$\newcommand{\realset}{\mathbb{R}}$ $\newcommand{\derivative}[1]{#1^{\prime}}$ $\newcommand{\domain}[1]{\operatorname{dom}\left(#1\right)}$
I am reading the following link about nascent delta functions: https://en.wikipedia.org/wiki/Nascent_delta_function. Based on the definition and properties, I would like to prove the following theorem:
Let $\eta$ be a function on $\realset$ and \begin{equation*} \int_{-\infty}^{\infty} \eta\left(x\right) \mathrm{d}x = 1 \end{equation*} and $\gamma$ be a function such that $\domain{\gamma} = \realset \setminus \left\{0\right\}$ and for any $\epsilon \in \domain{\gamma}$, $\gamma_{\epsilon}$ is a function such that $\domain{\gamma_{\epsilon}} = \realset$ and for any $x \in \realset$, \begin{equation*} \gamma_{\epsilon}\left(x\right) = \epsilon^{-1} \eta\left(\frac{x}{\epsilon}\right). \end{equation*} Then for any $\epsilon \in \domain{\gamma}$, \begin{equation*} \int_{-\infty}^{\infty} \gamma_{\epsilon}\left(x\right) \mathrm{d}x = 1. \end{equation*} Further, let $f$ be a function such that $\domain{f} = \realset$ and $f$ is absolutely integrable. Let $t \in \realset$ and $g$ be a function such that $\domain{g} = \realset \setminus \left\{0\right\}$ and for any $\epsilon \in \domain{g}$, \begin{equation*} g\left(\epsilon\right) = \left(f * \gamma_{\epsilon}\right)\left(t\right). \end{equation*} Then $g$ converges at $0$ and \begin{equation*} \lim_{\epsilon\to 0} g\left(\epsilon\right) = f\left(t\right). \end{equation*}
I have the following incomplete proof. In particular, I cannot find a way to prove the sifting property. The difficulty is that without a particular form of the nascent delta function, I cannot narrow down the integral to $\left[t - \epsilon, t + \epsilon\right]$, as in Proof of Dirac Delta's sifting property. Can someone help?
By definition, \begin{equation*} \begin{aligned} \int_{-\infty}^{\infty} \gamma_{\epsilon}\left(x\right) \mathrm{d}x &= \int_{-\infty}^{\infty} \epsilon^{-1} \eta\left(\frac{x}{\epsilon}\right) \mathrm{d}x. \end{aligned} \end{equation*} Let $h$ be a function such that for any $x \in \realset$, $h\left(x\right) = x / \epsilon$. Then \begin{equation*} \int_{-\infty}^{\infty} \epsilon^{-1} \eta\left(\frac{x}{\epsilon}\right) \mathrm{d}x = \int_{-\infty}^{\infty} \eta\left(h\left(x\right)\right) \derivative{h}\left(x\right) \mathrm{d}x = \int_{-\infty}^{\infty} \eta\left(x\right) \mathrm{d}x = 1. \end{equation*} Thus, we have \begin{equation*} \int_{-\infty}^{\infty} \gamma_{\epsilon}\left(x\right) \mathrm{d}x = 1. \end{equation*}
Next, by definition, we have \begin{equation*} \begin{aligned} g\left(\epsilon\right) &= \left(f * \gamma_{\epsilon}\right)\left(t\right)\\ &= \int_{-\infty}^{\infty} f\left(x\right) \gamma_{\epsilon}\left(x - t\right) \mathrm{d}x\\ &= \int_{-\infty}^{\infty} f\left(x\right) \epsilon^{-1} \eta\left(\frac{x - t }{\epsilon}\right) \mathrm{d}x \end{aligned} \end{equation*}
