Lighthills definition of the Delta distribution

Question

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

Answer 1

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

1. 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).
2. 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 $\epsilon = 1/n$.