Norm convergence of approximations to the identity 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.
 A: 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.
