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I am trying to understand dirac delta better. We know that $\int_{\mathbb{R}^n} \delta(x) \phi(x)=\phi(0)$ where $\phi(x)$ is a test function and $\delta_{0}(x)$ represents dirac distribution. I have a function $\lim \limits_{l\rightarrow 0 }f(l)=c(constant)$. So, I think $\lim \limits_{l\rightarrow 0 }\int_{\mathbb{R}^n} \delta(x) f(l) \phi(x)=c\phi(0)$. What does happen, if I have such a function $\lim \limits_{l\rightarrow 0 }f_{l}(x)=\infty$, i.e like $f(x)=\frac{x}{l^2}$

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What is $\delta(\cdot)$? – Davide Giraudo Nov 4 '12 at 21:11
@thanks for your remark, I edited my question. – pcepkin Nov 4 '12 at 21:23
I asked the question because it's known that Dirac distribution cannot be represented by a locally integrable function. – Davide Giraudo Nov 4 '12 at 21:25
@Davide Giraudo then I should have defined as a measure and rewrited $\int_{\mathbb {R}}\phi(x)\delta(dx)=\phi(0)$ or didn't I get again? – pcepkin Nov 4 '12 at 21:40
Yes. I don't understand what you do next: what is $f_l(x)$ (why do you allow $f$ to depend on a parameter)? – Davide Giraudo Nov 4 '12 at 21:41
up vote 3 down vote accepted

I think your notation is at the very least misleading: $\delta (x)$ makes little sense, since it's a distribution, i.e. a functional on $\mathcal{D} ( \mathbb{R}^n)$. Although it's indeed quite common notation, handy for some manipulations, I'd rather write $\delta ( \phi) = 0$ or $\langle \delta_0, \phi \rangle_{\mathcal{D}' ( \mathbb{R}^n)} = \phi ( 0)$ if you like verbosity. Now, if your function $f_l$ is in $\mathcal{D} ( \mathbb{R}^n) = C^{\infty}_0 ( \mathbb{R}^n)$, you can define for any distribution $T$ the product $f_l T$ via $\langle f_l T, \phi \rangle := \langle T, f_l \phi \rangle$ and it obviously follows $( f_l \delta) ( \phi) = ( f_l \phi) ( 0)$. If $f_l \phi$ is not in $C^{\infty}_0$ then it cannot be an argument to $\delta \in \mathcal{D}'$ (although you can extend the domain of definition of $\delta$, of course).

In your example, and for some fixed test function, you can consider the delta as the limit of bumps it is, and apply each of these to the product $f_l \phi$ then see what happens in the limit, which may not even exist depending on $f_l$ and $\phi$.

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thank you answering my stupid question, I finally got it. – pcepkin Nov 4 '12 at 22:19
You are welcome. Also, I don't think not understanding something is stupid (I'd be totally depressed by now if I did...). I must cite here my favorite maths quotation: "Young man, in mathematics you don't understand things. You just get used to them." (Von Neumann) – Miguel Nov 4 '12 at 22:26
Also: shouldn't you accept the answer, to mark this question as resolved? Not that I'm reputation-hungry or anything... ;) – Miguel Nov 4 '12 at 22:31
sorry, I was excited to understand something:)I did it with some delay. – pcepkin Nov 4 '12 at 22:35
Actually, it's usually better to wait a while before accepting an answer. Question with an accepted answer often get less attention, and people may be discouraged to post alternative answers to such questions. – mrf Nov 4 '12 at 22:53

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