Show those PSH functions belongs to Sobolev space Let u be a plurisubharmonic function defined on the unit ball $\mathbb{B}$ of $\mathbb{C}^{k}$ such that $u \ge 1$.
Question: why the partial derivates $\frac{\partial u}{\partial x_{i}}$ (which are defined in the weak sense of distributions) of $u$ belongs to $L^{2}_{loc}$? (Here I wright $z_{i} = x_{i} + iy_{i}$ the standards coordinates).
More generally, suppose $u$ is such that the partial derivates $\frac{\partial u}{\partial x_{i}}, \frac{\partial u}{\partial y_{i}}$ of u belongs to $L^{2}_{loc}$.
Question: if $(u_{j})$ are PSH and decrease to $u$ point-wise, then why do the partial derivates $\frac{\partial u_j}{\partial x_{i}}, \frac{\partial u_j}{\partial y_{i}}$ of $u_j$ belongs to $L^{2}_{loc}$ for $j$ large enough and converge to  $\frac{\partial u}{\partial x_{i}}, \frac{\partial u}{\partial y_{i}}$ in the $L^{2}_{loc}$ sense?
Here is what I've tried:
Let us recall $d = \partial + \bar{\partial}$ and $d^{c} = \frac{1}{2\pi i}(\partial - \bar{\partial})$. The result is local so it's enough to show it on a relatively open set of $\mathbb{B}$. Since $u \ge 1$, $u^{2}$ is also plurisubharmonic. One can then approximate $u$ by a non-increasing sequence $(u_{j})$ of smooth plurisubharmonic functions in the $L^{1}_{loc}$ sense. We have
$$
dd^{c}u^{2}_{j} = 2u_{j}dd^{c}u_{j} + 2 du_{j} \wedge d^{c}u_{j}.
$$
Let $l$ be an index and set
$$
T := dz_{1} \wedge d\bar{z}_{1} \wedge \ldots \wedge dz_{l-1} \wedge d\bar{z}_{l-1} \wedge dz_{l+1} \wedge d\bar{z}_{l+1} \wedge \ldots \wedge dz_{n} \wedge d\bar{z}_{n}.
$$
Then, for any compact $K$,
$$
\text{Constant}\times \int_{K}\left|\frac{\partial u_{j}}{\partial z_{l}}\right|^{2}dV = \int_{K}(dd^{c}u^{2}_{j} - 2u_{j}dd^{c}u_{j}) \wedge T
$$
which is uniformly bounded in $j$ by the Chern–Levine–Nirenberg inequalities. Then, by compactness in $L^{2}_{loc}$, one can suppose $(\frac{\partial u_{j}}{\partial z_{l}})$ converge in $L^{2}_{loc}$. But as it already converges in the sense of distributions to $\frac{\partial u}{\partial z_{l}}$, it follows $\frac{\partial u}{\partial z_{l}}$ (and thus $\frac{\partial u}{\partial x_{i}}, \frac{\partial u}{\partial y_{i}}$) belongs to $L^{2}_{loc}$. What do you think?
If $u$ is bounded from below, then I can use the previous part and results about Monge-Ampère operators to conclude. But in general, I don't know this is true.
I thank you in advance and wish you a good day.
 A: I think it's a partial answer when $u$ and its gradient are locally bounded. Let's first show that
$$
du \wedge d^{c}u = \frac{1}{2 \pi i}\sum_{j}\bigg(\frac{\partial u}{\partial z_{j}}dz_{j} + \frac{\partial u}{\partial \bar{z}_{j}}d\bar{z}_{j}\bigg) \wedge \bigg(\sum_{l}\frac{\partial u}{\partial z_{l}}dz_{l} - \frac{\partial u}{\partial \bar{z}_{l}}d\bar{z}_{l}\bigg).
$$
Let $T$ be a smooth positive current of bidimension $(n-1, n-1)$. The assumption is local so it's enough to show it on a relatively compact open set. On such an open $W$, $u$ is limit of a decreasing sequence $(u_{j})$ of smooth plurisubharmonic functions. Such functions can be construct by convolution between $u$ and smooth kernels. One knows (see e.g. Lemma 9.1. page 266 of the book Functional Analysis, Sobolev Spaces and Partial Differential Equations of Brezis) that those  $u_{j}$ are thus in the Sobolev space $W^{1, 2}$ and converge to $u$ in the $W^{1, 2}$ sens. Now one knows that
$$
du_{n} \wedge d^{c}u_{n} \wedge T = \frac{1}{2 \pi i}\sum_{j}\bigg(\frac{\partial u_{n}}{\partial z_{j}}dz_{j} + \frac{\partial u_{n}}{\partial \bar{z}_{j}}d\bar{z}_{j}\bigg) \wedge \bigg(\sum_{l}\frac{\partial u_{n}}{\partial z_{l}}dz_{l} - \frac{\partial u_{n}}{\partial \bar{z}_{l}}d\bar{z}_{l}\bigg) \wedge T.
$$
The left hand side converge (in the weak sense of currents) to $du \wedge d^{c}u \wedge T$ (by "continuity" of the Monge-Ampère operator) while the right hand side converge (again in the weak sense of currents) to
$$
\frac{1}{2 \pi i}\sum_{j}\bigg(\frac{\partial u}{\partial z_{j}}dz_{j} + \frac{\partial u}{\partial \bar{z}_{j}}d\bar{z}_{j}\bigg) \wedge \bigg(\sum_{l}\frac{\partial u}{\partial z_{l}}dz_{l} - \frac{\partial u}{\partial \bar{z}_{l}}d\bar{z}_{l}\bigg) \wedge T
$$
by what we have been saying above.
Let $l$ be a fixed index, and $K$ be a compact $W$, let
$$
T := idz_{1} \wedge d\bar{z}_{1} \wedge \ldots \wedge idz_{l-1} \wedge d\bar{z}_{l-1} \wedge idz_{l+1} \wedge d\bar{z}_{l+1} \wedge \ldots \wedge idz_{n} \wedge d\bar{z}_{n},
$$
and finally let $(u_{j})$ be a non-increasing sequence of plurisubharmonic functions (not necessarily smooth) which converges to $u$. Given these assumptions, there exists a certain constant $c > 0$ such that
$$
\begin{split}
c \int_{K}\left|\frac{\partial u_{j}}{\partial z_{l}} \right|^{2}dV -  c \int_{K}\left|\frac{\partial u}{\partial z_{l}} \right|^{2}dV &= \int_{K}\big[dd^{c}(u_{j})^{2} - 2(u_{j})dd^{c}(u_{j})\big] \wedge T \\
&\quad- \int_{K}\big[dd^{c}(u)^{2} - 2(u)dd^{c}(u-u)\big] \wedge T
\end{split}
$$
which converges to $0$ by the continuity properties of the Monge-Ampère operator.  We then have
$$
c \int_{K}\left|\frac{\partial u}{\partial z_{l}} - \frac{\partial u_{j}}{\partial z_{l}} \right|^{2}dV = c \int_{K} \left|\frac{\partial u}{\partial z_{l}}\right|^{2} + \left|\frac{\partial u_{j}}{\partial z_{l}}\right|^{2}dV - c \int_{K} \frac{\partial u_{j}}{\partial z_{l}}\frac{\partial  u}{\partial z_{l}}.
\label{1}\tag{1}
$$
But
$$
\bigg|\int_{K} \frac{\partial u_{j}}{\partial z_{l}}\frac{\partial u}{\partial z_{l}}  - \int_{K} \frac{\partial u}{\partial z_{l}}\frac{\partial u}{\partial z_{l}}\bigg|\leq \mathrm{const.}\times \left\| \frac{\partial u_{j}}{\partial z_{l}} -  \frac{\partial u}{\partial z_{l}}\right\|_{L^{1}(K)}
$$
The last member converges to $0$ (indeed, one know that if $u_{j}$ are PSH and decrease to $u$, then the convergence is $L^{p}_{loc}$ for all $p \ge 1$ and the gradient converge in the $L^{q}_{loc}$ sense for all $1 \leq p < 2$). We can then conclude \eqref{1} converges to $0$.
I know that we post an answer when we have a complete answer which is not my case since my arguments work for the locally bounded case only. I edit my first "answer" since I couldn't apply the Monge-Ampère operator to a difference of PSH functions.
