Issue with proof by contradiction of necessary second-order condition for convexity

Question

We wish to prove that if $f : K \to \Bbb R$ is a $\mathcal C^2$, convex function on a convex domain $K \subset \Bbb R^n$, then it has a positive semidefinite Hessian matrix on every point $x \in K$.

To do this, I have a proof by contradiction: we assume there's a point $\bar x \in K$ where the Hessian is not positive semidefinite. That should mean $$ \exists y \in \Bbb R^n : y^T \nabla^2f(\bar x) y < 0 $$ but then instead the proof seems to go on implying that $\bar x^T\nabla^2f(\bar x)\bar x < 0$, while this does not seem true in general.

Or is it?

And hence how would you write such a proof (by contradiction)?

Answer 1

I think the problem is a subtle issue with the meaning of "positive semidefinite Hessian matrix on every point $x \in K$".

The actual theorem should be:

Theorem. If $f : K \to \Bbb R$ is twice differentiable, convex function on the convex domain $K \subseteq \Bbb R^n$, then for every point $x \in K$, the Hessian matrix $\nabla^2 f(x)$ is positive semidefinite on every point $x \in K$: $$ \forall y \in K, \quad y^T\nabla^2f(x)y \geq 0 $$

So that the definiteness condition is stated for $K$. Now the negation of this fact is precisely that there exists an $\bar x \in K$ such that $$ \exists y \in K, \quad y^T\nabla^2(\bar x)y < 0. $$