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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.

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    $\begingroup$ Your argument is correct, yes. Be careful though that the convergence of the partial derivatives in $L^2$ is a weak convergence (that's all you need of course). If $u$ is unbounded, this is not true anymore though (take $u=\log |z|^2$ on $\mathbb C$ so that $\frac{\partial u}{\partial z}=1/z$ is not in $L^2_{\rm loc}$ near $0$. $\endgroup$
    – Henri
    Jul 12, 2022 at 15:19
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    $\begingroup$ Hello Henri, many thanks. Indeed it's a weak convergence. I realize I made a mistake when I state my second question... So suppose $(u_{j})$ are PSH and decrease to $u$ point-wise and the partial derivates of $u$ belong to $L^{2}_{loc}$. 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? $\endgroup$
    – Analyse300
    Jul 12, 2022 at 17:01
  • $\begingroup$ do you assume u bounded again for that last question? If so, the same arguments apply immediately. Also, are you looking for a strong L^2 convergence? $\endgroup$
    – Henri
    Jul 13, 2022 at 10:27
  • $\begingroup$ Hello. No I didn't assume $u$ is locally bounded (if it is, same argument work as I say in my first message). I'm looking for a $L^{2}_{loc}$ convergence of the partial derivates of $u_{j}$. $\endgroup$
    – Analyse300
    Jul 13, 2022 at 17:09
  • $\begingroup$ For the second question I realize there is a mistake in my "proof". Suppose $u$ is locally bounded (if not I still don't know). We can suppose $u \ge 1$. Normally, $du \wedge d^{c}u$ is defined as $du \wedge d^{c}u:= \frac{1}{2}dd^{c}u^{2} - udd^{c}u $. $\endgroup$
    – Analyse300
    Jul 15, 2022 at 17:42

1 Answer 1

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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.

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