Bounded test function sets have common compact support

In a seminorm-generated topology, boundedness is equivalent to ${\sup }_{B}q<\mathrm{\infty }$ for every defining seminorm $q$. Indeed absorption by $q<1$ bounds $q$ on $B$. Conversely, for finitely many constraints $q_i<{\epsilon }_i$, choose $t>{\max }_i({\sup }_B{q}_i\mathrm{/}{\epsilon }_i)$, with $t=1$ if there are no constraints, to absorb $B$ in their intersection. By F1, a bounded $B$ therefore satisfies ${\sup }_B{Q}_m<\mathrm{\infty }$ for every $m$.

Assume $B$ bounded and take the explicit compact exhaustion $K_j=\{x:\mathrm{\mid }x\mathrm{\mid }\le j,\mathrm{dist}(x,{\mathbb{R}}^n\setminus \mathrm{\Omega })\ge 1\mathrm{/}j\}$ for $j\ge 1$, with distance to the empty set infinity and $K_0=\mathrm{\varnothing }$. These closed bounded sets are compactly inside $\mathrm{\Omega }$, satisfy $K_j\subseteq \mathrm{int}{K}_{j+1}$, and their interiors cover $\mathrm{\Omega }$. Put $A_j=K_j\setminus K_{j-1}$. The disjoint sets $A_j$ cover $\mathrm{\Omega }$, and each compact subset meets only finitely many of them, by a finite subcover from the interiors. Define $a_j=\sup \{\mathrm{\mid }f(x)\mathrm{\mid }:f\in B,x\in A_j\}$, with empty supremum zero. Step 1.1 bounds every $a_j$ by the finite number ${\sup }_B{Q}_0$.

Set $w(x)=j\mathrm{/}{a}_j$ for $x\in A_j$ when $a_j>0$, and zero on shells with $a_j=0$. This finite nonnegative function is bounded on every compact subset, since only finitely many shells meet that subset. Continuity of $w$ is unnecessary. The seminorm $q(f)={\sup }_{x\in \mathrm{\Omega }}w(x)\mathrm{\mid }f(x)\mathrm{\mid }$ is finite for each test and restricts to a seminorm bounded by ${\sup }_{K}w{p}_0$ on each ${\mathcal{D}}_K$, hence belongs to the defining family of F1. For any occupied shell ($a_j>0$), ${\sup }_{f\in B}q(f)\ge (j\mathrm{/}{a}_j)a_j=j$ by the supremum definition. If no $K_N$ contains all supports, for every $N$ some member has a nonzero value outside $K_N$ (otherwise its support, the closure of nonzero values, would lie in the closed set $K_N$). Thus occupied indices are unbounded, forcing ${\sup }_{B}q=\mathrm{\infty }$, contrary to step 1.1. A common compact $K_N$ must exist, and its derivative bounds are the $Q_m$ bounds already obtained.

Conversely suppose the stated common support and derivative bounds hold. On ${\mathcal{D}}_K$, any defining seminorm $q$ is continuous. Its unit ball contains $p_m<\epsilon$ for some $m,\epsilon >0$; scaling as in step 1.1 gives $q(f)\le (2\mathrm{/}\epsilon )p_m(f)$, with the zero-seminorm case handled by arbitrarily large scaling. Hence $q$ is bounded on $B$, and step 1.1 proves LF boundedness. Finally, if $f_j\to f$, continuity of each seminorm gives $q(f_j-f)\to 0$, so $\{q(f),q(f_j):j\ge 1\}$ is bounded by a finite initial maximum and a bounded tail. Step 1.1 proves boundedness of the sequence and its limit. Empty $B$ uses $K=\mathrm{\varnothing }$ and zero suprema; empty $\mathrm{\Omega }$ has only the zero test. The weights in step 3.1 are defined from suprema, without any countable choice. $\mathrm{\square }$