  Fix $\lambda>0$. Consider the Hardy space $H^1(\mathbb{R}_+,dm_\lambda)$ in
the sense of Coifman and Weiss, where $\mathbb{R_+}:=(0,\infty)$ and
$dm_\lambda:=x^{2\lambda}dx$ with $dx$ the Lebesgue measure. Also consider the
Bessel operators $\Delta_\lambda:=-\frac{d^2}{dx^2}-\frac{2\lambda}{x} \frac
d{dx}$, and $S_\lambda:=-\frac{d^2}{dx^2}+\frac{\lambda^2-\lambda}{x^2}$ on
$\mathbb{R_+}$. The Hardy spaces $H^1_{\Delta_\lambda}$ and $H^1_{S_\lambda}$
associated with $\Delta_\lambda$ and $S_\lambda$ are defined via the Riesz
transforms $R_{\Delta_\lambda}:=\partial_x (\Delta_\lambda)^{-1/2}$ and
$R_{S_\lambda}:= x^\lambda\partial_x x^{-\lambda} (S_\lambda)^{-1/2}$,
respectively. It is known that $H^1_{\Delta_\lambda}$ and
$H^1(\mathbb{R}_+,dm_\lambda)$ coincide but they are different from
$H^1_{S_\lambda}$. In this article, we prove the following: (a) a weak
factorization of $H^1(\mathbb{R}_+,dm_\lambda)$ by using a bilinear form of the
Riesz transform $R_{\Delta_\lambda}$, which implies the characterization of the
BMO space associated to $\Delta_\lambda$ via the commutators related to
$R_{\Delta_\lambda}$; (b) the BMO space associated to $S_\lambda$ can not be
characterized by commutators related to $R_{S_\lambda}$, which implies that
$H^1_{S_\lambda}$ does not have a weak factorization via a bilinear form of the
Riesz transform $R_{S_\lambda}$.
