Let $f$ be a complex function on an open connected set in $\mathbb{C}$. Let $g:\mathbb{C}\rightarrow \mathbb{C}^2$ be such that $g(z)=(z,f(z))$. If for any plurisubharmonic function $u$ on $\mathbb{C}^2$, $u\circ g$ is subharmonic, then prove that $f$ is holomorphic.
I guess it is a deal of chain rule, but my tries never successfull. I really want to do it myself, so can somebody please give me hints.
You can't use the chain rule or talk about derivatives before you establish that $f$ is actually smooth. Here is a sketch of the argument: If $u$ is pluriharmonic, then both $u$ and $-u$ are plurisubharmonic, so $\pm u \circ g$ are both subharmonic, which implies that $u \circ g$ is harmonic. If $u(z,w) = v(w)$ with a harmonic function $v$, then $u$ is pluriharmonic, so $v \circ f$ is harmonic. This shows that $f$ is a harmonic morphism (i.e., it preserves harmonic maps), and by a classical result this implies that $f$ is either holomorphic or anti-holomorphic. Now you can make the same argument for functions of the form $u(z,w) = v(z+w)$ and conclude that $z+f(z)$ is either holomorphic or anti-holomorphic. Combine these and your result follows.
