Lighthills definition of the Delta distribution Lighthill (An introduction to Fourier analysis and generalized functions) defines $\delta(x)$ as a sequence of good functions $f_n(x)$ in the sense that
$$
\lim_{n\rightarrow\infty}\int_{-\infty}^\infty f_n(x)F(x)dx=F(0)
$$ for any function $F(x)$.
Lets assume you have a sequence of good functions $g_n(x)$ with
$$\lim_{n\rightarrow \infty}g_n(x)=0 \mbox{ for }x\neq0 \quad\mbox{ and } \quad\int_{-\infty}^\infty g_n(x)dx=1 \tag{1}$$
Follows from these assumptions that
$$
\lim_{n\rightarrow\infty}\int_{-\infty}^\infty g_n(x)F(x)dx=F(0) \tag{2} 
$$
or is there an additional restriction on the sequences $g_n(x)$ or $F(x)$ necessary?
To prove this I started along the lines of Lighthills book according to
$$
\left|\int_{-\infty}^\infty g_n(x)F(x)dx-\int_{-\infty}^\infty g_n(x)F(0)dx\right|=\left|\int_{-\infty}^\infty g_n(x)\frac{F(x)-F(0)}{x-0}xdx\right|\le \left|max(F'(x))\int_{-\infty}^\infty g_n(x)xdx\right| \tag{3}
$$
The last integral converges, but is the limit for $n\rightarrow\infty$ zero? Interganging limits might suggest that, as $xg_n(x)$ tends to zero for all $x$. But how to show that it is convergent (with respect to n) before interganging the limits.
I was thinking about that in the context of Bartons (Elements of greens functions) remarks on the strong definition of $\delta_L(x)$. From the above it would follow
for a sequence with
$$\lim_{n\rightarrow \infty}g_n(x)=0 \mbox{ for }x\neq0 \quad\mbox{ and } \quad\int_{-\infty}^0 g_n(x)dx=\alpha \quad\mbox{ and } \quad\int_0^{\infty} g_n(x)dx=1-\alpha$$ that
$$
\lim_{n\rightarrow\infty}\int_{-\infty}^0 g_n(x)F(x)dx=\alpha F(0)
$$
Any comments would be very welcome
 A: Yes, you need additional conditions. These conditions also depend on the allowable $F$, and will be different for different spaces of $F$.
To be precise,
Definition. Let $\Phi\subseteq C(\Omega)$ and $\xi\in\Omega$. A sequence $(f_\epsilon(x-\xi),x\in \Omega)_{\epsilon\to 0}$ (these are yours "good functions") such that
$$
f(\xi)=\lim_{\epsilon\to 0}\int_\Omega f(x)f_\epsilon(x-\xi)dx
$$
holds for any $f\in\Phi$ is called $\delta$-sequence on the space $\Phi$.
The main point here is that $\delta$-sequence definition depends on the given space (which has to be a subspace of continuous functions).
Try to convince yourself that

*

*Sequence
$$
f_y(x)=\frac{1}{\pi}\frac{y}{x^2+y^2}
$$
is a $\delta$-sequence on the space $C_b(\mathbb R)$ (the space of continuous functions with bounded derivative) but not a $\delta$-sequence on the space $C(\mathbb R)$ (since for many continuous functions the integral simply will not be determined).

*The sequence
$$
f_t(x)=\frac{1}{2\sqrt{\pi t}}e^{-x^2/(4t)}
$$
is not a $\delta$-sequence on $C(\mathbb R)$ but a $\delta$-sequence on $\Phi=\{\phi\in C(\mathbb R)\colon \exists a, |\phi(x)e^{-ax^2}|\to 0\text{ for }|x|\to\infty\}$
Remark: If you are confused that I replaced your $n$ with $\epsilon\to 0$, just make opposite substitution $\epsilon = 1/n$.
