# Product of distributions

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The distributions are generalized by limits of finite sums:

$$d(f)=\sum_{k=1}^n d_k(f^k)$$

where the $d_k$ are simple distributions. Then, we can define a product of distributions by the product of the Dirac functions (as they are dense in the distributions):

$$\{ \prod_{i=1}^n \delta_{t_i} \} (f)=f^n(\sum_{i=1}^n t_i)$$

What are the properties of such a product?

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The product of distributions is associative, commutative, and distributive.
I explain that:
In your definition of the product of distributions, you have a summation over all possible products of Dirac delta functions.  So, for example, in the case of two distributions A and B, you would have:
A(f) * B(f) = sum over all products of Dirac delta functions of A and B
= sum over all products of Dirac delta functions of B and A
= B(f) * A(f)
So the product of distributions is commutative.
Similarly, for the case of three distributions A, B, and C, you would have:
A(f) * (B(f) * C(f)) = sum over all products of Dirac delta functions of A, B, and C
= sum over all products of Dirac delta functions of (A * B), and C
= (A * B)(f) * C(f)
= A(f) * (B(f) * C(f))
So the product of distributions is associative.
Finally, for the case of four distributions A, B, C, and D, you would have:
A(f) * (B(f) + C(f)) = sum over all products of Dirac delta functions of A and (B + C)
= sum over all products of Dirac delta functions of A, B, and C
= A(f) * B(f) + A(f) * C(f)
= (A * B)(f) + (A * C)(f)
= A(f) * (B(f) + C(f))
So the product of distributions is distributive.

answered Sep 29, 2022 by anonymous

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