year I'll be taking some probability and statistics theory subjects and maybe functional analysis.

18. Jun 29, 2015

micromass

Staff Emeritus
The notions of distributions depend quite a bit on functional analysis. So you should probably study that first. You don't want the books which don't assume functional analysis (such as Strichartz).

19. Jul 11, 2015

Cruikshank

Read Lighthill's Introduction to Fourier Analysis and Generalized Functions
Make sure to review Taylor Series and the Remainder theorem

20. Jul 11, 2015

jasonRF

Does Lighthill discuss convolution at all? I don't remember reading about it there, and I can't find it in the index or a quick perusal, but perhaps it is in there somewhere.