Problem 6.1.4 Grafakos Classical Fourier Analysis I'm trying to approach a problem from Classical Fourier Analysis by Grafakos. The assumptions I have are the following. Suppose that $\psi$ is an integrable function on $\mathbb{R}^n$ with mean value zero that satisfies the following:
$$
|\psi(x)| \leq \frac{B}{(1+|x|)^{-n-\varepsilon}}
$$
and
$$
\int_{\mathbb{R}^n} |\psi(x-y)-\psi(x)| \, dx \leq B |y|^{\delta}
$$
for some constants $B, \varepsilon, \delta > 0$ and for all $y \neq 0$. Define $\psi_t(x) = t^{-n} \psi(x/t)$. I want to prove the bound
$$\sup_{y \in \mathbb{R}^n \setminus\{0\}} \int_{|x| \geq 2|y|} \left(\int_{0}^\infty |\psi_t(x-y)-\psi_t(x)|^2 \, \frac{dt}{t}\right)^{1/2} dx \leq c_n' B$$
where $c_n'$ is a constant that only depends on $n$.
So far, what I've done is follow the hint given in the problem. I've proven that
$$\int_{|x| \geq 2|y|} \left(\int_{0}^\infty |\psi_t(x-y)-\psi_t(x)|^2 \, \frac{dt}{t}\right)^{1/2} dx \leq A_n |y|^{-\varepsilon/2} \left(\int_{|x| \geq 2|y|} |x|^{n+ \varepsilon} \int_{0}^\infty |\psi_t(x-y) - \psi_t(x)|^2\frac{dt}{t}dx\right)^{1/2}.$$
Now, I'll try to deal with the term inside the parentheses on the right:
\begin{align*}
\int_{|x| \geq 2|y|} |x|^{n+ \varepsilon} \int_{0}^\infty |\psi_t(x-y) - \psi_t(x)|^2\frac{dt}{t}dx &= \int_{|x| \geq 2|y|} |x|^{n+ \varepsilon} \int_{t < |y|} |\psi_t(x-y) - \psi_t(x)|^2\frac{dt}{t}dx \\
&+ \int_{|x| \geq 2|y|} |x|^{n+ \varepsilon} \int_{t \geq |y|} |\psi_t(x-y) - \psi_t(x)|^2\frac{dt}{t}dx.
\end{align*}
For the second term on the right, use boundedness of $\psi$ to get
\begin{align*}
\int_{|x| \geq 2|y|} |x|^{n+ \varepsilon} \int_{t \geq |y|} |\psi_t(x-y) - \psi_t(x)|^2\frac{dt}{t}dx &\leq M_\psi \int_{|x| \geq 2|y|} |x|^{n+ \varepsilon} \int_{t \geq |y|} |\psi_t(x-y) - \psi_t(x)|\frac{dt}{t^{n+1}}dx \\
& \leq M_\psi \int_{|x| \geq 2|y|} |x|^{n+ \varepsilon} \int_{t \geq |y|} |\psi_t(x-y) - \psi_t(x)|\frac{dt}{|y|^{n+1}}dx, \\
\end{align*}
but I wasn't sure what to do from here. Is there a good way to approach the first term, and is my work for the second term on the right track?
 A: What you have so far is almost correct: For the second term, when you use the boundedness of $\psi$, you're throwing away powers of $|x|$ that you'll need in order to handle the term $|x|^{n+\varepsilon}$. More specifically we have (squiggly symbols mean there's some implicit constant depending at most on dimension)
$$\tag{1}
|\psi_t(x-y)-\psi_t(x)|\lesssim B\dfrac{t^\varepsilon}{(t+|x|)^{n+\varepsilon}},
$$
whenever $|x|\geq 2|y|$ (since $|x-y|\geq |x|-|y|\geq |x|/2$).
Therefore we have
\begin{equation}
\begin{split}
\int_{|x|\geq 2|y|} |x|^{n+\varepsilon} \int_{t\geq |y|} |\psi_t(x-y)-\psi(x)|^2\, \dfrac{dt}{t}\, dx & \lesssim B \int_{|x|\geq 2|y|} \int_{t\geq |y|}  |\psi_t(x-y)-\psi_t(x)| \dfrac{|x|^{n+\varepsilon}}{(t+|x|)^{n+\varepsilon}} \, \dfrac{dt}{t^{1-\varepsilon}}\, dx\\
&\leq B\int_{t\geq |y|}\int_{|x|\geq 2|y|} |\psi_t(x-y)-\psi_t(x)|\, dx\,  \dfrac{dt}{t^{1-\varepsilon}}\\
& \lesssim B^2\int_{t\geq 2|y|} |y|^\delta t^{-\delta} \, \dfrac{dt}{t^{1-\varepsilon}}(*)\\
& \lesssim B^2 |y|^{\varepsilon},
\end{split}
\end{equation}
where in the $(*)$ step we have extended the $x$ integral to $\mathbb{R}^n$ and do a change of variables. Also notice that to integrate $t^{-1+\varepsilon-\delta}$ we need $\delta>\varepsilon$, but we can always make $\varepsilon$ smaller if needed, so that's not an issue.
For the first piece, you want to use $(1)$ directly to get
\begin{equation}
\begin{split}
\int_{|x|\geq 2|y|} |x|^{n+\varepsilon} \int_{t\geq |y|} |\psi_t(x-y)-\psi(x)|^2\, \dfrac{dt}{t}\, dx & \lesssim B^2\int_{|x|\geq 2|y|}\int_{t<|y|} \dfrac{|x|^{n+\varepsilon}t^{2\varepsilon}}{(t+|x|)^{2(n+\varepsilon)}}\, \dfrac{dt}{t}\, dx \\
& \lesssim B^2\int_{t<|y|} \int_{|x|\geq 2|y|} |x|^{-n-\varepsilon} \, dx \, \dfrac{dt}{t^{1-2\varepsilon}}\\
&\lesssim B^2 |y|^\varepsilon.
\end{split}
\end{equation}
