Here are some background settings.
Let $K\in L^1$ s.t. $\int_{\mathbb R^n}K(x)dx=1$ and let $K_{\varepsilon}(x):=\varepsilon^{-n}K(x/\varepsilon)$ for all $\varepsilon\gt0.$
Then we will automatically have that $\int_{\mathbb R^n}K_{\varepsilon}(x)dx=1$ for all $\varepsilon\gt0$ and lim$_{\varepsilon\to 0}\int_{|x|\gt a}K_{\varepsilon}(x)dx=0$ for all $a\gt 0.$ And by the Minkoski's inequality for integrals, we will have $||f*K_{\varepsilon}-f||_{L^p}\to 0$ as $\varepsilon\to0.$
Now we suppose further that $K\in L^\infty$ and lim$_{|x|\to\infty}|x|^nK(x)=0,$
then my text claims that $K_{\varepsilon}*f(x)\to f(x)$ pointwisely for all $f\in L^1$ whenever $f$ is continuous at $x$.
My Attempt:
Let $x\in \mathbb R^n$.
$$\lvert\int K_\epsilon (y)f(x-y)dy - f(x)\rvert=\lvert\int K(y)f(x-\varepsilon y)dy-\int K(y)f(x)dy\rvert\,\quad\quad\quad\quad\quad\quad\quad\,\,\,\,\,=\lvert\int K(y)(f(x-\varepsilon y)-f(x))dy\rvert.$$
The integral domain is $\mathbb R$ and so even though $\varepsilon$ is small, the product $\varepsilon y$ may still be large, which prevents us from using the continuity of $f$ at $x$.
So basically my idea is to separate the integral domain into two parts. Let $B(r,0)$ be the open ball centered at the origin with radius $r$. Then we will consider the integral over $B(r,0)$ and its complement $B(r,0)^c.$
In the ball $B(r,0)$, we can make $f(x-\varepsilon y)-f(x)$ arbitrarily small by taking $\varepsilon$ small, together with the integrability of $K$, which makes this part small.
Outside the ball $B(r,0)$, I used triangle inequality to obtain this $$\lvert\int_{B(r,0)^c} K(y)(f(x-\varepsilon y)-f(x))dy\rvert\le\lvert\int_{B(r,0)^c} K(y)f(x-\varepsilon y)dy\rvert + \lvert \int_{B(r,0)^c} K(y)f(x)dy\rvert.$$
Because $f\in L^1$ and $f$ is continuous at $x$, we know $f(x)$ must be finite. (Otherwise $|f|$ will be $\infty$ on a set with positive measure, which makes $f\notin L^1.$ ) When $r$ is large enough, $\int_{B(r,0)^c} K(y)dy$ is small because $\int_{\mathbb R^n}K(x)dx=1\lt \infty$. So the second term can be small too.
But I don't know how to bound the first term, i.e. $\lvert\int_{B(r,0)^c} K(y)f(x-\varepsilon y)dy\rvert$, and so far I haven't used the assumption that $f\in L^1$, $K\in L^\infty$ and $K$ decays at some rate.
My question is how to bound $\lvert\int_{B(r,0)^c} K(y)f(x-\varepsilon y)dy\rvert$, or alternatively, how to prove that $K_{\varepsilon}*f(x)\to f(x)$ pointwisely for all $f\in L^1$ whenever $f$ is continuous at $x$.
I got stuck in this step for the entire afternoon...
Thanks for help :)