Distributions Test Functions and Differentiation

For each point of $K$, there is a rational center $q$ and positive rational $r$ such that the point belongs to $B(q,r)$ and $\bar{B}(q,2r)\subseteq \mathrm{\Omega }$. These inner balls cover $K$, so a finite list suffices by compactness. Put $b_j(x)=b((x-q_j)\mathrm{/}{r}_j)$ and $\chi =1-{\prod }_j(1-b_j)$. Then $\chi$ is smooth, lies in $[0,1]$, equals one on the union of the inner balls, and has support in the finite union of the compact outer balls inside $\mathrm{\Omega }$. For $K=\mathrm{\varnothing }$ take the empty product and $\chi =0$.

For nonempty $\mathrm{\Omega }$, set $K_j=\{x:\mathrm{\mid }x\mathrm{\mid }\le j,\mathrm{dist}(x,{\mathbb{R}}^n\setminus \mathrm{\Omega })\ge 1\mathrm{/}j\}$ for $j\ge 1$, interpreting distance to the empty set as infinity; set $K_0=K_{-1}=\mathrm{\varnothing }$. These are compact subsets of $\mathrm{\Omega }$, $K_j\subseteq \mathrm{int}{K}_{j+1}$, and their interiors cover $\mathrm{\Omega }$. Closedness follows from continuity of distance (its absolute difference is at most the distance of the two points), and boundedness gives compactness. The shell $H_j=K_j\setminus \mathrm{int}{K}_{j-1}$ is compact and lies in the open set $G_j=\mathrm{int}{K}_{j+1}\setminus K_{j-2}$.

Fix an enumeration of rational center/radius pairs and a coding of finite lists by natural numbers. For each $j$, consider pairs with $\bar{B}(q,2r)\subseteq G_j$ and with this closed ball contained in some $U\in \mathcal{U}$. Their inner balls cover $H_j$: at any shell point openness of $G_j$ and of one cover member gives a sufficiently small ball, then a rational center and radius. Compactness gives a finite subcover. Select the least code of a finite list that covers $H_j$, taking the empty list for an empty shell. This is a specified function of $j$, not Countable Choice.

Form the corresponding translated dilates $b_{j,k}$ of F1. Their supports lie in $G_j$, and each support lies in some cover member. This family is locally finite: a point has a neighborhood inside some $K_N$, and supports with $j\ge N+2$ miss $K_N$, while only finitely many balls occur at each of the finitely many earlier stages. Every point belongs to some shell, so $s={\sum }_{j,k}{b}_{j,k}$ is everywhere positive. The sum is locally finite and smooth; hence ${\eta }_{j,k}=b_{j,k}\mathrm{/}s$ are smooth, nonnegative, have compact support in the same outer balls, and sum to one. The double index is countable. For empty $\mathrm{\Omega }$ use the empty family. These constructions choose no cover labels and no arbitrary sequence of witnesses. $\mathrm{\square }$

Nonnegativity, symmetry and separation for $d$ follow from F1 and its $p_0$ term. The inequality $\min (1,a+b)\le \min (1,a)+\min (1,b)$ gives the triangle inequality term by term. For fixed $m$ and $0<\epsilon <1$, $d(f,g)<2^{-m-1}\epsilon$ implies $p_m(f-g)<\epsilon$. Conversely, given $\epsilon >0$, choose $M$ with ${\sum }_{m>M}{2}^{-m-1}<\epsilon \mathrm{/}2$; then $p_M(f-g)<\epsilon \mathrm{/}2$ gives $d(f,g)<\epsilon$. These bounds identify the two topologies and their Cauchy sequences. Seminorm balls are convex, and their triangle and homogeneity inequalities give continuity of vector operations.

Let $(f_j)$ be $d$-Cauchy and extend each function smoothly by zero to ${\mathbb{R}}^n$. For every multi-index $\alpha$, the functions ${\mathrm{\partial }}^{\alpha }{f}_j$ are uniformly Cauchy on all of ${\mathbb{R}}^n$: outside $K$ they vanish and on $K$ the bound is $p_{\mathrm{\mid }\alpha \mathrm{\mid }}(f_j-f_k)$. At each point their complex values have a unique limit $g_{\alpha }(x)$. Passing $k$ to infinity in the uniform Cauchy bound proves uniform convergence to $g_{\alpha }$. This definition uses unique limits, not a choice of subsequences. Each $g_{\alpha }$ is continuous: at a point, approximate it uniformly by one continuous derivative within $\epsilon \mathrm{/}3$ and use continuity of that derivative. It vanishes off $K$.

Fix a coordinate direction $e_i$, a point $x$ and a positive $h$. On $[-h,h]$, the functions $t\mapsto {\mathrm{\partial }}^{\alpha }{f}_j(x+t{e}_i)$ and their derivatives converge uniformly to $g_{\alpha }(x+t{e}_i)$ and $g_{\alpha +e_i}(x+t{e}_i)$ respectively. F2, componentwise, gives ${\mathrm{\partial }}_i{g}_{\alpha }(x)=g_{\alpha +e_i}(x)$. All these functions are continuous by step 2.1, so iterating this identity shows $g_0$ is smooth with every derivative $g_{\alpha }$. Its support is contained in the closed set $K$, hence $g_0{\mathrm{\mid }}_{\mathrm{\Omega }}\in {\mathcal{D}}_K$.

Uniform convergence of the finitely many derivatives of order at most $m$ gives $p_m(f_j-g_0)\to 0$ for each $m$, hence $d(f_j,g_0)\to 0$ by step 1.1. This proves completeness. If $K$ is empty or has empty interior, F1 makes the space zero and the same argument yields its sole element. No endpoint differentiation in $\mathrm{\Omega }$ was assumed: the coordinate segments in step 3.1 lie in the globally smooth zero extension. $\mathrm{\square }$

Finite intersections of translated seminorm balls form a topology: each ball is stable under sufficiently small translates, by $\mathrm{\mid }q(f)-q(g)\mathrm{\mid }\le q(f-g)$, and intersections refine intersections. Addition is continuous since $q(h+k)\le q(h)+q(k)$. Joint scalar multiplication is continuous at $(a,f)$ since $q(bg-af)\le \mathrm{\mid }b\mathrm{\mid }q(g-f)+\mathrm{\mid }b-a\mathrm{\mid }q(f)\mathrm{.}$ Near $a$, bound $\mathrm{\mid }b\mathrm{\mid }$ by $\mathrm{\mid }a\mathrm{\mid }+1$ and make the two terms small for each of finitely many seminorms. Balls at zero are convex and balanced by the seminorm inequalities. Thus the construction is a locally convex vector topology.

Each inclusion is continuous by the definition of $\mathcal{P}$. Define $Q_m(f)={\max }_{\mathrm{\mid }\alpha \mathrm{\mid }\le m}{\sup }_{x\in \mathrm{\Omega }}\mathrm{\mid }{\mathrm{\partial }}^{\alpha }f(x)\mathrm{\mid }$, with value zero on the empty domain. Compact support makes this finite and its restriction to ${\mathcal{D}}_K$ is $p_m$, so $Q_m\in \mathcal{P}$. In particular $Q_0$ separates points: if $f\mathrm{\ne }g$, the disjoint balls of radius $Q_0(f-g)\mathrm{/}3$ about them separate them. On ${\mathcal{D}}_K$ the induced topology is no finer than its prescribed topology by inclusion continuity, and no coarser because all $p_m$ are restrictions of $Q_m$.

Let $L$ be linear. If $L$ is continuous, composing with each continuous inclusion proves continuity of every restriction. Conversely suppose all restrictions are continuous. For each seminorm $r$ generating the topology of $E$, $r\circ L$ is a seminorm whose fixed-support restrictions are continuous. Thus $r\circ L\in \mathcal{P}$, and inverse images of translated finite seminorm balls are open, proving continuity of $L$. Applying this result to the identity into any other locally convex topology with continuous inclusions shows that topology is contained in the constructed one.

Let $(K_j)$ be increasing compact sets with interiors covering $\mathrm{\Omega }$. Every compact $K\subseteq \mathrm{\Omega }$ is contained in some $K_N$, by a finite subcover of these interiors and the maximum of its indices. The inclusion ${\mathcal{D}}_K\to {\mathcal{D}}_{K_N}$ is continuous since every derivative seminorm restricts to the same seminorm. Hence a seminorm continuous on all ${\mathcal{D}}_{K_j}$ is continuous on every ${\mathcal{D}}_K$, and the converse is immediate. The defining family of seminorms is therefore identical for any such exhaustion; step 3.1 also gives the stated map criterion using only exhaustion stages. By F2 these stages are Fréchet spaces. For $\mathrm{\Omega }=\mathrm{\varnothing }$ they and the limit are zero, and the same statements hold. No witnesses are selected for infinitely many compact sets: the containment argument is for one fixed $K$ at a time. $\mathrm{\square }$

Differentiation does not enlarge support, and F4 gives $p_m({\mathrm{\partial }}^{\alpha }\phi )\le p_{m+\mathrm{\mid }\alpha \mathrm{\mid }}(\phi )$. Thus it is continuous on each fixed-support space. For products the one-coordinate product rule follows by subtracting $a(x)\phi (x)$ from $a(x+t{e}_i)\phi (x+t{e}_i)$, inserting $a(x+t{e}_i)\phi (x)$ and dividing by $t$; continuity and the derivative limits give the two terms. Induction using F4 and Pascal's identity then gives [given, F2, F4, algebra] ${\mathrm{\partial }}^{\beta }(a\phi )={\sum }_{\gamma \le \beta }\left(\genfrac{}{}{0px}{}{\beta }{\gamma }\right)({\mathrm{\partial }}^{\gamma }a)({\mathrm{\partial }}^{\beta -\gamma }\phi )\mathrm{.}$ All derivatives of $a$ through order $m$ are bounded on compact $K$, so $p_m(a\phi )\le 2^m{A}_{m,K}{p}_m(\phi )$, where $A_{m,K}={\max }_{\mathrm{\mid }\gamma \mathrm{\mid }\le m}{\sup }_K\mathrm{\mid }{\mathrm{\partial }}^{\gamma }a\mathrm{\mid }$. Support again stays in $K$.

Translation takes support into $K+h$ and preserves each derivative supremum. Its inverse is $T_{-h}$. Composition takes support into $L=F^{-1}(K)$, compact because it is the image of $K$ under the continuous inverse. The first derivative formula is ${\mathrm{\partial }}_i(\phi \circ F)={\sum }_j({\mathrm{\partial }}_j\phi )\circ F{\mathrm{\partial }}_i{F}_j$, by F3. Inductively, each derivative of order $\mathrm{\ell }\le m$ is a finite sum of $({\mathrm{\partial }}^{\beta }\phi )\circ F$, $\mathrm{\mid }\beta \mathrm{\mid }\le \mathrm{\ell }$, times products of derivatives of $F$ of orders at most $\mathrm{\ell }$: differentiating a term either differentiates its coefficient by the product rule of step 1.1 or raises $\beta$ by a coordinate using the first derivative formula. All coefficients are bounded on $L$, so $p_{m,L}(\phi \circ F)\le C_{m,K,F}{p}_{m,K}(\phi )$ for a finite constant.

The bounds in steps 1.1 and 2.1 prove continuity from each source stage into the indicated target stage. Composing with its continuous inclusion and applying F1 proves all asserted LF continuities. Apply the composition argument to $F^{-1}$ and the translation argument to $-h$ for continuous inverses. Empty support gives the zero test and all bounds hold with zero left side; the zero multi-index gives the identity map. Only finitely many derivative bounds are used for each estimate, so no choice axiom is needed. $\mathrm{\square }$

Suppose $u$ is continuous, and fix $K$. By F2 its restriction is continuous at zero. Thus there are $m\ge 0$ and $\epsilon >0$ such that $p_m(\psi )<\epsilon$ implies $\mathrm{\mid }u(\psi )\mathrm{\mid }<1$: take the largest order and the smallest positive radius in a finite basic neighborhood contained in the inverse image of the open unit disk. If that intersection has no constraints, the whole space maps into the disk; linearity then makes $u$ zero on this stage, and $m=0,\epsilon =1$ works.

For $p_m(\phi )>0$, apply step 1.1 to $\epsilon \phi \mathrm{/}(2p_m(\phi ))$ to get $\mathrm{\mid }u(\phi )\mathrm{\mid }\le (2\mathrm{/}\epsilon )p_m(\phi )$. If $p_m(\phi )=0$, every positive real multiple of $\phi$ satisfies the same strict neighborhood inequality, so $t\mathrm{\mid }u(\phi )\mathrm{\mid }<1$ for every $t>0$, forcing $u(\phi )=0$. This proves the estimate for the fixed $K$, and the argument applies to every $K$ without selecting a family of pairs.

Conversely suppose the stated estimates hold. Fix $K$ and one witnessing pair. If $C_K=0$, the restriction is zero. Otherwise, for any $\delta >0$, the neighborhood $p_{m_K}<\delta \mathrm{/}{C}_K$ maps into $\mathrm{\mid }z\mathrm{\mid }<\delta$. Hence every restriction is continuous; F2 implies $u$ is continuous on $\mathcal{D}(\mathrm{\Omega })$, and F1 makes it a distribution. For empty $K$ the space is zero and $C_K=m_K=0$ suffice. $\mathrm{\square }$

Fix $i$ and a test $\phi$. The ordinary product rule gives $a{\mathrm{\partial }}_i\phi ={\mathrm{\partial }}_i(a\phi )-({\mathrm{\partial }}_{i}a)\phi$: subtract the product at a point from the product at its coordinate increment, insert the mixed product, divide by the increment and pass to the limit. Therefore [given, F1, F2, algebra] $⟨{\mathrm{\partial }}_i(au),\phi ⟩=-⟨u,a{\mathrm{\partial }}_i\phi ⟩=⟨{\mathrm{\partial }}_{i}u,a\phi ⟩+⟨u,({\mathrm{\partial }}_{i}a)\phi ⟩\mathrm{.}$ By F2 this is the first-order formula.

From F1, evaluating two consecutive derivative operations on a test gives the sign $(-1{)}^{\mathrm{\mid }\gamma \mathrm{\mid }+1}$ times $u({\mathrm{\partial }}^{\gamma }{\mathrm{\partial }}_i\phi )$. Commutation of the smooth test partials identifies this with $({\mathrm{\partial }}^{\gamma +e_i}u)(\phi )$. Hence ${\mathrm{\partial }}_i{\mathrm{\partial }}^{\gamma }u={\mathrm{\partial }}^{\gamma +e_i}u$. The asserted formula for $\alpha =0$ is $au=au$.

Suppose the formula holds for $\alpha$. Differentiate it by ${\mathrm{\partial }}_i$ and apply step 1.1 to each smooth coefficient times its distribution. Step 2.1 yields the two terms with indices $\beta +e_i$ and $\beta$, respectively. For each resulting index $\eta$, their coefficients add to $\left(\genfrac{}{}{0px}{}{\alpha }{\eta -e_i}\right)+\left(\genfrac{}{}{0px}{}{\alpha }{\eta }\right)=\left(\genfrac{}{}{0px}{}{\alpha +e_i}{\eta }\right)$, using zero for an out-of-range binomial coefficient. The equality is Pascal's identity in coordinate $i$ with all other coordinate factors unchanged. Thus the formula holds for $\alpha +e_i$, and induction on total degree proves every case. All sums are finite. For $a=0$ or $u=0$ every term is zero, and on the empty domain the identity is between zero functionals. $\mathrm{\square }$

For $V\subseteq \mathrm{\Omega }$ and $\phi \in \mathcal{D}(V)$, its support is compactly inside $V$, so extension by zero is smooth on $\mathrm{\Omega }$. On each compact $K\subseteq V$ its derivative seminorms are unchanged. The bound of F2 for $u$ on $K$ therefore gives that bound for its restriction; this proves restriction is a distribution. Testing successive zero extensions proves composition and identity of restrictions.

Take the partition $({\eta }_{\mathrm{\ell }})$ of F4. For each $\mathrm{\ell }$ and each test $\phi$, the test ${\eta }_{\mathrm{\ell }}\phi$ has compact support inside any member containing $\mathrm{supp}{\eta }_{\mathrm{\ell }}$. Its evaluation by that member's distribution is independent of the member: two such members overlap on its support, so compatibility applies. Denote this uniquely specified number by $v_{\mathrm{\ell }}(\phi )$; this definition selects no labels. Set $u(\phi )={\sum }_{\mathrm{\ell }}{v}_{\mathrm{\ell }}(\phi )$. Only finitely many partition supports meet $\mathrm{supp}\phi$: local finiteness provides neighborhoods meeting finitely many supports, and a finite subcover of the compact support suffices. Thus the sum exists. Applying the same finite set to the union of two test supports proves complex linearity.

Fix compact $K\subseteq \mathrm{\Omega }$. Only finitely many partition supports meet $K$. For these finitely many indices take containing cover members and their finite-order bounds on the compact supports of the corresponding ${\eta }_{\mathrm{\ell }}$. Finite choices are provable by finite induction in ZF. Let $m$ be the maximum of their orders, or zero if none occur. The finite product rule gives [step 2.1, F2, F3] $p_m({\eta }_{\mathrm{\ell }}\phi )\le 2^m{\max }_{\mathrm{\mid }\gamma \mathrm{\mid }\le m}{\sup }_{\mathrm{supp}{\eta }_{\mathrm{\ell }}}\mathrm{\mid }{\mathrm{\partial }}^{\gamma }{\eta }_{\mathrm{\ell }}\mathrm{\mid }{p}_m(\phi )(\phi \in {\mathcal{D}}_K),$ where derivatives of $\phi$ vanish off $K$. Summing the finitely many bounds yields $\mathrm{\mid }u(\phi )\mathrm{\mid }\le C_K{p}_m(\phi )$ with finite $C_K$. Hence $u$ is a distribution by F2.

If $\phi \in \mathcal{D}(U_j)$, every nonzero summand is evaluated on a test with compact support in $U_j$ and a containing cover member. Compatibility makes it $u_j({\eta }_{\mathrm{\ell }}\phi )$. The finite sum is $u_j(\phi )$ since $\sum {\eta }_{\mathrm{\ell }}=1$. Thus the restrictions are the prescribed ones. If a distribution $w$ restricts to zero on each $U_j$, the same finite decomposition gives $w(\phi )=\sum w({\eta }_{\mathrm{\ell }}\phi )=0$ for every test. Applying this to the difference of two glued distributions proves uniqueness, and therefore independence of the partition. For the empty cover of the empty domain the sum defines the zero distribution and uniqueness still holds. $\mathrm{\square }$

If $S=\mathrm{\varnothing }$, F2 gives $u=0$ and take $L=\mathrm{\varnothing }$, $m=C=0$, with the empty supremum zero. Otherwise take $\chi$ from F3 equal to one near $S$ and put $L=\mathrm{supp}\chi$. It is compactly inside $\mathrm{\Omega }$ and contains a neighborhood of $S$. For every test $\phi$, the test $(1-\chi )\phi$ vanishes near $S$, so $u(\phi )=u(\chi \phi )$ by F2.

Apply F1 on the single compact $L$ to obtain $C_0,m$ with $\mathrm{\mid }u(\psi )\mathrm{\mid }\le C_0{p}_m(\psi )$ on ${\mathcal{D}}_L$. The finite product rule gives ${\mathrm{\partial }}^{\alpha }(\chi \phi )={\sum }_{\beta \le \alpha }\left(\genfrac{}{}{0px}{}{\alpha }{\beta }\right){\mathrm{\partial }}^{\beta }\chi {\mathrm{\partial }}^{\alpha -\beta }\phi$; this follows by iterating the coordinate product rule, with coefficients combined by Pascal's identity. If $A={\max }_{\mathrm{\mid }\beta \mathrm{\mid }\le m}{\sup }_L\mathrm{\mid }{\mathrm{\partial }}^{\beta }\chi \mathrm{\mid }$, then $p_m(\chi \phi )\le 2^{m}A{\max }_{\mathrm{\mid }\gamma \mathrm{\mid }\le m}{\sup }_L\mathrm{\mid }{\mathrm{\partial }}^{\gamma }\phi \mathrm{\mid }$. The derivatives of $\chi$ are bounded on $L$, so $A$ is finite. Together with step 1.1 this gives the asserted estimate with $C=C_0{2}^{m}A$.

If $\phi \in {\mathcal{D}}_K$ for another compact $K$, all its derivatives vanish off $K$, so the maximum over $L$ is at most $p_{m,K}(\phi )$. Thus the same exponent gives the global finite-order property (indeed the same $C$ works here). The proof selected only one cutoff and one finite-order witness pair and used no choice axiom. $\mathrm{\square }$

Suppose its support is contained in $\{a\}$. Obtain an order $m$ and constant from F2. Take a fixed smooth $\chi$ equal to one near zero with support in a ball $B(0,R)$, using F3 in Euclidean space. For all sufficiently small $\epsilon >0$, ${\chi }_{\epsilon }(x)=\chi ((x-a)\mathrm{/}\epsilon )$ is supported inside $\mathrm{\Omega }$ and equals one near $a$. Hence $u(\phi )=u({\chi }_{\epsilon }\phi )$ for every test.

Suppose ${\mathrm{\partial }}^{\gamma }\phi (a)=0$ for every $\mathrm{\mid }\gamma \mathrm{\mid }\le m$. For fixed $\gamma$, Taylor's formula F4 on a ball about $a$ applied to ${\mathrm{\partial }}^{\gamma }\phi$ gives ${\mathrm{\partial }}^{\gamma }\phi (a+h)=o(\mathrm{\mid }h{\mathrm{\mid }}^{m-\mathrm{\mid }\gamma \mathrm{\mid }})$; when $\mathrm{\mid }\gamma \mathrm{\mid }=m$ this is continuity with value zero. The little-o bounds are uniform over $\mathrm{\mid }h\mathrm{\mid }\le R\epsilon$ by their definition: their suprema are $o({\epsilon }^{m-\mathrm{\mid }\gamma \mathrm{\mid }})$. The product rule expands a derivative of order $\beta$, $\mathrm{\mid }\beta \mathrm{\mid }\le m$, of ${\chi }_{\epsilon }\phi$ into finitely many terms [step 1.1, F2, F4, algebra] $\left(\genfrac{}{}{0px}{}{\beta }{\gamma }\right){\epsilon }^{-\mathrm{\mid }\beta -\gamma \mathrm{\mid }}({\mathrm{\partial }}^{\beta -\gamma }\chi )((x-a)\mathrm{/}\epsilon ){\mathrm{\partial }}^{\gamma }\phi (x),\gamma \le \beta \mathrm{.}$ Each has supremum $o({\epsilon }^{m-\mathrm{\mid }\beta \mathrm{\mid }})$, hence tends to zero, even when $\mathrm{\mid }\beta \mathrm{\mid }=m$. All derivatives vanish outside the shrinking support. Thus F2 and step 1.1 give $\mathrm{\mid }u(\phi )\mathrm{\mid }\le C{\max }_{\mathrm{\mid }\beta \mathrm{\mid }\le m}\sup \mathrm{\mid }{\mathrm{\partial }}^{\beta }({\chi }_{\epsilon }\phi )\mathrm{\mid }\to 0$, so $u(\phi )=0$.

Take a fixed cutoff $\eta$ equal to one near $a$ and compactly supported in $\mathrm{\Omega }$. For $\mathrm{\mid }\alpha \mathrm{\mid }\le m$ put $q_{\alpha }(x)=\eta (x)(x-a{)}^{\alpha }\mathrm{/}\alpha !$. Direct monomial differentiation gives ${\mathrm{\partial }}^{\beta }{q}_{\alpha }(a)=1$ for $\beta =\alpha$ and zero for every other $\mathrm{\mid }\beta \mathrm{\mid }\le m$. Therefore $\phi -{\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le m}{\mathrm{\partial }}^{\alpha }\phi (a)q_{\alpha }$ has zero jet through degree $m$. Step 2.1 makes its pairing zero, giving $u(\phi )=\sum u(q_{\alpha }){\mathrm{\partial }}^{\alpha }\phi (a)$. By F1 this is the claimed representation with $c_{\alpha }=(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}u(q_{\alpha })$.

Conversely, every test supported in $\mathrm{\Omega }\setminus \{a\}$ has all derivatives zero at $a$, so F1 makes every displayed combination vanish there. Its support is therefore contained in $\{a\}$. Evaluate a zero combination on the tests $q_{\alpha }$ constructed with the largest order occurring to see each coefficient is zero. This proves uniqueness. Finally a nonzero distribution cannot have empty support by locality (as used in F2), so a nonzero combination has support exactly $\{a\}$. The zero combination is allowed with $m=0,c_0=0$. All jets and sums are finite and no choice is used. $\mathrm{\square }$

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. $\mathrm{\square }$