Novikov convolution is a filtered ring operation

If a support $S$ is nonempty, choose one ${\gamma }_0\in S$. The set $S\cap \{\gamma :\nu (\gamma )<\nu ({\gamma }_0)+1\}$ is finite by [F1] and contains ${\gamma }_0$, so it has a least valuation ${\mathrm{\ell }}_S$. Every support element outside that finite set has valuation at least $\nu ({\gamma }_0)+1>{\mathrm{\ell }}_S$. Thus $S$ has a finite lower bound and its valuation infimum is attained; the empty support is the zero series.

If either factor has empty support, its product is zero. Otherwise let ${\mathrm{\ell }}_T$ be the attained minimum valuation of the second support from step 1.1. For fixed $\gamma \in \mathrm{\Gamma }$, a nonzero product coefficient at $q^{\gamma }$ can involve only ${\gamma }_1$ in the first support with $\nu ({\gamma }_1)\le \nu (\gamma )-{\mathrm{\ell }}_T$. This weak sublevel set is finite because it lies strictly below the cutoff $\nu (\gamma )-{\mathrm{\ell }}_T+1$ in [F1]; then ${\gamma }_2=\gamma -{\gamma }_1$ is fixed, so the coefficient is a finite sum. For each cutoff $C$, any supported product term with valuation below $C$ has $\nu ({\gamma }_1)<C-{\mathrm{\ell }}_T$; there are finitely many such first factors, and for each, finitely many second factors below $C-\nu ({\gamma }_1)$. Thus the product has finite support below $C$.

Support of a sum is contained in the union of the two supports, giving $v(a+b)\ge \min (v(a),v(b))$. Every supported product exponent is a sum of one exponent from each factor, so its valuation is at least $v(a)+v(b)$; cancellation only removes terms. If any of the three factors is zero, both bracketings are zero. Otherwise their supports $S_1,S_2,S_3$ have minimum valuations ${\mathrm{\ell }}_1,{\mathrm{\ell }}_2,{\mathrm{\ell }}_3$ by step 1.1. For a fixed exponent $\gamma$, every raw contributing triple ${\gamma }_1+{\gamma }_2+{\gamma }_3=\gamma$ satisfies $\nu ({\gamma }_1)\le \nu (\gamma )-{\mathrm{\ell }}_2-{\mathrm{\ell }}_3$. There are finitely many such ${\gamma }_1\in S_1$, using the strict cutoff one unit larger. For each, $\nu ({\gamma }_2)\le \nu (\gamma )-\nu ({\gamma }_1)-{\mathrm{\ell }}_3$ leaves finitely many ${\gamma }_2\in S_2$, and ${\gamma }_3=\gamma -{\gamma }_1-{\gamma }_2$ is determined. Thus the raw triple sum is finite before any cancellation, and both bracketings may be expanded and regrouped as that same sum; associativity of addition in $\mathrm{\Gamma }$ and multiplication in $k$ prove associativity of convolution. The same finite-coefficient argument proves distributivity, and $q^0$ is a unit. In the topology with zero-neighbourhoods $F^C$, a Cauchy net $(a_i)$ has an eventually constant coefficient at each $\gamma$: use a cutoff $C>\nu (\gamma )$, for which $a_i-a_j\in F^C$ forces the coefficients at $q^{\gamma }$ to agree on a tail. Let $a_{\gamma }$ be this unique eventual value. For any cutoff $C$, choose $i_0$ so that $a_i-a_j\in F^C$ for all $i,j\ge i_0$. Then $a$ and $a_{i_0}$ agree at every exponent of valuation below $C$, so the support of $a$ below $C$ is finite; hence $a\in {\mathrm{\Lambda }}_{\nu ,k}$. The same tail condition shows that every $a_i$ with $i\ge i_0$ agrees with $a$ below $C$, so $a_i-a\in F^C$ and the net converges. If the net lies in ${\mathrm{\Lambda }}_{\nu ,k}^0$, every stabilized coefficient with negative valuation is zero, so $a\in {\mathrm{\Lambda }}_{\nu ,k}^0$ and the same argument proves completeness in its induced topology.