Discontinuity of Dirac Delta distribution

Question

I know that the following holds for Step function, but not sure if it holds for the distribution too.

Does the following hold for the Dirac delta distribution too?

$\delta$ is a linear functional from a space of test functions. The space is here taken from Schwartz space $S$ or the space of all smooth functions of compact support $D$. The Dirac distribution is differentiable everywhere $(-\infty, \infty)$ except at the point $x = 0$ where the function has a nontrivial jump discontinuity. This can be solved by removing the discontinuity of $\delta'$ by setting \begin{equation} \delta' = 0 \end{equation} which is continuous now on the entire line even though $\delta$ is not differentiable on the real line.

Answer 1

The text you quoted messes with different notions of differentiability and I can't make any sense of it.

Since the delta distribution is a linear functional on a space of test functions it is not a function on the real line and hence, it does not make sense to say something like "$\delta$ is differentiable on $(-\infty,\infty)$ except at $x=0$" since $\delta$ is not defined on that set.

When viewing $\delta:\mathcal{S}\to\mathbb{R}$ (linear and continuous with respect to the usual semi-norms on the Schwartz-space – or similar on the space of test functions), it makes sense to say that $\delta$ is continuous. To say that it is differentiable, one has to define the notion of differentiability for such objects (which is done by duality).

Answer 2

Suppose you have $\delta_0\in D'(\Bbb R)$ - Dirac delta distribution. If you take a test function $\phi\in D(\Bbb R)$, then we write by definition $$\langle\delta_0,\phi\rangle:=\phi(0).$$

You can also use the Heaviside function $H(x)$, $\langle H,\phi\rangle:=\int_0^\infty\phi(x)dx$.

Then again, you define the derivative of any distribution $T\in D'(\Bbb R)$ by setting $$\langle T',\phi\rangle:=-\langle T,\phi'\rangle.$$

Then, it's easy to show that $H' =\delta_0$. Even further $\langle\delta_0',\phi\rangle=-\phi'(0)$.

To repeat, all distributions are differentiable (in the sense of distributions, of course) and not all distributions can be represented by $L^1_{loc}$ functions.

Does this answer your question?

Answer 3

It is entirely possible to talk about local behavior of distributions (which does require proof: smooth partitions of unity, etc.). So, yes, $\delta$ itself is locally identically $0$ away from $0$. Its (distributional) derivative is locally identically $0$ away from $0$. But this does not at all dodge the issue of what happens at $0$. For example, a classic fallacy is to think in terms of measure-theoretic "almost everywhere" stuff, so, supposedly, since $\delta$ is almost everywhere $0$ (?!?) it is $0$. Btw, distributions do not pay attention to "almost everywhere"... :)