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: Because $Y$ has bounded support $\mathcal S,$ the set $\mathcal S \Sigma^{-1}X = \{y\Sigma^{-1}X\mid y\in\mathcal S\}$ is bounded, too.   Let $c\gt 0$ be any upper bound of the norms of its elements.
For $\beta\in\mathbb{R}^m$ and $y\in\mathcal S,$ the Cauchy-Schwarz Inequality implies
$$|y \Sigma^{-1}X\beta| \le ||y\Sigma^{-1}X||\,||\beta|| \le c||\beta||.$$
We may therefore choose for $B$ any nonempty subset of ball of radius $1/c$ in $\mathbb{R}^m,$ for which $\beta\in B$ implies
$$||y\Sigma^{-1}X||\,||\beta|| \le (c)(1/c) = 1,$$
QED.

This demonstration began with the claim that when $\mathcal S \subset \mathbb{R}^n$ is bounded and $A = \Sigma^{-1}X = (a_{ij})$ is any $n\times m$ matrix, then $\mathcal S A \subset \mathbb{R}^n$ is bounded, too.  To see this, let $s$ be any upper bound for the norms of elements of $\mathcal S$ and let $a$ be an entry in $A$ with largest absolute value.  Then for any $y\in \mathcal S,$ write $y_k$ for an entry of largest absolute value and--noting that $|y_k|^2 \le \sum_i |y_i|^2 = ||y||^2$--observe that
$$||yA||^2 = \sum_{i=1}^m \left(\sum_{j=1}^n y_j a_{ji}\right)^2 \le \sum_{i=1}^m\left(\sum_{j=1}^n |y_k| |a|\right)^2 = mn||y||^2|a|^2  \le mns|a|^2 ,$$
demonstrating $mns|a|^2$ is an upper bound for $\mathcal S A.$
This should make it clear that $\mathbb{R}^m$ and $\mathbb{R}^n$ can be replaced by any dual normed linear spaces and $A$ need only be a bounded linear operator for the result to be true.  In other words, everything results from the fact that finite matrices represent bounded operators.

Finally, the converse is not always true.  A characteristic counterexample has $\mathcal S = \{(x,0)\} \subset \mathbb R^2$ and $B \subset \{(0,y)^\prime\}\subset \mathbb R^2$ (with $\Sigma^{-1}X$ the identity transformation).  By construction, $|y\beta|=0$ for all $y=\mathcal S$ and all $\beta\in B,$ even though $\mathcal S$ is unbounded.
This helps show that the point to the claim in the question is that a bounded set $\mathcal S$ is bounded in all possible directions (as represented by $\Sigma^{-1}X\beta$).  In the counterexample, $\mathcal S$ is bounded in some special directions (namely, those spanned by the vector $(0,1)^\prime$) but not in all of them.  When the image of the map $\beta \to \Sigma^{-1}X \beta$ does not span all possible directions (that is, $\Sigma^{-1}X$ is not of full rank), such counterexamples can always be constructed.
A: 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.
