Convolution and uniform continuity If $f\in L^{\infty}(\mathbb{R}^n)$ and $f$ is continuous at $x$, then $$\lim_{k\to\infty}(f*\phi_k)(x)=cf(x)$$ If $f\in L^{\infty}(\mathbb{R}^n)$ and is uniformly continuous, then $f*\phi_k\to cf$ uniformly; that is, $$\lim_{k\to\infty}||f*\phi_k-cf||_{\infty}=0$$

Here's a very similar result in my real analysis book:
Theorem: Let $\{\phi_k\}$ be a sequence of functions in $L^1(\mathbb{R}^n)$ such that


*

*$\lim_{k\to\infty}\int\phi_kdx=c$ exists;

*$\int|\phi_k|dx\le M$ for some finite constant $M$;

*for all $r>0$,$$\lim_{k\to\infty}\int_{|x|\ge r}|\phi_k(x)|dx=0$$
Then for every $f\in L^p(\mathbb{R}^n)$, $1\le p <\infty$, $$\lim_{k\to\infty}||f*\phi_k-cf||_p=0$$

I see this question is a special case with $p=\infty$, but unfortunately the theorem states that $1\le p <\infty$. Does this theorem help in establishing the case $p=\infty$? Thank you!
 A: It seem to me that your goal is to prove the case $p=\infty$. I don't know if the case $p\in [1,\infty)$ can be used, however, there is a straightforward proof for the case $p=\infty$. 
Let $B(\delta,0)=\{y\in \mathbb{R}^n:\ \|y\|\leq\delta\}$. Note that \begin{eqnarray}
 \left|\int_{\mathbb{R}^n}f(x-y)\phi_k(y)dy-\int_{\mathbb{R}^n} f(x)\phi_k(y)dy\right| &\le& \int_{\mathbb{R}^n}|f(x-y)-f(x)||\varphi_k(y)|dy      \nonumber \\
   \tag{1}\end{eqnarray}
and $$  \int_{\mathbb{R}^n}|f(x-y)-f(x)||\varphi_k(y)|dy\le\int_{B(\delta,0)}|f(x-y)-f(x)||\varphi_k(y)|dy+\\ +\int_{\mathbb{R}^n\setminus B(\delta,0)}|f(x-y)-f(x)||\varphi_k(y)|dy\tag{2}$$
Assume that $f$ is continuous at $x$. Fix $\epsilon>0$ and choose $\delta>0$ in such a way that $|f(x-y)-f(x)|\leq\epsilon$ for $y\in B(\delta,0)$. We conclude from $(2)$ that 
$$\int_{\mathbb{R}^n}|f(x-y)-f(x)||\varphi_k(y)|dy\le \epsilon M+2\|f\|_{\infty}\int_{\mathbb{R}^n\setminus B(\delta,0)}|\varphi_k(y)|dy \tag{3}$$
It follow from the property 3. and $(3)$ that $$\lim_{k}\int_{\mathbb{R}^n}|f(x-y)-f(x)||\varphi_k(y)|dy\le \epsilon M\tag{4}$$
We conclude from $(1)$ and $(4)$ that $$\lim_k \left|\int_{\mathbb{R}^n}f(x-y)\phi_k(y)dy-\int_{\mathbb{R}^n} f(x)\phi_k(y)dy\right| =0$$
or equivalently $$\lim_k \int_{\mathbb{R}^n}f(x-y)\phi_k(y)dy=\lim_k \int_{\mathbb{R}^n} f(x)\phi_k(y)dy=cf(x)$$
A careful look at inequality $(2)$ shows that if $f$ is uniformly continuous then  $$\lim_{k\to\infty}||f*\phi_k-cf||_{\infty}=0$$
