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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?

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Welcome to mathSE! Could you please tell us what have you tried? – rlartiga Apr 11 '14 at 18:19
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
up vote 4 down vote accepted

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: $\delta_n(x) \geq 0$, $\int_R \delta_n(x) = 1$, $\int_{-\epsilon}^{\epsilon} \delta_n(x) dx \to 1$ 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 $$E_n = \int_R \phi(x) \delta_n(x) dx - \phi(o) = \int_R (\phi(x)-\phi(0)) \delta_n(x) dx$$ using the second condition. Then $$|E_n| \leq \int_{\epsilon}^{\epsilon} |\phi(x)-\phi(0)|\delta_n(x)dx + \int_{R\setminus [-\epsilon, \epsilon]} |\phi(x)-\phi(0)| \delta_n(x) dx$$ or $$|E_n| \leq \sup_{x \in [-\epsilon, \epsilon]} |\phi(x)-\phi(0)| \int_{-\epsilon}^\epsilon \delta_n(x) dx + 2 \sup_{x \in R} |\phi(x)| \int_{R\setminus [-\epsilon, \epsilon]} \delta_n(x) dx.$$ 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.

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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. – nayrb 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. – nayrb Apr 11 '14 at 18:38
This sounds like a useful hint, thank you! – Ivica Smolić Apr 11 '14 at 18:52

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 $\phi$.

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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

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