Is the Kronecker Delta Equivalent to the Dirac Delta as h Tends to 0?

[Klaus_Hoffmann]

I do not know if it is true but is this identity true

\frac{\delta _{n}^{x} }{h} \rightarrow \delta (x-n)

as h tends to 0 ?, the first is Kronecker delta the second Dirac delta.

i suspect that the above it is true but can not give a proof

 

[matt grime]

It isn't true in any meaningful sense - of course, since you've not defined what you mean by convergent, then we're going to have to guess. What does convergence of generalized distributions even mean? The most meaningful, and traditional notion of a sequence of functions converging to the Dirac delta is as follows:

we have functions f_r(x) for r in the natural numbers, normally, we require the integral of f_r over the real line to be 1 for all r, and for the sequence of numbers

y_r:=\int f_r(x)g(x)dx

to converge to g(0).

These are all trivially false in your case. It is true for suitably normalized Gaussians, for instance.

 

[Klaus_Hoffmann]

ok, thanks i thought it was true by the similarity of the results

\sum_{n=0}^{\infty} f(n) \delta _{n}^{2} =f(2)

(due to Kronecker delta only the value f(2) is obtained in the end)

and \int_{0}^{\infty} f(x) \delta (x-2)dx =f(2)

in the first case f(n) takes only discrete values, whereas in the second f(x) involves a sum over the interval (0, infinity) although the sum and the series give the same result f(2)

 

[lukluk]

actually I think the relation between Kronecker and Dirac's delta is correct as the first poster says, at least for physicists. See
http://bado-shanai.net/Platonic Dream/pdDiracDelta.htm
for the proof.