The question at hand is the following:
Let $\psi \in \mathcal{S}(\mathbb{R}^d, \mathbb{C})$ be real–valued such that $\psi \ge 0$ and $\int_{\mathbb{R}^d} \psi(x)\,dx = 1$. Moreover, let $f \in L^1(\mathbb{R}^d)$ be continuous in $x_0 \in \mathbb{R}^d$ and \begin{equation} \psi_\varepsilon(x) := \frac{1}{\varepsilon^d}\psi\!\left(\frac{x}{\varepsilon}\right) \quad \text{for } \varepsilon > 0 \text{ and } x \in \mathbb{R}^d. \end{equation} Then, \begin{equation} \lim_{\varepsilon \to 0} (\psi_\varepsilon * f)(x_0) = f(x_0). \end{equation} Now, my approach looked very promising, up until one point. We let $\eta>0$ and choose $R>0$ such that \begin{equation} \int_{|x|>R} \psi(x) \text{d}x \leq \eta. \end{equation} Note that, by the transformation formula and the fact that $\psi$ integrates to one, \begin{equation} |(\psi_\varepsilon * f)(x_0)-f(x_0)|\leq \int_\mathbb{R^d}\psi(x)|f(x_0-\varepsilon x)-f(x_0)|\text{d}x. \end{equation} We now may split the integral into two parts: \begin{equation} A_\varepsilon:=\int_{|x|\leq R}\psi(x)|f(x_0-\varepsilon x)-f(x_0)|\text{d}x, \quad B_\varepsilon:=\int_{|x>R}\psi(x)|f(x_0-\varepsilon x)-f(x_0)|\text{d}x. \end{equation} I was indeed able to show, using the continuity of $f$ in $x_0$, that $A_\varepsilon < \eta$ for sufficiently small $\varepsilon$ (I can provide details how I achieved that, I did not want the question to be too long). However, I am struggling with the second integral $B_\varepsilon$. I tried using just the triangle inequality which would lead to \begin{equation} B_\varepsilon \leq |f(x_0)| \eta + \int_{|x|>R} \psi(x)|f(x_0-\varepsilon x)| dx. \end{equation} This does not seem promising. I also still have not used that $\psi$ is a Schwartz function! Any help/suggestion is greatly appreciated.
