My task is to prove that if $f$ is continuous on $\mathbb{R}^d$ with compact support and $\{K_\delta\}_{\delta>0}$ is a family of good kernels then $f \ast K_\delta \xrightarrow{\text{unif.}} f$ as $\delta \to 0$, where good kernels are defined to be satisfying the following:
$ (\text{i}) \quad \int_{\mathbb{R}^d} K_\delta(x) \text{d}x = 1 $
$ (\text{ii}) \quad \int_{\mathbb{R}^d} \vert K_\delta(x) \vert \text{d}x \leq A $
$ (\text{iii}) \quad \int_{\vert x \vert \geq \eta} \vert K_\delta(x) \vert \text{d}x \to 0 \quad \text{ as } \delta \to 0 \text{ for every } \eta > 0 $
Here is my attempt thus far:
Let $\varepsilon > 0$ and $E$ be the compact support of $f$, noting that $f$ is uniformly continuous on $E$ since $f$ is continuous everywhere. Using property (i), we may write \begin{align*} \vert (f \ast K_\delta)(x) - f(x) \vert &= \left\vert \int_{\mathbb{R}^d} f(x-y)K_\delta(y) \text{d}y - f(x) \right\vert \\[4pt] &= \left\vert \int_{\mathbb{R}^d} \left( f(x-y)K_\delta(y) - f(x)K_\delta(y) \right) \text{d}y \right\vert \\[4pt] &\leq \int_{\mathbb{R}^d} \left\vert f(x-y) - f(x) \right\vert \vert K_\delta(y) \vert \text{d}y \end{align*} We may "split" this integral into two parts in the following fashion: $$ \int_{\mathbb{R}^d} \left\vert f(x-y) - f(x) \right\vert \vert K_\delta(y) \vert \text{d}y = \int_{E} \left\vert f(x-y) - f(x) \right\vert \vert K_\delta(y) \vert \text{d}y + \int_{E^c} \left\vert f(x-y) - f(x) \right\vert \vert K_\delta(y) \vert \text{d}y $$ Using the uniform continuity of $f$ on $E$ and property (ii) above, (I think) we then have that for some constant $c$ $$ \int_{E} \left\vert f(x-y) - f(x) \right\vert \vert K_\delta(y) \vert \text{d}y \leq c \varepsilon \int_{E} \vert K_\delta(y) \vert \text{d}y \leq cA \varepsilon $$ This is where I get stuck. I am not sure how to bound the second integral over $E^c$. I think I need to use property (iii) by claiming that the integral over $E^c$ is somehow equivalent to integrating over $\vert y \vert \geq \eta$ but I am failing to make the connection. Any direction is greatly appreciated.