Suppose I have some $f(x) = g^{T}x + \frac{1}{2}x^{T}Hx$ where $g \in \mathbb{R}^{n}$ and $H$ is symmetric, positive semi-definite. This is the standard form for a quadratic minimization function.
It is just briefly mentioned in my textbook that unlike a positive definite $H$ which ensures $f(x)$ has a global minimiser, or an indefinite $H$ which ensures $f(x)$ is unbounded below, a positive semi-definite $H$ indicates that $f$ either has infinitely many solutions or is unbounded below. Considering that positive semi-definiteness of $H$ ensures $x^{T}Hx \geq 0 \ \forall \ x$, I don't see how this follows.
Suppose $x^THx=0$. Then certainly $\lambda x$ for any real $\lambda$ satisfies the same equation, and provided $g$ is nonzero, we may make the quantity $f(x)$ as small as we want. Otherwise if $g$ is zero, then by the same logic we have infinitely-many zeros of $f$, which in this case are minima.