So I am attempting to prove that the domain of a concave function is convex. The domain of $f$ is defined as
dom$f:=\{x\in\mathbb{R}^n\mid f(x)<+\infty\}$
and the definition of $f$ being concave is that $\forall x,y\in\mathbb{R}^n,\forall t\in[0,1]$, where $f(tx+(1-t)y)\geq tf(x)+(1-t)f(y)$.
And so to prove that the domain is convex means $\forall x,y\in$ dom$f, \forall t\in[0,1]:tx+(1-t)y\in$ dom$f$, which is equivalent to showing $f(tx+(1-t)y)< +\infty$.
However, I am having difficulty using the definition of $f$ being concave. Is there a different definition to use. We use some conventions handling operations with $\infty$, such as $\infty\cdot0=\infty$, $\infty-\infty=\infty$, and $(-\infty)\cdot0=0$. But I do not see how these are useful for the definition of $f$ being concave.
Best Answer
You seem to be using the extended-value extension for convex functions, which doesn't really makes sense for a concave function.
Typically a concave function is extended by defining it to be $-\infty$ outside its domain (see Boyd and Vandenberghe section 3.1.2). The idea is if you try to optimize a function $f$ outside of its domain, you get "infinitely bad" answers. Since we usually maximize a concave $f$, "infinitely bad" here means we define the extension $\tilde{f}(x)=-\infty$ for $x \notin \text{dom } f $.
If we define concavity, for the extended function $\tilde{f}$, as: $$ \forall x, y, 0 <\theta < 1, \tilde{f}(\theta x + (1-\theta)y) \ge \theta \tilde{f}(x) + (1-\theta)\tilde{f}(y)$$ then it's not hard to see why $\text{dom }f := \{ x |\tilde{f}(x) >-\infty \}$ is convex by our definition.