# Why does the solution to one system imply the other system has a negative solution?

Consider two systems A and B:

A$$: f_i(x) \lt 0, i=1, \dots, m, Ax = b$$

B$$: \min_{\{x,s\}} s$$ subject to $$f_i(x) - s \le 0, i = 1, \dots, m; Ax = b$$

Then B has optimal value $$p^* \lt 0$$ iff there exists a solution to A

I can prove $$\Rightarrow$$ but I'm having trouble showing $$\Leftarrow$$.

My original attempt was to assume $$x$$ is the solution of A, and therefore: IF $$x$$ is used in B then the smallest $$s$$ that satisfies the equation must be in $$[f_i(x), 0)$$ since $$f_i(x) \lt 0$$. But this is assuming that for $$B$$ the same value of $$x$$ is used. Which doesn't seem right because $$x$$ might not work for B.

Any ideas?

• can you show the first few steps for $\Leftarrow$? – LinAlg Feb 17 at 16:06
• A bit of pedantry here: if $B$ is unbounded below—and it certainly can be for certain instances of this—is it correct to say that $B$ has an optimal value $p^*<0$? After all, it has no finite optimal value. @LinAlg what do you think? – Michael Grant Feb 18 at 2:41
• And even if $p^*$ is finite, that doesn't mean there's an $x$ that attains it. (Though there must be such an $x$ for at least one $p$ satisfying $p^*\leq p<0$.) – Michael Grant Feb 18 at 2:43
• You really should combine your two questions together (the other being math.stackexchange.com/questions/3116263/…). – Michael Grant Feb 18 at 2:45
• @MichaelGrant not pedantic: you are right! – LinAlg Feb 18 at 2:46