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Consider $$F(z)=\min ae^{-x}+b e^{-y} s.t. x\ge 0, y\ge 0\text{ and } x+y=z$$

I checked if this function is continuous, but it is not at $z=0$. $F(z)=2\sqrt{ab}e^{-z/2}$ when $z\ne 0$, and $F(0)=a+b$.

I was wondering what the conditions on a program and constraints should be, so that the problem is continuous in its parameters. Is there some theorem which governs this? And does this problem not satisfy the requirements of such a theorem?

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So I think you're missing one key fact. When you do this minimization problem you should find that the points satisfying the minimization problem are $$x^* = \frac{z}{2} + \frac{1}{2} \log(\frac{a}{b})$$ $$y^* = \frac{z}{2} - \frac{1}{2} \log(\frac{a}{b})$$ Now, you require that $x^* \geq 0$ and $y^* \geq 0$ and so we necessarily have that $z$ simultaneously must satisfy both $$z \geq \log(\frac{a}{b})$$ $$z \geq \log(\frac{b}{a})$$ Now, as $z\rightarrow 0$ then it will not satisfy one of these conditions unless $a=b$ which gives you continuity in this case. When $a\neq b$ then eventually you'll require either $x=0$ and $y=z$ or $x=z$ and $y=0$. Either way, you get continuity in $z$.

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  • $\begingroup$ I see the function $F(z)$ is continuous and piecewise convex. Thanks for the clarification. $\endgroup$
    – Bravo
    Commented Apr 30, 2015 at 16:58

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