Stein Shakarchi Real Analysis Exercise 3.5.2 I don't follow a couple parts of the proof of this problem.
Problem 3.5.2
Suppose $\{K_\delta\}$ is a family of kernels that satisfies for all $\delta > 0$:

*

*$|K_\delta(x)| \le A\delta^{-d}$

*$|K_\delta(x)| \le A\delta/|x|^{d+1}$

*$\int_{\mathbb{R}^d} K_\delta(x)dx = 0$
Show that if $f$ is lebesgue-integrable on $\Bbb R^d$, then
$$
(f*K_\delta)(x) \rightarrow 0
$$
for a.e. $x$, as $\delta \rightarrow 0$.
Proof
We have
$$
(f*K_\delta(x) = \int f(x-y)K_\delta(y)dy
$$
For convenience, let
$$g(x,y,\delta) := |f(x-y)||K_\delta(y)|$$
Using the monotonicity of the integral, it suffices to show that
$$
\int f(x-y)K_\delta(y)dy \le \int g(x,y,\delta)dy \rightarrow 0
$$
as $\delta \rightarrow 0$
We break up this integral as follows
$$
\tag{1}
\int g(x,y,\delta)dy = \int_{|y| \le \delta}g(x,y,\delta)dy + \sum_{k=0}^\infty \int_{2^k \delta < |y| \le 2^{k+1} \delta} g(x,y,\delta)dy
$$
We use the property of good kernels to observe that
$ 
\tag{2} \int_{|y| \le \delta}|K_\delta(y)|dy \rightarrow 0
$
as $\delta \rightarrow 0$.

Issue 1:
For the second term, choose each term in the sum to be less than $\epsilon/2^k$ for any $\epsilon > 0$.


Issue 2:
For the first term, use $\int K_\delta = 0$ and the two estimates of good kernels to see that
$$
\int_{|y| \le \delta} g(x,y,\delta) \le CA\delta / \epsilon \int |f(x-y)|
$$

Two issues
Issue 1
For issue 1, I see that we can make the expression in (2) arbitrarily small for sufficiently small $\delta$, but it's not clear to me that this is true when you multiply the integrand by $|f(x-y)|$.
For each term in the sum, since $f$ is integrable, I could say that
$$
\int |f| = M < \infty
$$
But it does not follow that $f$ is even bounded a.e., so I can't say
$
\int g(x,y,\delta) \le M \int |K_\delta(y)|dy
$
So I'm not sure how to bound the terms in the sum of (2).
Issue 2
I don't follow this result, in particular how $\epsilon$ ends up in the denominator. It would help to see some intermediate steps.
 A: The solution is a slight modification of the strategy already outlined by the OP. Although I tried to cross all the t's and dot all the i's, I leave minor details to the OP.
In the course of answering this problem I noticed that the solution presented here proves something slightly more general:

Proposition: If $K_\delta$ satisfies assumptions (1) and (2) in the OP's problem, each $K_\delta\in L_1(\mathbb{R}^d,\lambda)$ and $a:=\int K_\delta$ for all $\delta$, then for any $f\in L_1$
$$\lim_{\delta\rightarrow0}|K_\delta*f(x)-f(x)a|=0$$
for all Lebesgue point $x$ of $f$. Conditions (1) and (2) may be substituted by the condition that there is an integrable  radial function $\psi:\mathbb{R}^d\rightarrow(0,\infty)$ such that

*

*$\psi(x)\leq \psi(x')$ whenever $|x'|\leq |x|$,

*$|K_\delta(x)|\leq \delta^{-d}\psi(\delta^{-1}x)=:\psi_\delta(x)$.


The case $a=0$ comes as a particular case of the statement above.
Proof: It follows from the assumption on $K_\delta$ that
$$\begin{align}
\Big|f*K(x)- af(x)\Big|&=\Big|\int_{\mathbb{R}^d}\big(f(x-y)-f(x)\big)K_\delta(y)\,dy\Big|\\
&\leq \int_{\mathbb{R}^d}\big|f(x-y)-f(x)\big||K_\delta(y)|\,dy\tag{1}\label{one}
\end{align}$$
Consider the radial function $\psi(x)=\phi_0(\|x\|)$ where
$\phi_0:(0,\infty)\rightarrow\mathbb{R}$ is the map $t\mapsto\mathbb{1}_{(0,1]}(t)+\frac{1}{t^{d+1}}\mathbb{1}_{(1,\infty)}(t)$.
Since $\phi_0$ is a positive monotone nonincreasing, $\psi$ satisfies $\psi(x)\leq\psi(x')$ whenever $|x'|\leq |x|$, and
$$\begin{align}
K_\delta(x)&\leq A\delta^{-d}\psi(\delta^{-1}x)=:A\psi_\delta(x)\tag{2}\label{two}
\end{align}$$
Moreover $\psi\in L_1(\mathbb{R}^d)$ and for any $r>0$,
$$\int_{\mathbb{R}^d}\psi(x)\,dy\geq\int_{r/2<|x|\leq r}\psi(x)\,dx=\sigma_d\int^r_{r/2}s^{d-1}\phi_0(s)\,ds\geq\frac{\sigma_d}{d}\phi_0(r)\frac{2^d-1}{2^d} r^d
$$
where $\sigma_d$ is the surface area of the sphere $\mathbb{S}_{d-1}$. In particular, for all $r>0$
$$\begin{align}
\phi_0(r)\leq C_d\frac{1}{r^d}\int_{r/2<|x|\leq r}\psi(x)\,dx\tag{3}\label{three}
\end{align}$$
for some constant $C_d>0$. Decomposing $\mathbb{R}^d$ as union of concentrical annular regions we obtain from \eqref{one} and \eqref{two} that
$$\begin{align}
|f*K_\delta(x)-a f(x)|&\leq\int_\mathbb{R}|f(x-y)-f(x)||K_\delta(y)|\,dy\\
&=\sum_{k\in\mathbb{Z}}\int_{2^k\delta<|y|\leq2^{k+1}\delta}|f(x-y)-f(x)||K_\delta(y)|\,dy\\
&\leq A \sum_{k\in\mathbb{Z}}\int_{2^k\delta<|y|\leq2^{k+1}\delta}|f(x-y)-f(x)|\psi_\delta(y)\,dy\
\end{align}
$$
For each $k\in\mathbb{Z}$, let $I_k:=\int_{2^k\delta<|y|\leq2^{k+1}\delta}|f(x-y)-f(x)|\psi_\delta(y)\,dy$. The monotonicity of $\phi_0$ along with \eqref{three} yields
$$
\begin{align}
I_k&\leq
\frac{B_d}{(2^{k+1}\delta)^d}\int_{2^k\delta<|y|\leq2^{k+1}\delta}|f(x-y)-f(x)|\Big(\int_{\tfrac{|y|}{2\delta}<|z|\leq \tfrac{|y|}{\delta}}\psi(z)\,dz\Big)\,dy\\
&\leq
\frac{B_d}{(2^{k+1}\delta)^d}\int_{|y|\leq2^{k+1}\delta}|f(x-y)-f(x)|\Big(\int_{2^{k-1}<|z|\leq 2^{k+1}}\psi(z)\,dz\Big)\,dy\\
&=\Big(\int_{2^{k-1}<|z|\leq 2^{k+1}}\psi(z)\,dz\Big)\frac{B_d\omega_d}{\lambda(B(x;2^{k+1}\delta))}\int_{|y|\leq2^{k+1}\delta}|f(x-y)-f(x)|\,dy
\end{align}
$$
where $\omega_d$ is the Lebesgue measure of the unit ball in $\mathbb{R}^d$ and $\lambda$ is Lebesgue measure in $\mathbb{R}^d$. Let
$$
a_k:=\int_{2^{k-1}<|z|\leq 2^{k+1}}\psi(z)\,dz, $$
and for any $\varepsilon>0$ define
$$
T_{\varepsilon}f(x):=\frac{1}{\lambda(B(x;\varepsilon))}\int_{B(0;\varepsilon)}|f(x-y)-f(x)|\,dy
$$
Clearly $\sum_{k\in\mathbb{Z}}a_k\leq 2\|\psi\|_1$. By Hardy-Littlewood's maximal theorem,
$$\lim_{\varepsilon\rightarrow0}T_\varepsilon f(x)=0$$
for any Lebesgue point $x$ of $f$. Adding over $k\in\mathbb{Z}$ yields
$$\begin{align}
|f*K_\delta(x)-a f(x)|\leq\sum_{k\in\mathbb{Z}}I_k
\leq B_d\omega_d\sum_{k\in\mathbb{Z}}a_kT_{2^{k+1}\delta}f(x)
\end{align}$$
As $\sup_{\varepsilon>0}T_{\varepsilon}f(x)\leq Mf(x)+|f(x)|$, where $Mf$ is the Hardy-Littlewood maximal function of $f$, an application of dominated convergence yields
$$\lim_{\delta\rightarrow0}\big|f*K_\delta(x)- af(x)\big|=0$$
whenever $x$ is a Lebesgue point of $f$.
