If $\forall y ≻ 0, \langle \lambda, y \rangle \geq \mu$, is it true that $\mu ≤ 0$ and $ λ \succcurlyeq 0$?

Question

Example 2.21, p. 50. of Boyd's Convex Optimization establishes conditions for the solvability of strict linear inequalities, where, at one point the book basically says that

the inequality $\forall y > 0, λ^\top y ≥ \mu$ implies that $\mu ≤ 0$, and $ λ \succcurlyeq 0$...

Can anybody help me see why?

Below attached is the whole context of my question:

Answer 1

How small can we make $\lambda^Ty$ by choosing $y$ subject to the restriction $y\succ 0$? Two cases:

1. $\lambda$ has a negative component. Then we can make $\lambda^Ty$ very negative by letting the corresponding component of $y$ be very large. Hence, there is no lower bound on $\lambda^Ty$ in this case.
2. All components of $\lambda$ are nonnegative, that is $\lambda\succeq 0$. Then there is no way for $\lambda^Ty$ to be negative but we can make it as close to zero as we wish by choosing very small positive numbers for the components of $y$. In particular, if $\mu$ were positive, we'd be able to make $\lambda^Ty$ less than $\mu$.

So, given that $\lambda^Ty\ge \mu$ for all $y\succ0$, we can conclude that

• Only Case 2 can occur; that is $\lambda\succeq 0$
• $\mu\le 0$