I feel this may be a stupid question, but when I look up things on convexity, all definitions are in $\mathbb{R}^n$. For example the definition of a simplex or Caratheodory's theorem. I can only find references of these in real spaces.
My question is: doesn't this also hold in complex spaces? I can't see why it wouldn't, but seeing as all definitions I find only mention the reals, I feel like I'm missing something.
We can extend the notion of convex sets to the complex domain without real worry if we keep our convex combinations $[x,y] = \{\theta x + (1-\theta)y : \text{real }0\leq\theta\leq1\}$ to use real weightings. This corresponds to thinking about $\mathbb{C}^n$ as its isomorphically isometric copy $\mathbb{R}^{2n}$.
The issue with extending things further to full generality lies with the lack of total ordering on the $\mathbb{C}$. For example, the standard definition of convexity $f(\theta x + (1-\theta)y) \leq \theta f(x) + (1-\theta) f(y)$ doesn't make much sense in a space like $\mathbb{C}$ without a notion of '$\leq$' that can compare every point.
Nevertheless, people have defined a notion of 'cone' convex functions on an arbitrary vector space $f : \mathcal{V}\to \mathcal{W}$ (such as $\mathbb{C}^n$ or even the set of positive operators $\mathbb{S}^n_+$ on $\mathbb{F}^n$) as follows: If we have a convex cone $K$ on $\mathcal{W}$, we define a partial ordering on $\mathcal{W}$ so that its elements $a\preceq_K b$ if and only if $b-a\in K$. From this we can define $f$ to be $K$-convex if and only if $$ f(\theta x + (1-\theta)y)\preceq_K \theta f(x) + (1-\theta)f(y). $$ It turns out that if $\mathcal{W}$ is a Banach space many properties of regular convex functions (like the linear lower bound, which now uses Frechet derivatives, or the epigraph definition) carry over uninhibited. There are wonderful results in this area from matrix analysis. See Bhatia's text for more.