Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $f(x)=\frac{x}{l^2}$

share|improve this question
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
show 8 more comments

1 Answer

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 $\phi$.

share|improve this answer
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
show 2 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.