Consider a function $f(x, \theta)$ where $\theta \in [0, \infty)$ and $f$ is an affine function of $\theta$ i.e. $f(x, \theta) = \theta g(x) + h(x)$.
Suppose $f$ is convex for $\theta \in [0,1]$ and there exists some finite value $v > 1$ such that $f$ is convex for $\theta\geq v$.
Does this automatically mean that $f$ is convex for all values of $\theta$ i.e. including the range $\theta\in (1, v]$? If so, how can this be proven and if not, is there an explicit counterexample showing that this doesn't hold?
My idea is relatively simple: The second derivative of $f$ with respect to $x$ shoud be non-positive for $\theta = 0$. This shows that $h$ is convex but I'm not sure if it necessarily means that $g$ is convex, which is the goal here.
A bit more generally, you can check that for $t\in [0,1]$ and any $a,b$ we have
$f(x,ta+(1-t)b)=tf(x,a)+(1-t)f(x,b)$
so convexity for $\theta=a$ and $\theta=b$ implies convexity for all $\theta\in[a,b]$ (linear combination of convex functions with positive coefficients is still convex).