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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 $\delta(\cdot)$?2012-11-04
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    @thanks for your remark, I edited my question.2012-11-04
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    I asked the question because it's known that Dirac distribution cannot be represented by a locally integrable function.2012-11-04
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    @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?2012-11-04
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    Yes. I don't understand what you do next: what is $f_l(x)$ (why do you allow $f$ to depend on a parameter)?2012-11-04
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    @Davide Giraudo thank you very much to answering my stupid questions:( Actually, I only try to understand it better from the definition. May I think $\int_{\mathbb{R}^n} c \phi(x) \delta(dx)=c \phi(0)$?2012-11-04
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    Yes, that's right.2012-11-04
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    @Davide Giraudo so, what does happen if I had another function instead of constant? Let me say $f(x)=l\,x$.2012-11-04
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    More generally, if $f$ is measurable, it's integral with respect to $\delta$ is $f(0)$.2012-11-04
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    @DavideGiraudo what does $f(0)$ mean if $f$ is just measurable? $0$ might not be a Lebesgue point, right?2012-11-04
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    @Davide Giraudo thank you very much for your help I finally got it:(2012-11-04
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    @DavideGiraudo But Borel measurability doesn't imply continuity either...2012-11-04
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    I don't understand, why would we need continuity?2012-11-04

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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.2012-11-04
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    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)2012-11-04
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    Also: shouldn't you accept the answer, to mark this question as resolved? Not that I'm reputation-hungry or anything... ;)2012-11-04
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    sorry, I was excited to understand something:)I did it with some delay.2012-11-04
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    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.2012-11-04
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    Hmm... hadn't thought of that. Can he un-accept the question?2012-11-04
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    a mathematician(candidate) does not have to be always a "he":)2012-11-04