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?



When is a convex program continuous in its constraint vectors?
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 $z$.


