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 $\|(f-g)\ast K_t\|_p\leq \|f-g\|_p\|K_t\|_1<\epsilon\|K_t\|_1=\epsilon\|K\|_1$.
Now I need to bound $\|g\ast K_t-g\|_p$.
I have $$\left|(g\ast K_t)(x)-g(x)\right|_p=\left|\int_\mathbb{R}(g(x-y)-g(x))K_t(y)dy\right|$$ Since $g$ is continuous and compactly supported, it is uniformly continuous. But what can I do with the $K_t(y)$?
In the last displayed integral, use the substitution $s:=y/t$. Then $$\lVert g\star K_t-g\rVert_p^p=\int_{\mathbb R}\left|\int_{\mathbb R}(g(x-ts)-g(x))K(s)\mathrm ds\right|^p\mathrm dx.$$ I assume $K\geqslant 0$ (not restrictive). Then by Jensen's inequality, it's enough to prove that $$\lim_{t\to 0}\iint_{\mathbb R^2}|g(x-ts)-g(x)|^pK(s)\mathrm ds\mathrm dx=0.$$ This can be done by an approximation argument: take $\varepsilon\gt 0$: then there is $R$ such that $\int_{\mathbb R\setminus [-R,R]}K(s)\mathrm ds\lt \varepsilon$. Then we have $$\int_{\mathbb R}\int_{\mathbb R\setminus [-R,R]}|g(x-ts)-g(x)|^pK(s)\mathrm ds\mathrm dx\leqslant 2\varepsilon\lVert g\rVert_p.$$ We have $$\int_{\mathbb R}\int_{[-R,R]}|g(x-ts)-g(x)|^pK(s)\mathrm ds\mathrm dx\leqslant 2R\sup|g'|t^p\lVert K\rVert_1.$$