From the definition of convexity:
Set $S$ is convex if $x, y \in S$ and the line segment $\theta x + (1-\theta) y$ is also in the set.
How would you find the convexity of something like the following?
$$S = \{(P_{mn}, Q_{mn}, v_m, l_{mn}):P^2_{mn}+Q^2_{mn}=v_ml_{mn}\}$$
where $v_m > 0$ and $l_{mn} \geq 0$. How would you apply the definition of convexity to this?
If $(P_{mn}, Q_{mn}, v_m, l_{mn})$ and $(P'_{mn}, Q_{mn}', v'_m, l'_m)$ are in $S$. Then for an arbitrary $\lambda\in (0,1)$, we need to check whether $$(\lambda P_{mn}+(1-\lambda)P'_{mn},\lambda Q_{mn}+(1-\lambda)Q'_{mn}, \lambda v_{m}+(1-\lambda)v'_{m} , \lambda l_{mn}+(1-\lambda)l'_{mn})\in S,$$ or more explicitly, whether $$[\lambda P_{mn}+(1-\lambda)P'_{mn}]^2+[\lambda Q_{mn}+(1-\lambda)Q'_{mn}]^2=[\lambda v_{m}+(1-\lambda)v'_{m}][ \lambda l_{mn}+(1-\lambda)l'_{mn}],$$ or equivalently, whether \begin{equation}2(P_{mn}P'_{mn}+Q_{mn}Q'_{mn})=v_ml'_{mn}+v'_ml_{mn}.\,\,\,\,\,\,\,\,(*)\end{equation} Then it is easy to see that $S$ is not convex, as for any $v_m,l_{mn}>0$, we have $$(\sqrt{v_ml_{mn}}, 0, v_m, l_{mn}), (0,\sqrt{v_ml_{mn}}, v_m, l_{mn})\in S.$$ But plug these two points back to $(*)$, it is easy to see that the LHS $=0$ but the RHS is strictly positive.