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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.

Anirudh Kalla
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Jeff Strom
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    Continuous is relative to a pair of topologies. It is continuous as a function from $C^\infty_c$ to $\mathbb{R}$. – Ian Sep 02 '15 at 00:04
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    In what context do people call it continuous? – littleO Sep 02 '15 at 00:11
  • Wikipedia does at the end of this section. I can see why it might be considered to be continuous, but it doesn't quite make sense when there's only one possible value in the range – Jeff Strom Sep 02 '15 at 00:14
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    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. – Ian Sep 02 '15 at 00:21
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    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.) – littleO Sep 02 '15 at 02:23
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    Isn't the constant zero function continuous? – Carl Mummert Sep 02 '15 at 02:24

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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.

Ud779
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    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. – Zach Stone Sep 02 '15 at 01:43
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    Thank you for clarifying. Yes I agree my answer was quite misinformed. – Ud779 Sep 02 '15 at 02:04
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    @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 http://math.stackexchange.com/a/289749/630 – Carl Mummert Sep 02 '15 at 02:25
  • @ZachStone That depends entirely on your choice of foundations. There are plenty of topoi where a perfectly normal smooth function with the properties of the Dirac delta exist.

    It is dishonest to claim that it is has to not be a function based on the law of the excluded middle, when most physicists intuitively work in a way that is more consistent with smooth topoi than with set theory. If you do not assume that the predicate (x=0 or x!=0) has to be true, then it is consistent to assume that a Heaviside step function and its derivatives are all smooth.

     – saolof Oct 20 '22 at 19:41
  • (This is related to the fact that the real line is connected. Basically, it is not obvious that the real number line can be written as a union of two disjoint subsets, if you work in foundations where every set is really a space of some kind, which is much more relevant to physicist. In such foundations, Heaviside functions can be smooth.) – saolof Oct 20 '22 at 19:55