Is there anyone could give me a hint how to find the distributional derivative of the delta function $\delta$? I don't know how to deal with the infinite point.
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4First of all $\delta$ isn't a function but a distribution. So you need to talk about distributional derivatives. – May 26 '16 at 01:43
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I wouldn't think that the delta function is differentiable--clearly it's derivative is $0$ for most of the function but when $x = 0$, I would think there is a discontinuity and thus it isn't differentiable at that point--I don't know though. One would have to actually define it for me first. I think that a delta function, $\delta(x - b)$ is a Guassian function such that $lim_{c \rightarrow 0}$. – Jared May 26 '16 at 01:46
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1If we go on https://en.wikipedia.org/wiki/Distribution_(mathematics)#Derivatives_of_distributions, we could probably find the distributional derivation in using integration by part. However, I see again the notation $\delta ' (0)$. What is it mean exatly? – May 26 '16 at 01:51
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@J.Doe It's not $\delta'(0)$ it's $\varphi'(0)$. See my answer. – May 26 '16 at 01:52
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$\delta$ is NOT a function. $\delta$ is what's called a distribution. A distribution is a "linear functional" $\delta\colon C_c^{\infty}(\Bbb{R})\to\Bbb{R}$. Where $C_c^{\infty}(\Bbb{R})$ are infinitely differentiable functions that vanish outside of some interval.
We typically write (although this is absolutely an abuse of notation, this is not really an integral):
$$\delta(f)=\int_{-\infty}^{\infty} \delta(x)f(x)~dx$$
We define the distributional derivative through "integration by parts" (note the quotes. Since this is not really an integral integration by parts is not totally correct. This is heuristics). Meaning:
$$\begin{align*} \delta'(f)&=\int_{-\infty}^{\infty} \delta'(x)f(x)~dx\\ &=-\int_{-\infty}^{\infty}\delta(x)f'(x)\\ &=-f'(0)\end{align*}$$
Where the first to second line is integration by parts (and noting the boundary terms vanish).
So the derivative of $\delta$ is the linear functional $f\mapsto -f'(0)$.
To truly understand distributions you need to learn functional analysis. Also see distributions. I did abuse a lot of notation and was quite informal. Distribution theory formally requires functional analysis. Please see comments below for further clarifications.
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You haven't really explained that the integration in your $\delta(f)$ definition isn't really integration, but rather notation, since $\delta(x)$ isn't a function. – layman May 26 '16 at 01:53
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Also, you should mention that your integration by parts comment is just a heuristic, not rigorous. We can't integrate by parts if $\delta(x)$ isn't really a function. – layman May 26 '16 at 01:54
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3I'd mention something to the effect of "there is a large class of distributions given by integration against locally integrable functions; accordingly we often abuse notation by writing the application of a distribution which is not given by integration against any locally integrable function in the same way". – Ian May 26 '16 at 01:54
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@user46944 That's why I put "integration by parts" in quotes but if you want, I'll make it more explicit. – May 26 '16 at 01:55
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1@ZacharySelk Yes, please do make it more explicit. I had a lot of trouble first learning about the dirac distribution because of these abuses of notation, so I think it's important to be extra clear. – layman May 26 '16 at 01:56
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4(Cont.) This abuse of notation is productive because most things that you might want to do can be justified by passing to a sequence of nice functions that converges in an appropriate sense to your distribution, doing your manipulations on the functions in this sequence (on which operations like differentiation etc. are defined in the familiar way), and then taking a limit of the final result. – Ian May 26 '16 at 01:57
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@Ian Could you explain explicitly and rigorously in a "Answer" what you mean in your two previous comments? I think it should be convenient, interesting to understand well te concept of integration and distribution. Knowing that I have already a little background in the study of distribution. – May 26 '16 at 03:07
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@J.Doe Check out this question I asked about the dirac $\delta$ function. It received some very good answers: http://math.stackexchange.com/questions/1372858/question-about-the-dirac-delta-function – layman May 27 '16 at 00:15
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Upon request in the comments:
There is a large class of distributions which are given by integration against locally integrable functions. Specifically, given a locally integrable $f$ and a smooth compactly supported $g$, one can define $T_f(g)=\int_{-\infty}^\infty f(x) g(x) dx$. This leads to a common abuse of notation, where we write the same thing for distributions which are not given by integration against a locally integrable function. Thus we write things like $\delta(f)=\int_{-\infty}^\infty \delta(x) f(x) dx$, even though "$\delta(x)$" is meaningless by itself.
This abuse of notation turns out to be productive, because we can often define operations in distribution theory by passing to an approximating sequence, performing the operation on the approximating sequence (where it is defined the way we want) and then taking a limit. For example, if $T_{f_n} \to T$ and $f_n$ are smooth, then we can define $T'$ (the distributional derivative) to be $\lim_{n \to \infty} T_{f'_n}$. But with $T'_{f_n}$ we can really integrate by parts, which gives the formula $T'(g)=-T(g')$. Note that the result does not involve the approximation scheme, so we can just call this formula the "distributional derivative" and forget about approximating it.
The abuse of notation hints at trying such manipulations. Often people write them out without mentioning any approximation scheme. In this case the manipulations are called "formal", because they only pay attention to "form" without worrying about semantics. (This is the same use of the word as in "formal power series".)
Anyway, in the case of the Dirac delta, this procedure winds up telling you that $\delta'(f)=-f'(0)$ since by definition $\delta(f)=f(0)$.
Ian
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1I don't 100% agree that this is an abuse of notation, but more a new notation in itself, asking for being defined rigorously (unfortunately this is rarely done, and many people say "you can't write $\int_a^b T(x) \phi(x) dx$ with $T$ being a distribution" even when the meaning is obvious, e.g. when $\phi$ is supported on $]a,b[$) – reuns May 26 '16 at 12:23
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@user1952009 While there is some substance there (indeed, what I wrote above hints at some of that substance), overall I do think it should be treated as an abuse of notation. For instance, while "$\delta(x)=0$ whenever $x \neq 0$" can be translated into something with rigorous meaning, "$\delta(0)=+\infty$" really cannot. Indeed strictly speaking the "physicists' definition" of the Dirac delta is self-contradictory: there is a function which is $0$ everywhere except $+\infty$ at one point and its integral is zero. – Ian May 26 '16 at 12:26
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I didn't talk of $\delta(0)$, which obviously has no meaning. what I mean is that it is exactly the same when you write $\int_{[a,b]} f d \mu$ or even $\int_a^b f(x) d \mu(x)$. why do you think $T(x) dx$ is so different of $d \mu(x)$ ? – reuns May 26 '16 at 12:28
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(and everywhere zero except at one point is uncorrect, but everywhere zero except in the neighborhood of one point is : the idea is that there are some limits hidden in the notation, finally not so different of $\int_a^b g(x) dx$ which also has an hidden limit, or $\sum_{n=1}^\infty a_n$ ) – reuns May 26 '16 at 12:30
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@user1952009 The first one is fine, nothing that isn't an ordinary function looks like an ordinary function. The second one isn't great, but something of the kind is necessary when talking about multiple integrals in the Lebesgue framework. You may be taking the phrase "abuse of notation" as I use it more negatively than I intend. To me an abuse of notation is a tool that we should use carefully, nothing more, nothing less. – Ian May 26 '16 at 12:32
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yes then we agree, "abuse of notation = notation that we use but that we didn't define rigorously in every case yet" – reuns May 26 '16 at 12:35
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1@user1952009 That's part of it, but there is also a concern about it being misleading. But that is not necessarily a reason to throw it out, one can be careful to avoid being misled by misleading notation. – Ian May 26 '16 at 12:38
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@Ian Thanks a lot for your answer! If I understand well your comment, that means when we say that $\int_{-\infty}^{\infty} f(x) \delta(x) dx = \int_{-\infty}^{\infty} f(0) \delta(x) dx =f(0) \int_{-\infty}^{\infty} \delta(x) dx = f(0)$ it is pure abuse of notation, because normally we begin to compute the integral and then we evaluated the frontiers points of the integral. To be rigorous, I think we have to define the function
$$ b(x,\epsilon) = \begin{cases} 0 & |x|>\frac{\epsilon}{2}\ \frac{1}{\epsilon} & |x| \leq \frac{\epsilon}{2}\ \end{cases} $$
– May 26 '16 at 14:30 -
When the Dirac delta function appears inside an integral, we can think of the delta function as a delayed limiting process.
$$\int_{-\infty}^{\infty} f(x) \delta(x) dx := \lim_{\epsilon \to 0} \int_{-\infty}^{\infty} f(x) b(x,\epsilon) dx$$
Let $f(x)$ be a continuous function and let $F'(x) = f(x)$. We compute the integral of $f(x)δ(x)$.
– May 26 '16 at 14:30 -
$$\int_{-\infty}^{\infty} f(x) \delta(x) dx = \lim_{\epsilon \to 0} \frac{1}{\epsilon}\int_{\frac{-\epsilon}{2}}^{\frac{-\epsilon}{2}} f(x)dx$$ $$ \lim_{\epsilon \to 0} \frac{1}{\epsilon}[F(x)]_{\frac{-\epsilon}{2}}^{\frac{-\epsilon}{2}}=F'(0)=f(0).$$
Is it what you mean?
– May 26 '16 at 14:30 -
@J.Doe This is a way of doing the manipulations, yes. But the result does not depend on the regularization, and after using the regularization to prove/define something, you should not think in terms of it after that. In other words, $\delta(f)=f(0)$, there is no regularization to be seen here. – Ian May 26 '16 at 14:40