$L^1$ approximate identities converge uniformly on compacta for bounded continuous functions

Let $\eta >0$. Choose $\delta >0$ so that [L2, L3, given, choose] $\mathrm{\mid }f(x-y)-f(x)\mathrm{\mid }<\eta$ whenever $x\in K$ and $\mathrm{\mid }y\mathrm{\mid }<\delta$; this is possible by [L2] on a compact neighborhood of $K$.

For $x\in K$, [L1, step 1.1, algebra] $\mathrm{\mid }(f\ast K_{\epsilon })(x)-f(x)\mathrm{\mid }\le \int \mathrm{\mid }{K}_{\epsilon }(y)\mathrm{\mid }\mathrm{\mid }f(x-y)-f(x)\mathrm{\mid }dy\mathrm{.}$ Split the integral into $\mathrm{\mid }y\mathrm{\mid }<\delta$ and $\mathrm{\mid }y\mathrm{\mid }\ge \delta$. The near part is at most $\eta \mathrm{\parallel }{K}_{\epsilon }{\mathrm{\parallel }}_1$, while the far part is bounded by $2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\mathrm{\mid }y\mathrm{\mid }\ge \delta }\mathrm{\mid }{K}_{\epsilon }(y)\mathrm{\mid }dy$.

The $L^1$ norms are uniformly bounded and the tail term tends to $0$ by [L1]. [L1, step 2.1] Hence first choose $\eta$ small enough and then $\epsilon$ small enough to make the right-hand side uniformly small for all $x\in K$. This is exactly the claimed uniform convergence on $K$.