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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

Emi
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  • I don't know the definition of a «good» function, but mass could slide at infinity. Take $g_n(x) = g_0(x-n)$. If these are «good» functions, it looks like a counterexample. – Gribouillis Feb 08 '22 at 12:39
  • Everywhere differentiable any number of times and such that it and all its derivatives are O(|x|^(-N)) as |x|->oo for all N – Emi Feb 08 '22 at 12:53
  • In that case what @Gribouillis said gives a counterexample. If you assume that the sequence is "uniformly good" (ie that $O(1/|x|^N$ holds uniformly) that's probably enough... – David C. Ullrich Feb 08 '22 at 13:05
  • There must be some condition on $F$ in the book! If $F$ can be "any function"then the integrals need not even exist... – David C. Ullrich Feb 08 '22 at 13:08
  • Is there any difference with the Schwartz space? If not, why call them «good functions»? – Gribouillis Feb 08 '22 at 13:11
  • As far as I see $F(x)$ is supposed to be a good function too. – Emi Feb 08 '22 at 13:43

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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 $\epsilon = 1/n$.

Artem
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  • I think the good functions in Lighthills book are more the one you mention in your example 2, $f_t(x)$ is also the function he uses in an example. – Emi Feb 09 '22 at 13:49
  • Maby I did not get your answer right, but I would be interested in the assumptions on my above $g_n(x)$ (see equation (1)) to conclude equation (2). Gribouillis pointed out that e.g. $\sqrt{n\pi}e^{-n(x−n)^2}$ would fullfill (1) but would not lead to (2), so what could be a further restriction on the $g_n(x)$. From (3) one may conclude that the last integral must converge to zero and the derivative of $F(x)$ must be bounded, but are there less strict restrictions possible too. – Emi Feb 09 '22 at 14:18
  • @Emi Yes, I think you did not get my answer. The short answer is "any functions for which the given definition holds." Note that definition depends on the type of functions $F$ you are using. Say, if want to consider all continuous $F$, then your $g_n$ must be required compactly supported, etc. – Artem Feb 09 '22 at 18:53