Show that a $C^2$-function $u$ is plurisubharmonic if and only if the Hessian matrix $H_u(z)(\omega, \omega)>0$

Question

I'm trying to show that the theorem in my book:

A $C^2$-function $u$ is plurisubharmonic if and only if the matrix (the complex Hessian) $$H_u(z)=\left( \dfrac{\partial^2 u}{\partial z_j \partial \bar z_k} \right)$$ is positive semidefinite. It means that $\forall \omega=(\omega_1, \ldots, \omega_n) \in \Bbb{C}^n$ then

$$H_u(z)(\omega,\omega)=\sum_{j,k=1}^{n}\dfrac{\partial ^2 u}{\partial z_j \partial \overline{z_k}}(z)\cdot \omega_j \overline{\omega_k}\ge 0$$

Solution

$\bigstar $ We'll prove that $\forall z \in \Omega, \omega \in \Bbb C^n,\xi \in \Bbb C $ then $$\dfrac{1}{4}\Delta_\xi u(z+\xi\omega)\mid_{\xi=0}=\sum_{j,k=1}^{n}\dfrac{\partial ^2 u}{\partial z_j \partial \overline{z_k}}(z)\cdot \omega_j \overline{\omega_k}$$

• Since $\xi=x+iy \in \Bbb C \implies \left\{\begin{matrix} & x=\dfrac{\xi+\overline{\xi}}{2}\\ & y=\dfrac{\xi-\overline{\xi}}{2i} \end{matrix}\right.$

• Since $\dfrac{\partial u}{\partial \overline{\xi}}=\dfrac{1}{2}\dfrac{\partial u}{\partial x}-\dfrac{1}{2i}\dfrac{\partial u}{\partial y} \ \text{and} \ \dfrac{\partial u}{\partial {\xi}}=\dfrac{1}{2}\dfrac{\partial u}{\partial x}-\dfrac{i}{2}\dfrac{\partial u}{\partial y}$ we have
  $$ \begin{align*}\dfrac{\partial^2 u}{\partial \xi \partial \overline{\xi}} &=\dfrac{1}{4}\left (\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2} \right )\\ &=\dfrac{1}{4}\Delta _\xi u \tag{I}\end{align*}$$

• Now, I have stuck when I want to show that $$\dfrac{\partial^2 u (z+\xi\omega)}{\partial \xi \partial \overline{\xi}}\mid_{\xi=0}=\sum_{j,k=1}^{n}\dfrac{\partial ^2 u}{\partial z_j \partial \overline{z_k}}(z)\cdot \omega_j \overline{\omega_k} \tag{II}$$

$\bigstar $ Since $(I)$ & $(II)$ we're done.

But How can we show that $(II)$?

Any help will be appreciated! Thanks!

Answer 1

(II) is a computation with the multivariable chain rule. Applied once, the rule yields $$\dfrac{\partial u (z+\xi\omega)}{\partial \xi }=\sum_{j=1}^{n}\dfrac{\partial u}{\partial z_j }(z)\cdot \omega_j \tag1$$ Now, if the right side is denoted $g(z)$, we have the analog of (1), $$\dfrac{\partial g (z+\xi\omega)}{\partial \overline\xi }=\sum_{k=1}^{n}\dfrac{\partial g}{\partial \overline z_k }(z)\cdot \overline \omega_k \tag2$$ Plug the formula for $g$ into the right side of (2), and conclude.

Answer 2

Show that $(II)$:

• We have: $$\forall z \in \Omega \subset \Bbb C^n, \omega \in \Bbb C^n, \xi \in \Bbb C \implies z+\xi \omega =\left (z_1+\xi \omega_1,\ldots, z_n+\xi \omega_n \right ) \in \Bbb C^n$$

• Applying the multivariable chain rule, we have: $$\dfrac{\partial u (z+\xi\omega)}{\partial \xi }\mid_{\xi=0}=\sum_{j=1}^{n}\dfrac{\partial u}{\partial z_j }(z)\cdot \omega_j:=g(z) $$

• Thus, LHS of $(I)$ becomes:$$\begin{align*}\\ \dfrac{\partial g (z+\xi\omega)}{\partial \overline\xi }\mid_{\xi=0}&=\dfrac{\partial^2 u (z+\xi\omega)}{ \partial \overline{\xi} \partial \xi}\mid_{\xi=0}\\ &=\dfrac{\partial }{\partial \overline{\xi} }\left ( \sum_{j=1}^{n}\dfrac{\partial u}{\partial z_j }(z+\xi \omega)\cdot \omega_j \right )\mid_{\xi=0}\\ &=\sum_{k=1}^{n}\dfrac{\partial^2 u}{\partial z_j \partial \overline z_k }(z)\cdot \omega_j \overline{\omega_k} \blacksquare \end{align*} $$ We're done!