The Distribution of Delta Functions

[mhill]

let be the distribution

D^{m} \delta (x-a) D^{k} \delta (x)

my questions are , what happens whenever x=a or x=a ??

is this identity correct

\delta (x-a) = e^{-a D} \delta (x)= \sum_{n=0}^{\infty}(-a)^{n} \frac{D^{n}}{n!}\delta (x)

 

[jostpuur]

I don't understand the first question, but this

is this identity correct

\delta (x-a) = e^{-a D} \delta (x)= \sum_{n=0}^{\infty}(-a)^{n} \frac{D^{n}}{n!}\delta (x)

is correct if you using such test functions that the Taylor series work for them. It can be checked with integration by parts. If the Taylor series don't converge for the test functions, then this distribution thing stops working as well.

 

[Hurkyl]

what happens whenever x=a or x=a ??

This doesn't make sense to me. What meaning did you intend?

 

[Redbelly98]

Probably meant to say x=a or x=0, since the expression contains the terms δ(x-a) and δ(x-0).

 

[Hurkyl]

I meant that the idea of "plugging in" a value for x doesn't appear to make sense in this context.

 

[mhill]

the idea is

D^{m} \delta (x-a) D^{k} \delta (x) however the value

D^{m} \delta (-a) D^{k} \delta (0) and D^{m} \delta (0) D^{k} \delta (a)

is not defined since delta functions are just oo

another question , how would we define \int_{-\infty}^{\infty}dx D^{m} \delta (x-a) D^{k} \delta (x)

also , under suitable test function f , then

< f | \delta (x-a) > = \sum_{n=0}^{\infty} (-a)^{n} \frac{ < \delta | D^{n} f>}{n!}

although it would make no sense , i think we could say

(2\pi ) i^{m}D^{m}\delta (0) = \int_{-\infty}^{\infty}dx x^{m} which is divergent... although in Cauchy's principal value the integral should be 0 for m Odd