Why does continuity of the Hessian matrix allow us to assume a point is an interior point in the following proof?

Question

Background

I am studying Optimisation and I encounter the following statement to be proved:

Let $f$ be a twice continuously differentiable function. If $f$ is a convex function over a convex set $\Omega$ containing an interior point, then the Hessian matrix of $f$ is positve semidefinite throughout $\Omega$.

Problem

In the proof, the following statement appears: "If $\nabla^2 f$ is not positive semi-definite at some $x\in\Omega$, then the continuity of the Hessian allows us to assume that $x$ is an interior point of $\Omega$.'' Why is this statement true?

Answer 1

It is an odd way to prove the result, but there you are.

If the convex set $\Omega$ has an interior point $x$, then for any $p \in \Omega$ and any $t \in (0,1]$ the point $tx+(1-t)p $ also lies in the interior.

If $H(p)$ is not positive semidefinite then by continuity and the previous result we see that for some nearby $x$ in the interior then $H(x)$ is also not positive semidefinite.