A sequence of functions $\rho_\epsilon \in L^1(\mathbb{R^n})$ ) is called an approximate identity if $\rho_\epsilon \ge0$ , $\int \rho_\epsilon(x)dx=1$ and $\int_{|x|>\delta} \rho_\epsilon(x)dx \to 0 , \forall \delta>0$ as $\epsilon\to0$.
prove that if $f$ is continuous and bounded then $$f*\rho_\epsilon \to f , \text{pointwise}$$
I guess I can show that $\|f*\rho_\epsilon -f\|_1\to 0$. but how do I show the pointwise convergence here? (once I look at it I cannot see a way to bring the function $f$ itself out of the intergral)
Let $x \in \mathbb R^{n}$. Then $|\int f(x-y)\rho_{\epsilon} (y) dy -f(x)|\leq \int |f(x-y)-f(x)|\rho_{\epsilon} (y) dy$. Now $\int |f(x-y)-f(x)|\rho_{\epsilon} (y) dy =\int_{|y| <\delta} |f(x-y)-f(x)|\rho_{\epsilon} (y) dy+\int_{|y| \geq \delta} |f(x-y)-f(x)|\rho_{\epsilon} (y) dy$
Given $\eta >0$ choose $\delta >0$ such that $|f(x-y)-f(x)| <\eta$ for $|y|<\delta$. Then use boundedness of $f$ and the fact that $\int_{|y| \geq \delta} \rho_{\epsilon} (y) dy \to 0$ to make the second term $<\eta$ for $\epsilon$ sufficiently small.