  We prove a weak-type (1,1) inequality for square functions of non-commutative
martingales that are simultaneously bounded in $L^2$ and $L^1$. More precisely,
the following non-commutative analogue of a classical result of Burkholder
holds: there exists an absolute constant $K>0$ such that if $\cal{M}$ is a
semi-finite von Neumann algebra and $(\cal{M}_n)^{\infty}_{n=1}$ is an
increasing filtration of von Neumann subalgebras of $\cal{M}$ then for any
given martingale $x=(x_n)^{\infty}_{n=1}$ that is bounded in $L^2(\cal{M})\cap
L^1(\cal{M})$, adapted to $(\cal{M}_n)^{\infty}_{n=1}$, there exist two
\underline{martingale difference} sequences, $a=(a_n)_{n=1}^\infty$ and
$b=(b_n)_{n=1}^\infty$, with $dx_n = a_n + b_n$ for every $n\geq 1$, \[ |
(\sum^\infty_{n=1} a_n^*a_n)^{{1}/{2}}|_{2} + | (\sum^\infty_{n=1}
b_nb_n^*)^{1/2}|_{2} \leq 2| x |_2, \] and \[ | (\sum^\infty_{n=1}
a_n^*a_n)^{{1}/{2}}|_{1,\infty} + | (\sum^\infty_{n=1}
b_nb_n^*)^{1/2}|_{1,\infty} \leq K| x |_1. \] As an application, we obtain the
optimal orders of growth for the constants involved in the Pisier-Xu
non-commutative analogue of the classical Burkholder-Gundy inequalities.
