# 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.

• 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 $0$. Jul 12, 2022 at 15:19
• 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? Jul 12, 2022 at 17:01
• 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? Jul 13, 2022 at 10:27
• 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 $u_{j}$. Jul 13, 2022 at 17:09
• 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 $du \wedge d^{c}u:= \frac{1}{2}dd^{c}u^{2} - udd^{c}u$. Jul 15, 2022 at 17:42

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 $$0$$.