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Is it true that any sequence of real functions $(\delta_n)_n$, such that $$\lim_{n\to\infty} \delta_n(x) = 0 \qquad \forall\,x\ne 0$$ and $$\int_{-\infty}^\infty \delta_n(x)\,dx = 1 \ ,$$ tends to a delta function, $$\lim_{n\to\infty} \delta_n(x) = \delta(x)$$ in a sense that $$\lim_{n\to\infty} \int_{-\infty}^\infty \phi(x)\,\delta_n(x) \, dx = \phi(0)$$ for any real test function $\phi$? If no, what else should one assume so that the sequence $(\delta_n)_n$ necessarly tends to a delta function?

Deeponjit Bose
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Ivica Smolić
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    Welcome to mathSE! Could you please tell us what have you tried? – rlartiga Apr 11 '14 at 18:19
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    It is unkind to the answers already given to change the question in a way that invalidates them. It would be more polite to append the question asking what else one would need to assume. – robjohn Apr 11 '14 at 18:42
  • You're right, I apologize! I've reformulated the question again, so that it respects the answers already given. – Ivica Smolić Apr 11 '14 at 18:50

2 Answers2

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No. For instance, take the function $\delta_n(x) = 1$ when $x \in [n, n+1]$ and $0$ otherwise.

EDIT: The following conditions will give the conclusion you want: $\delta_n(x) \geq 0$, $\int_\mathbb{R} \delta_n(x) = 1$, $\int_{-\epsilon}^{\epsilon} \delta_n(x) dx \to 1$ for all $\epsilon > 0$. I'm pretty sure other combinations of conditions will work, but I'm not sure what necessary and sufficient conditions are. That's actually an interesting question.

Here's the argument: Let $$ E_n = \int_\mathbb{R} \phi(x) \delta_n(x) \, dx - \phi(0) = \int_\mathbb{R} (\phi(x)-\phi(0)) \delta_n(x) \, dx $$ using the second condition. Then $$ |E_n| \leq \int_{\epsilon}^{\epsilon} |\phi(x)-\phi(0)|\delta_n(x) \, dx + \int_{\mathbb{R} \setminus [-\epsilon, \epsilon]} |\phi(x)-\phi(0)| \delta_n(x) \, dx $$ or $$ |E_n| \leq \sup_{x \in [-\epsilon, \epsilon]} |\phi(x)-\phi(0)| \int_{-\epsilon}^\epsilon \delta_n(x) \, dx + 2 \sup_{x \in \mathbb{R}} |\phi(x)| \int_{\mathbb{R} \setminus [-\epsilon, \epsilon]} \delta_n(x) \, dx. $$ By the last (and second) condition, $$ \limsup_{n \to \infty} |E_n| \leq \sup_{x \in [-\epsilon, \epsilon]} |\phi(x)-\phi(0)|. $$ Take $\epsilon \to 0$ and you're done.

Jon Warneke
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abnry
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  • OK, nice example! So, is there any way to "remedy" the attempt of the definition from the question, by putting some further constraint on the sequence $\delta_n$? – Ivica Smolić Apr 11 '14 at 18:25
  • My example shows one possible problem: the "mass" of $\delta_n$ can escape to infinity. What it is important is that $\delta_n$ becomes more and more concentrated near $0$. Another thing to consider is whether $delta_n$ needs to be positive. – abnry Apr 11 '14 at 18:28
  • There are examples which are not positive definite, such as $\delta_n(x) = \sin(nx)/\pi x$, so it seems that this "concentration near 0" is the property that has to be encapsulated in the definition of the sequence $(\delta_n)_n$. – Ivica Smolić Apr 11 '14 at 18:33
  • It turns out that for any $f(x)$ with $\int_R f(x)dx = 1$ that is nice enough (maybe just $L^1$?), $\delta_n(x) = (1/n) f(xn)$ works out. In fact, your example with $\sin x$ fits this property. Look up mollifiers. – abnry Apr 11 '14 at 18:38
  • This sounds like a useful hint, thank you! – Ivica Smolić Apr 11 '14 at 18:52
  • I was also wondering why one only finds that statement with positivity of the sequence, so much that I started to think that it was really necessary. In fact it is NOT: in your penultimate inequality, without positivity, you'll need to put absolute values. Then for the first term, you don't really need that the integral of the absolute value goes to 1, but simply that it is bounded. For the second, you will however need that the integral of the absolute value outside a neighborhood of 0, goes to 0 (which is not the case for the sinc function = Dirichlet kernel). – Noix07 Sep 06 '19 at 14:39
  • For the mollifier, the point is usually to have regularity properties of the convolution $\delta_n \ast f$. – Noix07 Sep 06 '19 at 14:40
  • nice answer .............. +1 – TShiong Dec 24 '23 at 10:37
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You could set have $ \delta_n(x) = n/2 $ if $x\in (0,1/n]$ or $x\in (1,1+1/n]$. Then $$ \lim \int_{-\infty}^\infty \phi(x)\delta_n(x)dx = \frac12(\phi(0)+\phi(1)) $$ for all continuous functions $\phi$.

daw
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    Thank you for all these nice counterexamples! I've edited the original version of the question in order to put focus on the quest for the successful version of the definition. – Ivica Smolić Apr 11 '14 at 18:41