If there are two sets of linear constraints in different variables, Ax <= b with x_l <= x <= x_u and Cy <= d with y_l <= y <= y_u, and a set of equality constraints of a specific non-linear but convex form relating the two sets of variables e^y - x = 0, is the feasible space convex, and what is the proof behind the answer? The difficulty seems to me that the equality constraint would be equivalently formulated as e^y - x <= 0 and e^y - x >= 0, which means that one of these inequalities would be convex and the other would be concave, which may somehow create concavities in the feasible space.
2026-04-28 20:12:58.1777407178
convexity of two linear spaces connected by a convex equality constraint
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No. All equality constraints has to be linear -- this is almost your observation.
As an example, take $-1 \le x \le 1$, $-1 \le y \le 1$ and $\exp(y) = x$. Then, the feasible set is a non-convex part of the graph of $\exp$.