# Show an identity - distribution

1. Apr 11, 2010

### fluidistic

1. The problem statement, all variables and given/known data
Show the following identity (in the sense of distribution): $$g(\bold x)\delta (\bold x)=g(\bold 0) \delta (\bold x)$$ for a function g.

2. Relevant equations
No idea.

3. The attempt at a solution
I don't have a concrete idea about what a distribution is (It's an assignment in my electromagnetism course and we've not been introduced the required math so I guess I should search in books, but I don't know which ones and the article http://en.wikipedia.org/wiki/Distribution_(mathematics) in wikipedia doesn't seem easy to understand for me).
From the identity, I get the idea that $$\delta (\bold x)$$ is 0 everywhere except in $$\bold x=0$$ but I'm not sure and even if it were so, I don't really understand why the equality holds.
Any help is greatly appreciated.

2. Apr 11, 2010

### Dick

If d1(x) and d2(x) are equal as distributions, that means the integrals of d1(x)*f(x)dx and d2(x)*f(x)dx are equal for any test function f(x). Can you show that?

3. Apr 11, 2010

### gabbagabbahey

Two distributions are considered equal if you get the same result integrating either over any region. What do you get when you integrate $g(\textbf{x})\delta(\textbf{x})$ over any volume not enclosing the origin? How about if you integrate it over any volume that does enclose the origin? Compare that to what you get when integrating $g(0)\delta(\textbf{x})$ in both cases.

4. Apr 11, 2010

### Dick

Not quite. I think you want to integrate against an arbitrary test function. Not just f(x)=1.

5. Apr 11, 2010

### fluidistic

Thanks to both,
Ah ok, I wasn't aware of this. I do not even know what a distribution is. I have in mind the Dirac's delta function. The integral (what kind of integral is required?!) not containing the origin of d(x)f(x)dx would be equal to 0 because d(x)=0 for all x different from x=0, right?
And the integral of d(x)f(x)dx would be equal to a constant if the origin is contained in the domain of integration... right?
I never dealt with this mathematically before. So if you know some books (for physicists better) explaining what a distribution is and how to manipulate them mathematically, I'd be glad.

6. Apr 11, 2010

### gabbagabbahey

I don't see why. If it is true for f(x)=1, it will be true for any other test function as well. Won't it?

In terms of probability distributions, my statements amounts to saying that if you calculate the probability of measuring $\textbf{x}$ within any region, you will get the same value for both distributions. How could this be true if the two distributions weren't equal?

7. Apr 11, 2010

### gabbagabbahey

When you write $\delta(\textbf{x})$, can I assume that you are referring to the 3-dimensional Dirac delta which is zero everywhere except the origin? If so, then you need to perform a volume integral, and yes, you will get zero for any volume that does not enclose the origin.

Right, but you can do better than just "a constant";

$$\int_{\mathcal{V}}g(\textbf{x})\delta(\textbf{x})d^3\textbf{x}=g(\mathbf{0})$$

for any volume $\mathcal{V}$ that encloses the origin.

For this type of problem, you may like to think of probability distributions. Integrating a probability distribution $p(\textbf{x})$ over a given region gives the probability of measuring $\textbf{x}$ to be within that region. If two such distributions give the same probability for any region, they can be considered equal.

8. Apr 11, 2010

### fluidistic

Ok thank you both for all.
No, the exercise doesn't specify the Dirac's delta function in 3 dimensions but only a "distribution function" I believe. I posted the exercise as it appears in the problem we've been given.

9. Apr 11, 2010

### gabbagabbahey

The identity you are asked to prove isn't true for just any old distribution $\delta(\textbf{x})$. It is true for the 3d Dirac Delta distribution however, and if this is for an EM course I'd go ahead and assume that it is indeed the Dirac Delta. (Are you by chance studying from Jackson? He uses the same notation.)

10. Apr 11, 2010

### fluidistic

Ok, I'll take note of this.
I just started the upper undergrad level electromagnetism course and I'm suffering a lot. I should be using a course I'll be taking next semester (mathematical methods used in physics). We're studying the course at Jackson's level and I've heard many of our exercises are taken from Jackson's book but I didn't check it out. So it's likely the same notation as Jackson, yes.

11. Apr 11, 2010

### gabbagabbahey

Section 1.2 in Jackson covers the Dirac Delta fairly succinctly, and he gives references for further reading on it too.

12. Apr 11, 2010

### fluidistic

Thank you so much, I'll take a look.

13. Apr 11, 2010

### Dick

Integrating over regions is sort of the same as saying they are the same for step functions. I'd rather do it for the usual class of test functions, where f(x) is CONTINUOUS with compact support. Otherwise you get into awkwardness like what is the integral of delta(0)*f(x), where f(x)=1 for x in [0,1] and 0 otherwise. For the same reason, I'd also be happier if the statement of the problem had said g(x) is continuous at 0.

14. Apr 11, 2010

### gabbagabbahey

I guess, but the regions can be made arbitrarily small.

Both of our methods are easy to apply to this particular distribution, I just thought mine might be more intuitively obvious/easy to picture.

15. Apr 11, 2010

### Dick

If f(x)=1 at x=0 and f(x)=0 otherwise, are you ok with saying integral f(x)*delta(0)dx=1?

16. Apr 11, 2010

### gabbagabbahey

Do you mean, am I okay with saying $\int_{-\infty}^{\infty} f(x)\delta(x)dx=1$? If so, yes. Should I not be okay with it?

17. Apr 11, 2010

### Dick

No, I don't think you should. $\int_{-\infty}^{\infty} f(x)\delta(x)dx$ is completely undefined. The only point I'm trying to make here is $\int_{-\infty}^{\infty} f(x)\delta(0)dx=f(0)$ only if f is continuous at 0. It's the continuous bit I'm missing here.

18. Apr 11, 2010

### gabbagabbahey

I don't see why. Could you show me a proof of that?

Keep in mind that this problem is posed on a physics assignment, so it is naturally lacking a certain amount of mathematical rigor. In context, I would assume that by "function", the questioner is referring to a continuous function.

19. Apr 11, 2010

### fluidistic

I just checked once again the exercise and it doesn't mention the word "continuous". Maybe it's ill posed.
Edit: and about the rigorousness of the mathematics, I believe I should be very rigorous. One of the professors to help us in this course is very rigorous. He proves everything, puts $$\vec 0$$ instead of just 0 when we're dealing with the null vector, etc.

20. Apr 11, 2010

### Dick

$\int_{-\infty}^{\infty} f(x)\delta(x)dx$ is just plain sloppy. You know that right? I don't need to prove it's undefined. delta(x) is 'infinite' at x. You are integrating an 'infinite' thing times f(x) for all x. You meant to write delta(0).