Let $\varphi \in L^1(\mathbb{R}^d)$ be such that $$\int_{\mathbb{R}^d} \varphi(x) \, dx = 1.$$
For each $\varepsilon>0$, let $\varphi_\varepsilon:= \varepsilon^{-d} \varphi\left( \dfrac x \varepsilon \right)$.
Then for any $f\in C^0(\mathbb{R}^d)$ (the set of bounded and continuous function), show $$f*\varphi_\varepsilon\to f$$ in $L^\infty$ norm.
I have no any idea how to do this problem but it seems that it is not true for $f$ not continuous.
The following proof only works for uniformly continuous functions $f$ (note that uniform continuity implies continuity). Introduce a neighbourhood $X\subset\mathbb{R}^n$ around the origin. Let $X^c=\mathbb{R}^d\setminus X$ denote the complement of $X$, and $\mathbb{R}^n=X+X^c$. The idea is to examine the convergence problem over $X$ and $X^c$, and use an $\epsilon -\delta$ argument. It suffices to show that \begin{equation*} \sup |(\varphi_{\epsilon}\ast f)(x)-f(x)|=0 \end{equation*} holds. We can prove that the two estimates \begin{align*} & |(\varphi_{\epsilon}\ast f)(x)-f(x)|<\frac{\delta}{2c}\to 0 \\ & \int_{X^c}|\varphi_{\epsilon}(y)|d\mu<\frac{\delta}{4||f||_{L^{\infty}}}\to 0 \end{align*} hold, where $c$ bounds the dilation operator $\varphi_{\epsilon}$ under $L^1$ norm. Therefore \begin{align*} & \sup|(k_{\epsilon}\ast f)(x)-f(x)| \\ & \leq \int_{X}|\varphi_{\epsilon}(y)|\sup|\varphi_{\epsilon}(y)f(x-y)-f(x)|d\mu+\int_{X^c}|\varphi_{\epsilon}(y)|\sup|\varphi_{\epsilon}(y)f(x-y)-f(x)|d\mu \\ & \leq \frac{\delta}{2}+\frac{\delta}{2}=\delta \end{align*}
For details, see classical Fourier analysis by Loukas Grafakos p. 25-27.