Convolutions with $L^\infty$ functions

Question

I read the following theorem about convolutions with $L^p$ functions in real analysis:

Let ${\phi_n \in C^\infty_c({\bf R}^d)}$ be a sequence of approximations to the identity. If ${f \in L^p({\bf R}^d)}$ for some ${1 \leq p < \infty}$, show that ${f*\phi_n}$ converges in ${L^p({\bf R}^d)}$ to ${f}$. (One can use the density of ${C_c({\bf R}^d)}$ in ${L^p({\bf R}^d)}$, and the Young’s inequality.)

Would anybody come up with a counterexample for the case $p=\infty$?

Answer 1

Try $d=1$, $f(x) = \sin(nx)$ on $[n\pi, (n+1)\pi]$. For any $\phi \in C_c^\infty$ supported in, say, $(-\epsilon,\epsilon)$ with $\epsilon < \pi/2$, $f \star \phi$ will be small on $(n\pi+\epsilon, (n+1)\pi - \epsilon]$ for large $n$, so $\|f - f \star \phi\|_\infty \ge 1$.

Answer 2

Take $d=1$, $f=1_{[0,\infty)}$, then since $f \star \phi_n$ is continuous, we must have $\|f-f \star \phi_n\| \ge { 1\over 2}$.

Why ${ 1\over 2}$?

Let $c=(f \star \phi_n)(0)$.

If $c \ge {1 \over 2}$ then $|f(0)-(f \star \phi_n)(0)| \ge {1 \over 2}$.

Otherwise, we have $\lim_{x\uparrow 0} (f(x)-(f \star \phi_n)(x)) = 1-c \ge {1 \over 2}$.

Since $f,f \star \phi_n$ are continuous on $[0,\infty)$ and $(-\infty, 0)$, we see that $\|f- f \star \phi_n\| = \operatorname{ess} \sup_x |f(x)- (f \star \phi_n)(x)| = \sup_x |f(x)- (f \star \phi_n)(x)| \ge {1 \over 2}$.