If we consider the Gaussian or the Lorentzian representation of $\delta$, then we obtain $\newcommand\dif{\mathop{}\!\mathrm{d}}$ \begin{equation} \lim_{\epsilon\to 0}\int_0^\infty \delta_\epsilon(x)\dif x=1/2 \end{equation} but as far as I understand the problem is that the equation is not true for other nascent dirac deltas (i.e. sequences of functions such that the associated sequences of distributions converges to $\delta$). Can we find a nascent dirac delta such that 1) the limit does not exist and 2) the limit exists and does not equal $1/2$?
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1For starters on 2), make the function not balanced around $0$, or not even. P.S. Distribution theory won't help, as the $\delta$ distribution is not "integrable." – Ted Shifrin Aug 07 '22 at 19:03
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1@filippo Pleased to see your question. – Mark Viola Aug 08 '22 at 21:17
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Hello Filippo ! Regarding your second question, consider the case of the Dirac delta in polar or spherical coordinates. By excluding contributions from negative $r$ (for obvious reasons), all contributions are from positive values. So the limit is equal to $1$. – M. Wind Aug 10 '22 at 15:03
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@M.Wind See THIS ANSWER, which provides a rigorous expose of the Dirac Delta in spherical coordinates in $n$-dimensions. – Mark Viola Aug 10 '22 at 18:32
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1@Filippo Thank you! I've edited with the correct reference. – Mark Viola Aug 10 '22 at 18:33
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1And (+1) for the posted question – Mark Viola Aug 10 '22 at 18:56
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The sequence $f_n$ of piecewise-linear tent functions of height $n$ and base width $2/n$, but centered at $1/n$ rather than $0$, approximates $\delta$, but they all have integral $\int_0^\infty$ equal to $1$. Symmetrically, centering the tents at $-1/n$ have those integrals all $0$. If we interleave the two sequences, they approach $\delta$ distributionally but their integrals oscillate between $0$ and $1$, so have no limit.
paul garrett
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Oh hah, I was making a comment along these lines. Deleted it for redundancy. Great answer as always! – Cameron Williams Aug 07 '22 at 19:15
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(+1) Hi Paul! It's always great to see you here. This answer is useful to those who have misconceptions about the meaningless funtional $\langle \delta, H\rangle$ which some wirite with abusive notation as $\int_{0}^\infty \delta (x),dx$. – Mark Viola Aug 08 '22 at 21:16
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By the same logic one may argue that in two dimensions $\delta(x)\delta(y)$ is meaningful, but if one changes from Cartesian coordinates to polar, there is a problem. Because the integral over $\delta(r)$ runs over the positive real axis. Which is abuse of notation. Apparently because it is possible that negative values of $r$ contribute to the integral. Even though negative values of $r$ are not defined. – M. Wind Aug 10 '22 at 14:36
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@M.Wind You continue to discuss "integrals" in the context of distribution theory. Distributions are linear functionals, not integrals. See THIS, which provides a rigorous expose of the Dirac Delta in spherical coordinates in $n$-dimensions. – Mark Viola Aug 10 '22 at 18:34
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@Mark Viola. Then why do you praise Paul Garrett when he speaks about "integrals" ? – M. Wind Aug 10 '22 at 18:43
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@M.Wind Paul is talking about the integrals involved in a nascent Dirac Delta and was addressing the questions in the OP. You are conflating things quite a bit. And I bet you never even read the answer in the link, did you? You are fixated on maintaining your flawed paradigm about the Dirac Delta, despite all evidence to the contrary. – Mark Viola Aug 10 '22 at 18:52
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@Mark Viola. When I speak about the Dirac delta I always mean the limit of a (suitable) nascent Dirac delta. That was precisely what the OP asked about and how Paul Garrett treated it. – M. Wind Aug 10 '22 at 19:01
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@M.Wind No, that is not always the case. In your comment 5 hours ago your wrote about a transformation of coordinates. I gave you a reference that discusses what this means in terms of the Dirac Delta. And any nascent Dirac Delta is "suitable" by definition of a nascent Dirac Delta. But the Dirac Delta operates on test functions that are in $C_C^\infty$. The prevailing issue is that $H$ is not in that space. Hence, one can take two nascent Dirac Deltas and produce different results when operating on non-test functions such as $H$. Paul gave examples. I hope this makes sense. – Mark Viola Aug 10 '22 at 19:47
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Nascent Dirac deltas are usually chosen to be symmetric. Sometimes that is not feasible. For example when one changes from Cartesian coordinates to polar, cylindrical or spherical coordinates. Then the domain of a variable gets restricted to the positive real axis. Obviously this affects the Dirac delta function.
For example consider the case of (normalized) Gaussian nascent deltas in two dimensions. We have $\delta_{\epsilon}(x) = 1/(\epsilon\sqrt(2*\pi))exp(-x^2/(2\epsilon^2)$ and $\delta_{\epsilon}(y) = 1/(\epsilon\sqrt(2*\pi))exp(-y^2/(2\epsilon^2)$. The double integral over the product of the two deltas is equal to unity. Now introduce polar coordinates. We obtain the following nascent radial Dirac delta: $\delta_\epsilon(r) = (r/\epsilon^2)exp(-r^2/(2\epsilon^2)$, valid for $r \ge 0$. It satisfies:
$$\int_0^\infty \delta_\epsilon(r)\space dr = 1$$
M. Wind
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