Convolution convergent in $L^\infty$

Question

Suppose $f\in L^\infty(\mathbb{R})$ and $K\in L^1(\mathbb{R})$ with $\int_\mathbb{R}K(x)dx=1$. Define $$K_\epsilon(x)=\dfrac{1}{\epsilon}K\left(\dfrac{x}{\epsilon}\right)$$ Is it always true that $\lim_{\epsilon\rightarrow 0}\|f\ast K_\epsilon-f\|_\infty=0$?

I think it's always true for $L^p$, where $1\leq p<\infty$, but what about for $L^\infty$?

Answer 1

It is a well-known fact that smooth functions are not dense in $L_\infty$. Take $f(x)=\mathrm{sign}\,x$ and $K$ be a smooth non-negative even function with a compact support. Then $f*K_\epsilon(0)=0$ and $\|f\ast K_\epsilon-f\|_\infty=1$.