2
$\begingroup$

It seems like the Dirac Delta function is discontinuous as it has a value of $\infty$ at $x=0$ and $0$ everywhere else. It looks to be same as the Kronecker Delta Function, which we know to be discrete.

$\endgroup$
6
  • 5
    $\begingroup$ Continuous is relative to a pair of topologies. It is continuous as a function from $C^\infty_c$ to $\mathbb{R}$. $\endgroup$ – Ian Sep 2 '15 at 0:04
  • 1
    $\begingroup$ In what context do people call it continuous? $\endgroup$ – littleO Sep 2 '15 at 0:11
  • 3
    $\begingroup$ You should not think of the Dirac delta as a function of a real variable at all; the oft-quoted formula "$\delta(0)=+\infty,\delta(x)=0$ otherwise" has no real mathematical content. The valid arguments of the Dirac delta are test functions, not points. $\endgroup$ – Ian Sep 2 '15 at 0:21
  • 1
    $\begingroup$ Wikipedia is using the term "continuous" very loosely in that example. They are using it just to mean, like, "defined for all real numbers, as opposed to just defined on the integers". (But even that statement is not really correct, because technically the delta function is a distribution.) $\endgroup$ – littleO Sep 2 '15 at 2:23
  • 1
    $\begingroup$ Isn't the constant zero function continuous? $\endgroup$ – Carl Mummert Sep 2 '15 at 2:24
3
$\begingroup$

I think it has to do with the fact that continuity is implied by differentiability and integrability, and since the Dirac-Delta function is differentiable and integrable, it is continuous.

$\endgroup$
3
  • 1
    $\begingroup$ The Dirac delta 'function' is not a function on the reals. So calling it differentiable (or integrable really) is not honest. As a measure, it's a different story. Also, for real functions, differentiable implies continuous implies integrable (by whatever integral you want) is how the smoothness hierarchy looks. $\endgroup$ – Zach Stone Sep 2 '15 at 1:43
  • 1
    $\begingroup$ Thank you for clarifying. Yes I agree my answer was quite misinformed. $\endgroup$ – Ud779 Sep 2 '15 at 2:04
  • 1
    $\begingroup$ @Zach Stone: from another point of view, the Dirac delta function is indeed a function, albeit one that does not exist. I wrote about this at math.stackexchange.com/a/289749/630 $\endgroup$ – Carl Mummert Sep 2 '15 at 2:25

Not the answer you're looking for? Browse other questions tagged or ask your own question.