TITLE: Good Kernel's Properties
QUESTION [7 upvotes]: I am recently studying properties about a good kernel, and came across a problem.
Definition: A kernel $K_\delta$ is 'good' if they are Lebesgue integrable and satisfy the following conditions for $\delta>0$:

$\int_{\mathbb R^d}K_\delta(x)dx=1$
$\int_{\mathbb R^d}|K_\delta(x)|dx\le A$
For every $\eta>0$,
$$\int_{|x|\ge\eta}|K_\delta(x)|dx\to0\text{  }\text{ }\text{ as }\delta\to0$$

Here $A$ is a constant independent of $\delta$
The question is to prove that for every point of continuity of an integrable function $f$, $(f\star K_\delta)(x)\to f(x)$ as $\delta \to 0$, where we define $$(f\star K_\delta)(x)=\int_{\mathbb R^d}f(x-y)K_\delta(y)dy$$

REPLY [6 votes]: I am sorry you had to wait almost a year for someone to react!
For the difference we have
$$
(f\ast K_\delta)(x) - f(x) = \int_{\mathbb R^d} f(x-y) K_\delta(y)\ \text{d}y -  \int_{\mathbb{R}^d} f(x) K_\delta(y) \ \text{d}y,
$$
where we used property 1. Hence the difference equals
$$
\int_{\mathbb{R}^d} \left(f(x-y) - f(x)\right) K_\delta(y) \ \text{d}y,
$$
and we wish to see why this tends to zero. Let $\epsilon>0$ be given. By continuity of $f$ in $x$ there exists $\eta>0$ such that $|y|<\eta$ implies $|f(x-y)-f(x)|<\frac{\epsilon}{2A}$.
Now split the integral in two parts and use property 3 for the second part.