Dirac delta sequences 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?
 A: 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.
A: 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 $\phi$.
