Implications of bounded support on regression equation I am reading an academic paper right now, and the following claim is made without much explanation:

Suppose that the joint distribution $F(\textbf y)$ of the $n \times 1$ random vector  $\textbf Y$ has a density  $f(\textbf y )$ with bounded support $\mathscr Y$. Without loss of generality assume that the true coefficient equals $\beta_0=0$ and that  $\sigma^2=1.$
Because $\textbf Y$ has bounded support $\mathscr Y$ there is a set $B \subset \mathbb R^m$ such that  $|\textbf y' \Sigma^{-1}\textbf X \beta|<1$ for all $\beta \in B $ and $\textbf y \in \mathscr Y$.

I  don't understand how we are able to conclude that $|\textbf y' \Sigma^{-1}\textbf X \beta|<1$. Can anyone explain how this follows from the assumption of bounded support?

A try a proof using the notation of the linked paper A Modern Gauss-Markov Theorem.  As this is a finite-sample result, $n$ is a fixed integer $\Sigma>0$ is a known, fixed, positive-definite (posdef) matrix, and the design matrix $X$ is assumed known, so a constant. The range of $Y$, $\mathscr{Y}$, is a bounded set, so contained in a compact set, so we can assume it is compact.
Interest is in the expression $y^T \Sigma^{-1} X\beta$ which we can write as the continuous function
$$ T(y,\beta)=y^T \Sigma^{-1} X\beta $$ it is not necessary to include the constants as arguments. For each $\beta$, the image set $T(\mathscr{Y},\beta)$ is a compact set, by continuity of $T$. It is clear that $T(\mathscr{Y}, 0)=\{ 0 \}$ so the condition $ | T | < 1$ is fulfilled at least for one $\beta$, namely $\beta=0$.
Now write
$$ \sup_{\beta\in B,~ y \in \mathscr{Y}} = | T(\mathscr{Y}, B) |
$$ Now fix one nonzero $\beta$ and observe that
$$
| T(\mathscr{Y}, c\beta)| = |c| |T(\mathscr{Y}, \beta) |
$$ so is increasing linearly along $\beta$-rays from the origin. If this last value is $| c | G >0$, the condition $ |T | <1 $ will be fulfilled for $ |c| <  G^{-1}$.
We can conclude that the sought-for set $B$ is non-empty, larger than just $\{ 0 \}$, and it is in fact a cone.
