“Smallness” of Fourier transform of atoms A bounded measurable function $\mathfrak{a}$ on $\mathbb{R}^d$ is an atom associated to a ball $B \subset \mathbb{R}^d$, if:

*

*$\mathfrak{a}$ is supported in $B$, with $|\mathfrak{a}(x)| \le 1/m(B)$, for all $x$; and,

*$\int_{\mathbb{R}^d} \mathfrak{a}(x) dx = 0$.

I want to verify that $$\int_{\mathbb{R}^d} |\hat{\mathfrak{a}}(\xi)| \frac{d\xi}{|\xi|^d} \le A,$$
where $\hat{\mathfrak{a}}$ is the Fourier transform of $\mathfrak{a}$ and $A$ is independent of $\mathfrak{a}$.
Once this is shown, we can draw a similar conclusion about the "smallness" of the Fourier transform of functions in the real Hardy space $\mathbf{H}^1_r$, where norms are defined based on those of atoms.
 A: I have figured out the answer myself.
WLOG, we can assume $\mathfrak{a}$ is associated to the unit ball (centered at the origin), because neither translations nor scalings, defined as $\mathfrak{a}_r(x) = r^d \mathfrak{a}(rx)$, would change the integral. (A translation does not change the magnitude of $\hat{\mathfrak{a}}(\xi)$, while a scaling would result in $\hat{\mathfrak{a}_r}(\xi) = \hat{\mathfrak{a}}(\xi/r)$.)
Noticing
$$\hat{\mathfrak{a}}(\xi) = \int_{|t|\le 1} \mathfrak{a}(t) e^{-2\pi i \xi t} dt = \int_{|t|\le 1} \mathfrak{a}(t) \left(e^{-2\pi i \xi t} - 1\right) dt, $$
we have for $|\xi| \le 1$
$$\int_{|\xi| \le 1} |\hat{\mathfrak{a}}(\xi)| \frac{d\xi}{|\xi|^d} = \int_{|\xi|\le 1} \left| \int_{|t|\le 1} \mathfrak{a}(t) \frac{e^{-2\pi i \xi t} - 1}{|\xi|^d} dt \right| d\xi < c.$$
The inequality holds because the integrand is bounded, since $e^{-2\pi i \xi t} - 1 = O(|\xi|^d)$ as $|\xi| \rightarrow 0$.
For $|\xi| \ge 1$, we have
$$
\begin{equation}
\int_{|\xi| \ge 1} |\hat{\mathfrak{a}}(\xi)| \frac{d\xi}{|\xi|^d} \le \left(\int_{|\xi| \ge 1} |\hat{\mathfrak{a}}(\xi)|^2 d\xi\right)^{1/2} \left(\int_{|\xi| \ge 1} \frac{1}{|\xi|^{2d}} d\xi\right)^{1/2} \le \left(\int_{\mathbb{R}^d} |\hat{\mathfrak{a}}(\xi)|^2 d\xi\right)^{1/2} c' \le \left(\int_{\mathbb{R}^d} |\mathfrak{a}(t)|^2 dt\right)^{1/2} c' \le c''.
\end{equation}$$
Combined the two inequalities above, we get the desired result.
