A set of mixture distributions $Q$ is defined as,
$Q = \{f | f(.) = \sum_{i=1}^{k} q_{i} f_{i}(.) \}$,
where each $f_{i}$ is a probability density and $\sum_{i=1}^{k} q_{i}=1$ and $q_{i}>0$.
Is set of all mixture distributions, $Q$ a convex set?
To check if it is convex or not, I took two members $f_{1}$ and $f_{2}$ and tried to check if $\lambda f_{1} + (1-\lambda)f_{2} \in Q$ where $\lambda \in [0,1] $.
But not sure, how to prove it.
Well let $g,h \in Q$ and $g = \sum_{i=1}^kq_i f_i$ and $h = \sum_{i=1}^kp_i f_i$ the representations given by definition of $Q$. Then \begin{align*} \lambda g+(1-\lambda)h &= \lambda \sum_{i=1}^kq_i f_i+(1-\lambda)\sum_{i=1}^kp_i f_i\\ &= \sum_{i=1}^k(\lambda q_i+(1-\lambda)p_i)f_i\\ &= \sum_{i=1}^kr_if_i \end{align*} where I defined $r_i:= \lambda q_i+(1-\lambda)p_i$. So it remains to check, that $\sum_{i=1}^kr_i = 1$ and $r_i>0$. The latter is trivial, for the former we calculate $$\sum_{i=1}^kr_i = \lambda \sum_{i=1}^kq_i+(1-\lambda)\sum_{i=1}^kp_i = \lambda +(1-\lambda) = 1$$ Hope that helps.