Let $\Omega\subseteq\Bbb C$ open and $\varphi:\Omega\to\Bbb R$ strongly subharmonic, $\varphi\in\mathcal{C}^2$ such that $\partial\varphi\neq0$.
My problem is to prove that $\bar\partial\partial\varphi>0$.
My attempt: being $\varphi$ strongly subharmonic, by definition, there exists $\chi:\Bbb R\to\Bbb R$ with $\chi,\dot\chi,\ddot\chi>0$ such that $\psi:=\chi^{-1}\circ\varphi$ is subharmonic (that is exactly the definition of strongly subharmonic function I have to use).
From this we get $\varphi=\chi\circ\psi$, hence $$ \partial\varphi=\dot\chi(\psi)\cdot\partial\psi\;\;, $$ from which we know that $\partial\psi\neq0$. Then $$ \bar\partial\partial\varphi=\ddot\chi(\psi)\cdot\bar\partial\psi\partial\psi+\dot\chi(\psi)\cdot\bar\partial\partial\psi\;\;. $$ Next $$ \partial\psi=\frac1{\dot\chi(\chi^{-1}(\varphi))}\partial\varphi= \frac{\partial\varphi}{\dot\chi(\psi)} $$ hence $$ \bar\partial\partial\psi=\frac{\bar\partial\partial\varphi\cdot\dot\chi(\psi)-\frac{\ddot\chi(\psi)}{\dot\chi(\psi)}\bar\partial\varphi\partial\varphi}{[\dot\chi(\psi)]^2} $$ and substituing this last one in the above expression for $\bar\partial\partial\varphi$ we obtain $$ \bar\partial\partial\varphi=\ddot\chi(\psi)\cdot\bar\partial\psi\partial\psi+ \bar\partial\partial\varphi-\bar\partial\varphi\partial\varphi\frac{\ddot\chi(\psi)}{[\dot\chi(\psi)]^2} $$ i.e. $$ [\dot\chi(\psi)]^2\bar\partial\psi\partial\psi= \bar\partial\varphi\partial\varphi $$ which tells me only that $\bar\partial\varphi\partial\varphi $ and $\bar\partial\varphi\partial\psi$ has same sign, but I can't go further.
Any help would be really appreciated! Thank you!
Note that for (real-)differentiable $f\colon \Omega \to \mathbb{C}$, you have
$$\overline{\partial} f = \overline{\partial \overline{f}}.$$
Thus for $g \colon \Omega \to \mathbb{R}$, you have $\overline{\partial} g = \overline{\partial g}$, and hence
$$\overline{\partial}g \partial g = \lvert \partial g\rvert^2,$$
which gives you
$$\overline{\partial}\partial\varphi = \underbrace{\ddot{\chi}(\psi)}_{> 0}\cdot \lvert \partial\psi\rvert^2 + \underbrace{\dot{\chi}(\psi)}_{> 0}\cdot \underbrace{\overline{\partial}\partial \psi}_{\geqslant 0}.$$
Since $\partial\varphi = \dot{\chi}(\psi)\cdot \partial\psi$, the assumption $\partial\varphi \neq 0$ implies $\partial\psi \neq 0$ and hence $\lvert \partial\psi\rvert^2 > 0$, which shows that $\overline{\partial}\partial\varphi > 0$.