For $a \in \mathbb{R}_{\ge 0}^d$ and $b \in \mathbb{R}_{\ge 0}$, we can define a half-space as the set of points $x \in \mathbb{R}^d$ such that $a \cdot x \le b$, namely,
$$\mathcal{H}(a,b) = \{x \in \mathbb{R}^d \mid a \cdot x \le b\}.$$
Fix $\mu \in \mathbb{R}_{\ge 0}^d$. Let $X$ be a random variable on $\mathbb{R}^d$ with a Gaussian distribution, in particular, with the following distribution: $X=(X_1,X_2,\dots,X_d)$ where $X_1,X_2,\dots,X_d$ are mutually independent and $X_i \sim \mathcal{N}(\mu_i,1)$ (i.e., $X_i$ has a Gaussian distribution with mean $\mu_i$ and variance $1$).
We can define a function $f:\mathbb{R}_{\ge 0}^{d+1} \to \mathbb{R}$ as
$$f(a,b) = \Pr[X \in \mathcal{H}(a,b)].$$
Of course, $f$ implicitly depends on $\mu$.
My question: is $f$ convex/concave? Does it have at most one local maximum on closed, convex subsets of $\mathbb{R}_{\ge 0}^{d+1}$? If not, does it have any other nice properties that would make it easy to computationally find a maximum of $f$, given $\mu$?
(This arose while examining an application, explained here: Approximating a convex polyhedron, with fewer inequalities.)
The function $f$ does not have a maximum. You can take a sequence of $(a_k,b_k)$ such that $f(a_k,b_k)\to 1$ while $f(a,b)<1$ for all $a,b$.