Suppose $f\in L^p(\mathbb{R})$ and $K\in L^1(\mathbb{R})$ with $\int_\mathbb{R}K(x)dx=1$. Define $$K_t(x)=\dfrac{1}{t}K\left(\dfrac{x}{t}\right)$$ I'm trying to prove that $\lim_{t\rightarrow 0}\|f\ast K_t-f\|_p=0$.
I choose a compactly supported function $g\in C^\infty$ such that $\|f-g\|_p<\epsilon$ (possible because this class is dense in $L^p(\mathbb{R})$). Then I want to bound $$\|f\ast K_t-f\|_p\leq \|f\ast K_t-g\ast K_t\|_p+\|g\ast K_t-g\|_p+\|g-f\|_p$$
We have of course $\|g-f\|_p<\epsilon$. We have $\|(f-g)\ast K_t\|_p\leq \|f-g\|_p\|K_t\|_1<\epsilon\|K_t\|_1$. But how can we bound the term $\|K_t\|_1$?
Okay, I found that $\|K_t\|_1=\|K\|_1$ for all $t$, so that gives the necessary bound.