# Shae's Ramblings

Stuff I find slightly meaningful and close to achievable

## Minimum of a parametric convex function

Suppose a parametric function $f_\theta \colon A \to \mathbb{R}$ is convex in both $\theta$ and $x\in A$. Then one can prove the function $g(x) = \min_\theta f_\theta(x)$ is convex in $x$ (where $\theta$ is constrained to a compact set $\Theta$).
Indeed, take any two $x_1, x_2 \in A$ and $t\in (0,1)$, we have

\begin{align}
f_\theta(tx_1 + (1-t)x_2) &\leq tf_\theta(x_1) + (1-t)f_\theta(x_2) \\
& \leq t\min_\theta f_\theta(x_1) + (1-t)f_\theta(x_2) \\
& \leq t\min_\theta f_\theta(x_1) + (1-t)\min_\theta f_\theta(x_2) \\
\end{align}

simply because the inequality holds for any $\theta, x_1, x_2$ and thus holds for the parameters which minimize the function. Finally as $\min_\theta f_\theta(x) \leq f_\theta(x)$ for every $x$, we get the final result.
More formally, this argument can be extended to the infimum $\inf_\theta f_\theta(x)$ using the fact the infimum is the greatest lower bound (and must thus be greater than the LHS).

A similar derivation gives that $h(x) = \max_\theta f_\theta(x)$ (and the $\inf$ equivalent) is convex as well.