Let $f^\theta(x)$ be a convex function parametrized by $\theta$ in a convex set $\Theta$.
(toy example to illustrate: $f^\theta(x) = x^2 + \theta$, with $\Theta = [0,1]$.)
In general, is set of functions $\mathcal{F} = \{ f^\theta(\cdot), \theta \in \Theta\}$ convex?
The set $\mathcal{F}$ would be convex iff for any $f,g \in \mathcal{F}$, the "line segment" $tf + (1-t)g$ is in $\mathcal{F}$, for $t\in [0,1]$.
No. Let $f^\theta = 1$ for $\theta \in (0,1]$ and $f^0 = 0$. There is no $\theta$ such that $f^\theta = \frac12 (f^0+f^1)$.