Let $f_{\epsilon} = f * K_{\epsilon}$, where $f\in L^{p}(\mathbb R^n)$, $p>1$, $K\in L^{1}(\mathbb R^n)\cap L^{\infty}(\mathbb R^n)$,$ \int_{\mathbb R^n} K = 1$, and $K(x) = o(|x|^{-n})$ as $|x|\rightarrow +\infty$.
Prove that $f_{\epsilon}\rightarrow f$ as $\epsilon \rightarrow 0$ at each point of contimuity of $f$.
($K_\epsilon(x) := \epsilon^{-n}K(x/\epsilon)$, $f*K(x) = \int f(t)K(x-t)\, dt$ is its convolution.)
For what I'm trying, let $x$ be any continuous point of $f$, then given $\eta>0$, there exists $\delta>0$ so that $|f(x-t)-f(x)|<\eta$ whenever $|t|<\eta$.
Then $|f_\epsilon(x)-f(x)|= |\int f(x-t)K_{\epsilon}(t)-f(x)K_{\epsilon}(t) \, dt| \leq |\int_{|t|<\delta}|+ |\int_{|t|\geq \delta}| = A+B$.
Now $A\leq \eta\|K\|_1$.
By Holder inequality, $B \leq \int_{|t|\geq\delta} |f(x-t)-f(x)||K_{\epsilon}(t)|\, dt\leq \|g(t)\|_p\|K_{\epsilon}(t)\|_q$ where $g(t) = f(x-t)-f(t)$, $1/p+1/q = 1$.
We might need to use $K(x)= o(|x|^{-n})$ to make $\|K_\epsilon\|$ small ($<\eta$ or something) but I don't know how.
Help please.