In most books or papers, the convex problems are usually defined on the Real field. However, there are also a lot of works focus on the problem defined on the Complex field but also use the KKT conditions to solve the close-form solution.
My problem is how to judge that if a function with complex variable is convex function. As is known to all, for real variables, the first order condition is given by $f(y) \geqslant f(x)+\nabla f(x)^{T}(y-x)$
The second order condition is given by $\nabla^{2} f(x) \succeq 0$.
So, what's the conditions of the convex problem when vairable is complex? Thanks a lot in advance.
Typically, most of the functions map the complex field to real field, i.e., $f: \mathbb{C}^n \mapsto \mathbb{R}$.
The first order condition for the complex field reads $f(y) \geq f(x) + 2 \Re\left\{ \nabla f(x)^H \left( y-x \right) \right\}$, where $(\cdot)^H$ is Hermitian transpose.
The second order condition for the complex field (and real field) reads same, i.e., $\nabla^2 f(x) \succeq 0$ for all $x$.
Reference:
Ali H. Sayed (2014), "Adaptation, Learning, and Optimization over Networks", Foundations and Trends® in Machine Learning: Vol. 7: No. 4-5, pp 311-801. http://dx.doi.org/10.1561/2200000051.