[proofplan] We must show that every signature $S(x) \in \mathcal{S}$ lies in the weighted tensor algebra $T_\phi((V))$, i.e. has finite $\phi$-weighted norm. Fix an arbitrary path $x \in \mathcal{C}_p$ defined on an interval $[a,b]$. The factorial-decay estimate for signature levels — a consequence of the iterated-integral construction — bounds the level-$k$ tensor norm by $\|x\|_{1,[a,b]}^k / k!$. Substituting this into the definition of the weighted norm yields a series of the form $\sum_k C^k \phi(k) / (k!)^2$ with $C = \|x\|_{1,[a,b]}^2$. The hypothesis that this series converges for every $C > 0$ then provides the bound, completing the proof. [/proofplan] [step:Recall the definition of the weighted tensor algebra $T_\phi((V))$ and the signature space $\mathcal{S}$] Let $V$ be a finite-dimensional normed vector space (the state space). For $k \geq 0$, equip the tensor power $V^{\otimes k}$ with the projective tensor norm $\|\cdot\|_{V^{\otimes k}}$. The weighted tensor algebra is the set \begin{align*} T_\phi((V)) := \left\{ a = (a_k)_{k \geq 0} \in \prod_{k=0}^\infty V^{\otimes k} : \|a\|_\phi^2 := \sum_{k=0}^\infty \phi(k)\, \|a_k\|_{V^{\otimes k}}^2 < \infty \right\}. \end{align*} The signature space $\mathcal{S}$ is the set of signatures of paths $x$ of finite one-variation on a compact interval. By the [Definition of the Signature](/theorems/???), for $x \in \mathcal{C}_1([a,b]; V)$ the signature is the sequence of iterated integrals \begin{align*} S(x)_{a,b} = (S(x)_{a,b}^{(k)})_{k \geq 0}, \qquad S(x)_{a,b}^{(k)} \in V^{\otimes k}. \end{align*} Our task is to verify $\|S(x)_{a,b}\|_\phi^2 < \infty$ for every such signature. [/step] [step:Apply the factorial-decay estimate to bound each signature level] By the [Factorial Decay of Signature Levels](/theorems/???), for every $x \in \mathcal{C}_1([a,b]; V)$ and every $k \geq 0$, \begin{align*} \|S(x)_{a,b}^{(k)}\|_{V^{\otimes k}} \leq \frac{\|x\|_{1,[a,b]}^k}{k!}, \end{align*} where $\|x\|_{1,[a,b]}$ denotes the one-variation of $x$ over $[a,b]$. The hypothesis of the cited theorem — that $x$ has finite one-variation — holds by the standing assumption $x \in \mathcal{C}_1([a,b]; V)$. Squaring, \begin{align*} \|S(x)_{a,b}^{(k)}\|_{V^{\otimes k}}^2 \leq \frac{\|x\|_{1,[a,b]}^{2k}}{(k!)^2}. \end{align*} [guided] The whole proof reduces to a single asymptotic estimate: how fast does $\|S(x)^{(k)}\|$ decay as $k \to \infty$? The answer — given by the [Factorial Decay of Signature Levels](/theorems/2493) — is that the level-$k$ tensor norm decays at least factorially fast. **Statement of the cited theorem.** For every $x \in \mathcal{C}_1([a,b]; V)$ and every $k \geq 0$, \begin{align*} \|S(x)_{a,b}^{(k)}\|_{V^{\otimes k}} \leq \frac{\|x\|_{1,[a,b]}^k}{k!}. \end{align*} The hypotheses of /theorems/2493 are: (i) $V$ is a normed vector space; (ii) $x: [a,b] \to V$ is continuous of finite one-variation, i.e. $x \in \mathcal{C}_1([a,b]; V)$. Both hold by the standing assumptions of this theorem ($V$ is a finite-dimensional normed vector space, and $x \in \mathcal{C}_p([a,b]; V) \subseteq \mathcal{C}_1([a,b]; V)$ since $1 \le p$ and the spaces are nested with $\mathcal{C}_p \subseteq \mathcal{C}_1$ for $1 \le p \le \infty$, but in any case the hypothesis $x \in \mathcal{C}_1$ is direct from $x \in \mathcal{C}_p$ via the path-space inclusions in the standing setup). Applying the theorem at level $k$: \begin{align*} \|S(x)_{a,b}^{(k)}\|_{V^{\otimes k}} \leq \frac{\|x\|_{1,[a,b]}^k}{k!}. \end{align*} **Why squaring?** The weighted-norm definition $\|a\|_\phi^2 = \sum_k \phi(k) \|a_k\|^2$ uses **squared** level norms. To match this we square the level-$k$ bound: \begin{align*} \|S(x)_{a,b}^{(k)}\|_{V^{\otimes k}}^2 \leq \left( \frac{\|x\|_{1,[a,b]}^k}{k!} \right)^2 = \frac{\|x\|_{1,[a,b]}^{2k}}{(k!)^2}. \end{align*} The squaring is monotone-preserving on $[0, \infty)$: if $0 \le a \le b$, then $a^2 \le b^2$. We use this with $a = \|S(x)^{(k)}\|$ and $b = \|x\|_{1,[a,b]}^k / k!$, both non-negative, to obtain the squared inequality. **The geometric content.** Factorial decay reflects the simplex-volume structure of iterated integrals: $\|S(x)^{(k)}\|$ is bounded by the product of $k$ "speed-1" derivatives integrated over the $k$-simplex of side $L = \|x\|_1$, whose volume is $L^k / k!$. The $k!$ is *not* a numerical accident — it is the volume of the chronological simplex $\{0 < t_1 < \cdots < t_k < L\}$ inside the cube $[0, L]^k$, since the cube partitions into $k!$ such simplices, one per permutation of coordinates. **Why-not without factorial decay?** If we only had a bound $\|S(x)^{(k)}\| \le L^k$ (no $k!$), the resulting series $\sum \phi(k) L^{2k}$ would converge for only mild $\phi$ — say, $\phi(k) = 1/2^k$. The factorial improvement $L^k/k!$ buys huge convergence margin: even rapidly growing $\phi$ (like $\phi(k) = (k!)^{2-\varepsilon}$) gives a convergent series. This explains why the hypothesis on $\phi$ in the theorem statement is so generous — most reasonable $\phi$ satisfy it. [/guided] [/step] [step:Bound the $\phi$-weighted norm of $S(x)$ by a numerical series] Apply the level-wise bound from Step 2 inside the definition of $\|\cdot\|_\phi$: \begin{align*} \|S(x)_{a,b}\|_\phi^2 &= \sum_{k=0}^\infty \phi(k)\, \|S(x)_{a,b}^{(k)}\|_{V^{\otimes k}}^2 \\ &\leq \sum_{k=0}^\infty \phi(k)\, \frac{\|x\|_{1,[a,b]}^{2k}}{(k!)^2} \\ &= \sum_{k=0}^\infty \frac{\bigl(\|x\|_{1,[a,b]}^2\bigr)^k\, \phi(k)}{(k!)^2}. \end{align*} The termwise inequality $\phi(k)\, \|S(x)_{a,b}^{(k)}\|_{V^{\otimes k}}^2 \leq \phi(k)\, \|x\|_{1,[a,b]}^{2k} / (k!)^2$ between non-negative summands extends to the sum by [Tonelli's theorem](/theorems/???) on the counting measure on $\mathbb{N} \cup \{0\}$ (equivalently, by direct comparison for series of non-negative real terms): both sides are non-negative, and the supremum of partial sums on the left is dominated by the supremum of partial sums on the right. [guided] We assemble the weighted norm of the signature from the level-by-level bound. The definition is \begin{align*} \|S(x)_{a,b}\|_\phi^2 = \sum_{k=0}^\infty \phi(k)\, \|S(x)_{a,b}^{(k)}\|_{V^{\otimes k}}^2. \end{align*} Substituting the factorial-decay bound from Step 2 into each summand: \begin{align*} \phi(k)\, \|S(x)_{a,b}^{(k)}\|_{V^{\otimes k}}^2 \leq \phi(k)\, \frac{\|x\|_{1,[a,b]}^{2k}}{(k!)^2}. \end{align*} To sum this termwise inequality we invoke [Tonelli's theorem](/theorems/???) for the counting measure on $\mathbb{N} \cup \{0\}$. Tonelli requires non-negativity of the integrands; here both $\phi(k)\, \|S(x)_{a,b}^{(k)}\|_{V^{\otimes k}}^2 \geq 0$ (since $\phi$ takes values in $\mathbb{R}_+$ and squared norms are non-negative) and $\phi(k)\, \|x\|_{1,[a,b]}^{2k} / (k!)^2 \geq 0$, so the hypothesis is verified. Tonelli then gives that the integral (sum) of a non-negative function is the supremum of integrals (sums) over finite truncations, and termwise inequalities pass to the limit: \begin{align*} \sum_{k=0}^\infty \phi(k)\, \|S(x)_{a,b}^{(k)}\|_{V^{\otimes k}}^2 \leq \sum_{k=0}^\infty \phi(k)\, \frac{\|x\|_{1,[a,b]}^{2k}}{(k!)^2}. \end{align*} The right-hand side has the structure \begin{align*} \sum_{k=0}^\infty \frac{C^k\, \phi(k)}{(k!)^2}, \qquad C := \|x\|_{1,[a,b]}^2, \end{align*} which matches exactly the series the hypothesis controls. The choice $C = \|x\|_{1,[a,b]}^2$ is forced by squaring the path-dependent factor $\|x\|_{1,[a,b]}^k$ when we squared the level-$k$ bound. Why does the hypothesis require summability for *every* $C > 0$, when this proof only consumes summability at the single value $C = \|x\|_{1,[a,b]}^2$? Because the conclusion $\mathcal{S} \subset T_\phi((V))$ asserts that *every* signature is in $T_\phi((V))$, and $\|x\|_{1,[a,b]}$ ranges over $[0, \infty)$ as $x$ varies over $\mathcal{C}_1$. To handle every path simultaneously we need summability for every value of $C$ that any path could produce — i.e. for every $C > 0$. [/guided] [/step] [step:Apply the summability hypothesis with $C := \|x\|_{1,[a,b]}^2$] Set $C := \|x\|_{1,[a,b]}^2$. Since $x \in \mathcal{C}_1([a,b]; V)$ has finite one-variation on the compact interval $[a,b]$, $\|x\|_{1,[a,b]} \in [0, \infty)$ and hence $C \in [0, \infty)$. If $C = 0$ then $x$ is constant, $S(x)_{a,b}^{(k)} = 0$ for $k \geq 1$ and $S(x)_{a,b}^{(0)} = 1$, so $\|S(x)_{a,b}\|_\phi^2 = \phi(0) < \infty$ directly. If $C > 0$, the hypothesis of the theorem applies to this specific value of $C$: \begin{align*} \sum_{k=0}^\infty \frac{C^k \phi(k)}{(k!)^2} < \infty. \end{align*} Combining with Step 3 yields \begin{align*} \|S(x)_{a,b}\|_\phi^2 \leq \sum_{k=0}^\infty \frac{C^k \phi(k)}{(k!)^2} < \infty. \end{align*} Hence $S(x)_{a,b} \in T_\phi((V))$. [guided] We have arrived at a numerical series whose summability is known by hypothesis. We now consume that hypothesis at the specific value of $C$ produced by the path $x$. **Setting $C$.** Define $C := \|x\|_{1,[a,b]}^2$. Since $x \in \mathcal{C}_1([a,b]; V)$ has finite one-variation on the compact interval $[a,b]$, $\|x\|_{1,[a,b]} \in [0, \infty)$, and squaring gives $C \in [0, \infty)$. Two cases. **Case 1: $C = 0$.** Then $\|x\|_{1,[a,b]} = 0$, which forces $x$ to be constant on $[a, b]$ (a path of zero one-variation has all increments zero, hence is constant). For a constant path $x \equiv x_a$: - The level-$0$ signature is by convention $S(x)_{a,b}^{(0)} = 1 \in V^{\otimes 0} = \mathbb{R}$ (the unit of the tensor algebra). - For $k \ge 1$, the level-$k$ signature is the iterated integral over the $k$-simplex of $dx \otimes \cdots \otimes dx$. Since $dx = 0$ for a constant path (in the Stieltjes sense), $S(x)_{a,b}^{(k)} = 0$ for $k \ge 1$. Substituting into the weighted norm: \begin{align*} \|S(x)_{a,b}\|_\phi^2 = \phi(0) \cdot 1^2 + \sum_{k \ge 1} \phi(k) \cdot 0^2 = \phi(0). \end{align*} Since $\phi: \mathbb{N}_0 \to \mathbb{R}_+$ takes values in the non-negative reals, $\phi(0) \in [0, \infty)$ is finite. Hence $\|S(x)_{a,b}\|_\phi^2 = \phi(0) < \infty$, so $S(x)_{a,b} \in T_\phi((V))$ in this degenerate case. **Case 2: $C > 0$.** This is the main case. The hypothesis of the theorem states: for every $C > 0$, \begin{align*} \sum_{k=0}^\infty \frac{C^k \phi(k)}{(k!)^2} < \infty. \end{align*} We instantiate this at $C = \|x\|_{1,[a,b]}^2$. The hypothesis is universal in $C$, so plugging in any specific positive value is legal. Conclusion: \begin{align*} \sum_{k=0}^\infty \frac{\|x\|_{1,[a,b]}^{2k}\, \phi(k)}{(k!)^2} < \infty. \end{align*} Combining with the bound from Step 3, \begin{align*} \|S(x)_{a,b}\|_\phi^2 \le \sum_{k=0}^\infty \frac{\|x\|_{1,[a,b]}^{2k}\, \phi(k)}{(k!)^2} < \infty. \end{align*} By the [Definition of Weighted Tensor Algebra](/theorems/???) (recalled in Step 1), $a \in T_\phi((V))$ iff $\|a\|_\phi^2 < \infty$. Therefore $S(x)_{a,b} \in T_\phi((V))$. **Why split into cases $C = 0$ and $C > 0$?** The hypothesis of the theorem is summability for $C > 0$, with no constraint at $C = 0$. So we cannot directly invoke the hypothesis when $\|x\|_{1,[a,b]} = 0$. Fortunately the case $C = 0$ is immediate by direct computation: a constant path has degenerate signature $(1, 0, 0, \dots)$, and the weighted norm collapses to $\phi(0)$, which is finite by definition of $\phi$. **Why is the hypothesis stated for "every $C > 0$" rather than for one specific $C$?** Because the theorem we are proving — $\mathcal{S} \subseteq T_\phi((V))$ — must hold for *every* signature, hence every path. As the path varies over $\mathcal{C}_p$, $\|x\|_{1,[a,b]}$ ranges over $[0, \infty)$. To handle every path simultaneously we need the summability to hold uniformly across this range. The "every $C > 0$" hypothesis is exactly the universal quantification needed to reduce the path-uniform conclusion to a single summability check, applied path-by-path. **Connection to weight-function design.** The hypothesis is in practice satisfied by a wide range of weight functions $\phi$: any $\phi$ growing at most exponentially fast (e.g. $\phi(k) = M^k$ for some $M > 0$) directly satisfies the bound, since $\sum C^k M^k / (k!)^2$ converges as a Bessel-type series. Even $\phi(k) = (k!)^\alpha$ for $\alpha < 2$ works. The factorial-squared denominator $(k!)^2$ is the engine making this so. [/guided] [/step] [step:Conclude $\mathcal{S} \subset T_\phi((V))$] The argument of Steps 1–4 was carried out for an arbitrary signature $S(x)_{a,b} \in \mathcal{S}$ — i.e. an arbitrary path $x \in \mathcal{C}_p([a,b]; V)$ on an arbitrary compact interval $[a,b]$. Therefore every element of $\mathcal{S}$ lies in $T_\phi((V))$, that is, \begin{align*} \mathcal{S} \subset T_\phi((V)), \end{align*} which is the desired inclusion. This completes the proof. [/step]