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I'm trying to show that if $(x^*,y^*)$ is a local minimum for

$ \min_{x,y} f(x) \\ s.t. \; c(x) = y^2 $

then $x^*$ is a local minimum for

$ \min_{x} f(x) \\ s.t. \; c(x) \geq 0 $

If we consider globally optimal solutions then the result is easy to see:

$c(x)\geq0 \Rightarrow \exists y : (x,y)\in\{c(x)=y^2\} \Rightarrow f(x^*)\leq f(x)$

However, for some reason I haven't managed to show the corresponding result for local minima. Intuitively, for small enough $\varepsilon$ one should have

$x \in \{x:c(x)\geq0\}\cap B_{\varepsilon}(x^*) \Rightarrow \exists y : (x,y)\in\{(x,y) : c(x)=y^2\} \cap B_{\delta}(x^{*},y^*) \Rightarrow f(x^*)\leq f(x)$,

where the $B$:s are norm balls, but I haven't been able to prove the first implication here. Any ideas?

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    First note that $c$ must be continuous. Then use the epsilon-delta definition of continuity to relate a change in $x$ to a change in $y$. – 2017-01-26
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    Ah, of course! The continuity of c ensures that $|c(x)-c(x^*)|=|y^2-y^*2|$ can be made arbitrary small. I'll have to think a little to get the details straight, but thanks a lot! – 2017-01-26

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