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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.

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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 $\exp$.

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  • $\begingroup$ Thanks for the reply. Does that example still work if x is restricted to be non-negative? $\endgroup$ – DanZ Jun 7 '13 at 19:33
  • $\begingroup$ Yes. Note that $x$ is always non-negative if $x = \exp(y)$. $\endgroup$ – gerw Jun 9 '13 at 19:38

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