TITLE: Counterexample for approximation by convolution
QUESTION [1 upvotes]: I'm currently reading these lecture notes. In the remark to theorem 3.7, it is explicitly stated that the convergence doesn't hold for $p= \infty$. I'm not really sure about that, as I am not able to find a counterexample.
Is there an easy counterexample?

REPLY [2 votes]: Let $f=1$ on $[0, 1]$ and $f=0$ elsewhere, let $\varphi=1$ on $(-\frac 12, \frac 12)$ and $\varphi=0$ elsewhere. Thus $\varphi$ has integral $1$. We see for all $\epsilon<\frac 12$, $\varphi_\epsilon* f$ is continuous: for example near $0$ we have $\varphi_\epsilon*f=0$ for $x\leq -\frac{\epsilon}{2}$, and then increases linearly to $1$ at $x=\frac{\epsilon}{2}$. Thus for all $\epsilon<\frac 12$ we have
$$
\|\varphi_\epsilon*f-f\|_\infty=\frac 12.  
$$.
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