Pseudo convex quadratic forms are quasi convex

Question

I have been trying to show: A Pseudoconvex quadratic form is Quasiconvex in the same set X, by using the following definitions. These definitions are equivalent to the standard definitions of Pseudo and Quasiconvexity of the quadratic forms. In the proof lemma 8, it is written that this follows from continuity. I am not able to understand how to use continuity to prove this.

Thanks in advance!

Answer 1

Suppose $Q$ is Pseudoconvex and let $x$ and $y$ be in $X$.

If $x'Cx-y'Cy >0$, then by pseudoconvexity of $Q$, it follows that $(x-y)'Cx > 0$.

Suppose $x'Cx-y'Cy =0$. Assume on the contrary that $(x-y)'Cx < 0$, i.e $x'Cx-y'Cx<0$. So

$$(x-y)'C(x-y) = x'Cx+y'Cy-2y'Cx = 2x'Cx-2y'Cx <0$$ Therefore $$0'C0-(x-y)'C(x-y)>0$$ By the pseudoconvexity of $Q$, it follows that $$(0-x+y)'C0>0$$ i.e $$0>0$$ Which is absurd. Hence $(x-y)'Cx \ge 0$ i.e $Q$ is Quasiconvex.