A family of functions $(K_\delta)_{\delta>0}$ is called an approximate identity on $\mathbb{R}^d$ if
(i) $\int_{\mathbb{R}^d}K_\delta(x)\ dx=1$.
(ii)$\int_{\mathbb{R}^d}|K_\delta(x)|\ dx\leq A$ for some $A>0$ and all $\delta>0$.
(iii)$\int_{|x|\geq\eta}|K_\delta(x)|\ dx\to 0$ for all $\eta>0$ as $\delta\to 0$.
I want to prove the following two statements:
a) If $f\in L^\infty(\mathbb{R}^d)$ is continuous on some open $U\subseteq \mathbb{R}^d$, then $f\ast K_\delta\to f$ uniformly on every compact $V\subset\subset U$ as $\delta\to 0$.
b) If $f\in L^p(\mathbb{R}^d)$, $1\leq p<\infty$, then $||f\ast K_\delta-f||_p\to 0$ as $\delta\to 0$.
For a) I've gotten this far: Take $W\subset U$ open, such that $V\subset\subset W\subset\subset U$. The reason I introduce this set $W$ is to get some regularity, since we know nothing of $V$ itself other than that it's compact, and since uniform convergence on $W$ would imply uniform convergence on $V$ I figured it's a bit nicer to work with. Now since $f$ is continuous on $U$, $f$ is uniformly continuous on $\overline{W}$. Now let $\varepsilon>0$. By uniform continuity there exists $\delta_0>0$ such that $|x-y|<\delta_0$ implies $|f(x)-f(y)|<\varepsilon$ for all $x,y\in\overline{W}$. Take $x\in W$. Then
$$ |f\ast K_\delta(x)-f(x)|\leq \int_{\mathbb{R}^d}|f(x)-f(y)||K_\delta(x-y)|\ dy =\int_{|x-y|<\delta_0}|f(x)-f(y)||K_\delta(x-y)|\ dy+2\sup|f|\int_{|x-y|\geq\delta_0}|K_\delta(x-y)|\ dy.$$
Here's where I get a problem. I'd like to argue,that since $W$ is open, we can choose $\delta_0$ so small, that $B_{\delta_0}(x)\subset W$, and hence we can estimate the first integral. But this makes the choice of $\delta_0$ dependent on $x$, which means that the second integral doesn't necessarily vanish as $\delta\to 0$ after we take the $\sup$ over all $x\in W$. But on the other hand, if I don't shrink by $\delta_0$-ball, I'm not guaranteed that it necessarily already lies in $W$, hence I can't estimate the first integral at all. Does someone have an idea how to fix this? Any input is appreciated, tanks.
And for b) I just wanted to know if there's a way to prove this by showing it first for $C_c^\infty$-functions using a) and then proceeding with a density argument. I've already done the density argument, but I don't know how to show this result for $C_c\infty$-functions without using the Minkowski-integral-inequality, which I'd like to avoid, since we haven't covered it in class.