# Why can,t we multiply two distributions?

1. Jun 29, 2006

### eljose

I think Schwartz proved that 2 distributions couldn,t be multiplied..but why?..if we had 2 delta functions then their "product" is:

$$\delta (x-a) \delta(x-b)=f(a,b,x)$$

so i have obtained the product of 2 Dirac,s delta considering that delta is a distribution is not this a contradiction to Schwartz,s proof.

2. Jun 29, 2006

### matt grime

Here's one heuristic as to why it is not valid. What is the integral of d(x-a)d(x-b)? It ought to be 1, possibly after scaling, but at the same time it ought to be, in some sens d(b-a) and d(a-b) 'cos deltas ought to pick out values.

You have defined a symbol that is the product of two deltas, you have not at all shown that this symbol even begins to make sense, anymore than I have shown how to divide 0 by o when I define a symbol X=0/0.

You might also want to integrate f(x)d(x-a)d(x-b) by parts as well, that might produce an intereseting result.

Last edited: Jun 29, 2006
3. Jun 29, 2006

### Hurkyl

Staff Emeritus
No you haven't.

You haven't given all the details, so I can't tell what mistake you made. So I'll list the two most likely possibilities:

(1) You simply wrote down a formal product. You've not said what that product is, or even given a reason why it should exist!

(2) You've not multiplied distributions; you've simply convolved them.

4. Jun 29, 2006

### eljose

I have used the "usual" multiplication...2x3=6 and

$$\delta (x-a) x \delta(x-b)$$

The product is:

1 if x=a and a<b or b<a, or if x=b a<b or b<a
0 if x is different from a or b
oo if x=a=b

5. Jun 29, 2006

### matt grime

And what do you think that symbol does? Why has an extraneous x appeared in it?

So, your now object is either:

The dirac delta, probably, if a=b, and if a=/=b then it is a function that that is zero everywhere except at a and b when it is 1.

Is that consistent?

Have you tried to understand the exact statement of Schwartz's theorem?

Further you should understand the even if this is acceptable, one example that shows it can be done does not prove that it can always be done. You are being disingenuous by not stating Schwartz's result in full.

Last edited: Jun 29, 2006
6. Jun 29, 2006

### Hurkyl

Staff Emeritus
Wait, let's start from the very beginning...

Just what do you think the dirac delta distribution is? Your posts make it sound like you're talking about something different: you seem to be talking about the function $\delta:\mathbb{R} \rightarrow \mathbb{R}$ defined by:

$$\delta(x) := \begin{cases} 0 & x \neq 0 \\ 1 & x = 0 \end{cases}$$

but this function is not the dirac delta distribution.

Last edited: Jun 29, 2006