It's well-known that $x \mapsto x^\top A x$ is a convex function of $x \in \mathbb R^n$ when the matrix $A$ is (real) positive semidefinite (PSD). Is this still true when $x \in \mathbb C^n$ and $A$ is Hermitian PSD? More precisely, is $x \mapsto x^*Ax$ a convex function of $x$ when we view $x$ as a variable in $\mathbb R^{2n}$, i.e., $$\left[\begin{array}{c} \mathrm{Re}(x)\\ \mathrm{Im}(x)\end{array}\right]?$$
EDITS 2023/2/16:
Thanks to user550103's comments below, the answer seems to be yes, and a simple proof may be:
Let $f(x)\triangleq x^*Ax=g(h(x))$, where $g(x)=\Vert x\Vert^2$ and $h(x)=\sqrt{A}x$. Since $g$ is clearly convex, and $h$ is a linear transformation, it follows that $f=g\circ h$ is convex, whether we view $x$ as a variable in $\mathbb C^n$ or $\mathbb R^{2n}$, c.f. $g(h(ax+by))=g(ah(x)+bh(y))\le ag(h(x))+bg(h(y))$.
Please correct me if I'm mistaken somewhere above.