[proofplan] We reduce to a Lipschitz path via arc-length reparameterization. Setting $L := \|x\|_{1;[a,b]}$, the [Reparameterization Invariance](/theorems/2492) of the signature lets us replace $x$ with its arc-length parameterization $x^\tau: [0, L] \to V$, which is $1$-Lipschitz. Absolute continuity of $x^\tau$ supplies a Radon–Nikodym derivative $(x^\tau)' \in L^1([0,L], V)$ with $\|(x^\tau_r)'\| \le 1$ for $\mathcal{L}^1$-a.e. $r$. Fubini's theorem then expresses $S(x^\tau)^{(k)}$ as a Lebesgue integral over the $k$-simplex $\{0 < t_1 < \cdots < t_k < L\}$ of the tensor product of derivatives. Submultiplicativity of the tensor norm bounds the integrand by $1$, and the simplex has Lebesgue volume $L^k / k!$. Reverting to $x$ gives the claim. [/proofplan] [step:Reduce to the arc-length parameterization] Set $L := \|x\|_{1;[a,b]} \in [0, \infty)$. If $L = 0$, then $x$ is constant on $[a,b]$, all increments vanish, and $S(x)^{(k)} = 0$ for $k \ge 1$, so the inequality is trivial. Assume $L > 0$. Define the arc-length function \begin{align*} s: [a, b] &\to [0, L] \\ t &\mapsto \|x\|_{1;[a,t]}. \end{align*} Since $x$ is continuous and of bounded variation, $s$ is continuous and non-decreasing with $s(a) = 0$ and $s(b) = L$, hence a continuous non-decreasing surjection $[a,b] \to [0, L]$. Let $\sigma: [0, L] \to [a, b]$ be a right-continuous selection of $s^{-1}$ (when $s$ has a flat plateau, $\sigma$ chooses the right endpoint of the plateau; when $s$ has a jump, $\sigma$ is constant across the jump). Then $\sigma$ is a continuous non-decreasing surjection (it is monotone with values fixed at the endpoints; flat plateaus of $s$ correspond to single values of $\sigma$, and jumps of $s$ correspond to flat plateaus of $\sigma$). Define the **arc-length parameterization** \begin{align*} x^\tau: [0, L] &\to V \\ r &\mapsto x_{\sigma(r)}. \end{align*} By construction, $x^\tau = x \circ \sigma$. Apply [Reparameterization Invariance](/theorems/2492) (the hypothesis $p = 1 < 2$ is satisfied, $\sigma$ is a continuous non-decreasing surjection between the relevant intervals) to conclude \begin{align*} S(x)_{[a,b]} = S(x^\tau)_{[0, L]}. \end{align*} In particular, level by level, \begin{align*} S(x)^{(k)} = S(x^\tau)^{(k)} \in V^{\otimes k}. \end{align*} [guided] The factorial-decay bound $\|S(x)^{(k)}\| \le L^k / k!$ is fundamentally a consequence of integrating against a single-variable measure of total mass $L$. But the path $x$ is parameterized by an arbitrary interval $[a, b]$ — its parameterization may slow down or speed up. Could a clever reparameterization (e.g. one that compresses all the variation into a thin sub-interval) inflate the iterated-integral bound? It cannot — and the precise statement of this is **Reparameterization Invariance** of the signature. We exploit this freedom to choose the parameterization that simplifies the analysis the most: **arc length**. In an arc-length parameterization, the path moves at unit speed almost everywhere, so the level-$k$ iterated integral decomposes naturally over a $k$-simplex of total length $L$. Concretely: Set $L := \|x\|_{1;[a,b]} \in [0, \infty)$. The boundary case $L = 0$ is degenerate: the one-variation vanishes only when $x$ is constant on $[a,b]$, in which case all increments $x_t - x_s$ are zero, every iterated integral is zero, and the bound $\|S(x)^{(k)}\| = 0 \le 0/k!$ holds with both sides zero. Assume $L > 0$ from now on. The arc-length function \begin{align*} s: [a, b] &\to [0, L] \\ t &\mapsto \|x\|_{1;[a,t]} \end{align*} is non-decreasing (one-variation is monotone in the upper limit) and continuous (one-variation is continuous in the upper limit when $x$ is continuous), with $s(a) = 0$ and $s(b) = L$. Thus $s$ is a continuous non-decreasing surjection. But $s$ is generally **not injective** — it has flat plateaus wherever $x$ is locally constant in length (e.g. on intervals where $x$ doesn't move). So $s$ has no two-sided inverse. We must construct a one-sided inverse $\sigma: [0, L] \to [a, b]$ with $s \circ \sigma = \mathrm{id}_{[0, L]}$, taking care to handle plateaus consistently. We define $\sigma(r) := \sup\{t \in [a, b] : s(t) \le r\}$, the right endpoint of the level set $\{s = r\}$. (Equivalently, $\sigma$ is the right-continuous generalized inverse of $s$.) Then $\sigma$ is non-decreasing, $\sigma(0) = a$, $\sigma(L) = b$, and crucially $s(\sigma(r)) = r$ for all $r \in [0, L]$ (because $s$ is continuous and the supremum is attained on a closed level set). The map $\sigma$ is continuous at every $r$ where $s$ has no plateau and constant on each plateau of $s$; conversely, jumps of $s^{-1}$ correspond to plateaus of $s$ — but $s$ has no plateaus on the image, so $\sigma$ is continuous on $[0, L]$. Define \begin{align*} x^\tau: [0, L] &\to V \\ r &\mapsto x_{\sigma(r)}. \end{align*} By construction $x^\tau = x \circ \sigma$. We now invoke [Reparameterization Invariance](/theorems/2492). That theorem requires (i) a path of finite $p$-variation with $1 \le p < 2$, and (ii) a continuous non-decreasing surjection between the parameterization intervals. We verify both: $x \in \mathcal{C}_1([a,b]; V)$ has finite $1$-variation, and $1 < 2$, so (i) holds; $\sigma: [0, L] \to [a, b]$ is continuous non-decreasing with $\sigma(0) = a$ and $\sigma(L) = b$, so (ii) holds. The conclusion of the theorem is \begin{align*} S(x)_{[a,b]} = S(x \circ \sigma)_{[0, L]} = S(x^\tau)_{[0, L]}, \end{align*} and componentwise $S(x)^{(k)} = S(x^\tau)^{(k)}$ in $V^{\otimes k}$. Why-not: One might be tempted to use $s^{-1}$ directly, but $s^{-1}$ is set-valued (a multifunction) wherever $s$ has plateaus. Right-continuous selection gives a single-valued continuous monotone $\sigma$ — exactly the data Reparameterization Invariance demands. The choice of right vs left selection is immaterial for the conclusion (the signature is unchanged), but a definite choice is needed to invoke the theorem. [/guided] [/step] [step:Show $x^\tau$ is $1$-Lipschitz and apply Radon–Nikodym] For any $0 \le r_1 \le r_2 \le L$, \begin{align*} \|x^\tau_{r_2} - x^\tau_{r_1}\|_V = \|x_{\sigma(r_2)} - x_{\sigma(r_1)}\|_V \le \|x\|_{1;[\sigma(r_1), \sigma(r_2)]} = s(\sigma(r_2)) - s(\sigma(r_1)) = r_2 - r_1, \end{align*} where the inequality is the triangle-inequality bound on a single increment by the total variation, the next equality is the additivity $s(t_2) - s(t_1) = \|x\|_{1;[t_1, t_2]}$ for $t_1 \le t_2$, and the final equality uses $s(\sigma(r)) = r$ for all $r \in [0, L]$ (which holds because $s \circ \sigma$ is the identity on $[0, L]$ by construction of $\sigma$ as a section of $s$). Thus $x^\tau$ is $1$-Lipschitz on $[0, L]$. In particular, $x^\tau$ is absolutely continuous on $[0, L]$ (every Lipschitz function on a closed bounded interval is absolutely continuous). By the [Fundamental Theorem of Calculus for Absolutely Continuous Functions](/theorems/???) (a [Radon–Nikodym](/theorems/???)-type result on the real line), $x^\tau$ is differentiable $\mathcal{L}^1$-a.e. on $[0, L]$, the derivative \begin{align*} (x^\tau)': [0, L] &\to V \end{align*} is a representative of an element of $L^1([0, L], V; \mathcal{L}^1)$, and \begin{align*} x^\tau_t - x^\tau_s = \int_s^t (x^\tau_r)' \, d\mathcal{L}^1(r) \qquad \text{for all } 0 \le s \le t \le L. \end{align*} The Lipschitz bound $\|x^\tau_{r_2} - x^\tau_{r_1}\|_V \le r_2 - r_1$ implies, at any point $r$ of differentiability, \begin{align*} \|(x^\tau_r)'\|_V = \lim_{h \to 0^+} \left\| \frac{x^\tau_{r+h} - x^\tau_r}{h} \right\|_V \le 1, \end{align*} so $\|(x^\tau_r)'\|_V \le 1$ for $\mathcal{L}^1$-a.e. $r \in [0, L]$. [guided] We have replaced the original $x$ by an arc-length parameterization $x^\tau$. Why is $x^\tau$ better than $x$? The key feature is that $x^\tau$ is **$1$-Lipschitz**, hence absolutely continuous, hence differentiable a.e. with derivative in $L^1$, hence integrable in the Lebesgue sense — converting the iterated Stieltjes integral defining the signature into a manageable iterated Lebesgue integral. Without this, we would be stuck with abstract Stieltjes integration against a path of bounded variation, and Fubini-style swaps would be much harder. **Step 2.1 — $1$-Lipschitz.** For $0 \le r_1 \le r_2 \le L$, \begin{align*} \|x^\tau_{r_2} - x^\tau_{r_1}\|_V &= \|x_{\sigma(r_2)} - x_{\sigma(r_1)}\|_V \\ &\le \|x\|_{1;[\sigma(r_1), \sigma(r_2)]} \\ &= s(\sigma(r_2)) - s(\sigma(r_1)) \\ &= r_2 - r_1. \end{align*} Each line: - First equality: definition $x^\tau_r = x_{\sigma(r)}$. - Second line: the increment $x_{\sigma(r_2)} - x_{\sigma(r_1)}$ is bounded in norm by the total variation of $x$ over $[\sigma(r_1), \sigma(r_2)]$ — a single-increment-by-total-variation inequality, equivalent to $\|x_t - x_s\| \le \mathrm{Var}(x; [s,t])$ for $s \le t$. - Third line: by definition $s(t_2) - s(t_1) = \|x\|_{1;[a,t_2]} - \|x\|_{1;[a,t_1]} = \|x\|_{1;[t_1,t_2]}$ for $t_1 \le t_2$ (additivity of one-variation). - Fourth line: $s \circ \sigma = \mathrm{id}_{[0,L]}$ by construction of $\sigma$ in Step 1. Hence $x^\tau$ is $1$-Lipschitz on $[0, L]$. **Step 2.2 — Absolute continuity to Radon–Nikodym derivative.** Every Lipschitz function on a closed bounded interval is absolutely continuous (the Lipschitz constant $K$ provides the modulus: given $\varepsilon > 0$, choose $\delta = \varepsilon / K$; then any disjoint family of intervals of total length $< \delta$ has total oscillation $< \varepsilon$). By the [Fundamental Theorem of Calculus for Absolutely Continuous Functions](/theorems/???) (the vector-valued Lebesgue differentiation theorem; the Radon–Nikodym theorem applied to the Stieltjes measure $dx^\tau$ with respect to $\mathcal{L}^1$): 1. $x^\tau$ is differentiable $\mathcal{L}^1$-a.e. on $[0, L]$. 2. The pointwise derivative $(x^\tau)'$, defined a.e., admits a representative in $L^1([0, L], V; \mathcal{L}^1)$. 3. The fundamental theorem holds: \begin{align*} x^\tau_t - x^\tau_s = \int_s^t (x^\tau_r)' \, d\mathcal{L}^1(r) \qquad \text{for all } 0 \le s \le t \le L. \end{align*} In particular, the Stieltjes measure $dx^\tau$ is absolutely continuous with respect to $\mathcal{L}^1$ with Radon–Nikodym density $(x^\tau)'$. **Step 2.3 — Pointwise derivative bound.** At any point $r \in [0, L)$ where $x^\tau$ is differentiable, the Lipschitz bound from Step 2.1 gives \begin{align*} \left\| \frac{x^\tau_{r+h} - x^\tau_r}{h} \right\|_V \le 1 \qquad \text{for all } h \in (0, L - r]. \end{align*} Taking $h \to 0^+$ and using continuity of the norm, \begin{align*} \|(x^\tau_r)'\|_V = \lim_{h \to 0^+} \left\| \frac{x^\tau_{r+h} - x^\tau_r}{h} \right\|_V \le 1. \end{align*} Hence $\|(x^\tau_r)'\|_V \le 1$ for $\mathcal{L}^1$-a.e. $r \in [0, L]$. Why-not: A general bounded-variation path $x$ need not be differentiable a.e. as a vector-valued function — its Stieltjes measure can have a singular part (e.g. a jump or Cantor-type singular continuous component). Lipschitz, however, **rules out** singular components: by Rademacher's theorem (the special case for $\mathbb{R} \to V$ with $V$ finite-dimensional), Lipschitz implies a.e. differentiability with bounded derivative, equivalently absolute continuity. This is the technical reason arc-length parameterization is essential — without it we would be working with a possibly-singular Stieltjes measure. Connection to technique: The bound $\|(x^\tau_r)'\| \le 1$ a.e. is the analog of "unit-speed parameterization" from elementary differential geometry. It is the engine that turns the factorial-decay computation into a clean volume calculation in the next step. [/guided] [/step] [step:Express $S(x^\tau)^{(k)}$ as a Lebesgue integral over the $k$-simplex via Fubini] We claim \begin{align*} S(x^\tau)^{(k)} = \int_{\Delta_k(L)} (x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_k})' \, d\mathcal{L}^k(t_1, \dots, t_k), \end{align*} where $\Delta_k(L) := \{(t_1, \dots, t_k) \in [0, L]^k : 0 < t_1 < t_2 < \cdots < t_k < L\}$ is the open $k$-simplex. We prove this by induction on $k$. For $k = 1$, \begin{align*} S(x^\tau)^{(1)}_{[0, L]} = x^\tau_L - x^\tau_0 = \int_0^L (x^\tau_t)' \, d\mathcal{L}^1(t), \end{align*} the second equality being the integral representation from Step 2. The simplex $\Delta_1(L) = (0, L)$ has full $\mathcal{L}^1$-measure in $[0, L]$, so this is the claim at level $k = 1$. Inductive step: assume the formula holds at level $k$. For each $t \in [0, L]$, the level-$k$ iterated integral on $[0, t]$ is \begin{align*} S(x^\tau)^{(k)}_{[0, t]} = \int_{\Delta_k(t)} (x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_k})' \, d\mathcal{L}^k(t_1, \dots, t_k). \end{align*} By definition of the level-$(k+1)$ iterated integral and the Stieltjes/Lebesgue identity $dx^\tau_t = (x^\tau_t)' \, d\mathcal{L}^1(t)$ (which holds because $x^\tau$ is absolutely continuous, so its Stieltjes measure has Radon–Nikodym density $(x^\tau)'$ with respect to $\mathcal{L}^1$), \begin{align*} S(x^\tau)^{(k+1)} = \int_0^L S(x^\tau)^{(k)}_{[0, t_{k+1}]} \otimes (x^\tau_{t_{k+1}})' \, d\mathcal{L}^1(t_{k+1}). \end{align*} Substituting the inductive expression, \begin{align*} S(x^\tau)^{(k+1)} = \int_0^L \int_{\Delta_k(t_{k+1})} (x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_k})' \otimes (x^\tau_{t_{k+1}})' \, d\mathcal{L}^k(t_1, \dots, t_k) \, d\mathcal{L}^1(t_{k+1}). \end{align*} We apply [Fubini's Theorem](/theorems/???) to combine the iterated integral into a single Lebesgue integral on $\mathbb{R}^{k+1}$. The integrability hypothesis is verified: the integrand is bounded in norm by $\prod_{i=1}^{k+1} \|(x^\tau_{t_i})'\| \le 1$ a.e. (using the bound from Step 2), so it lies in $L^\infty(\Delta_{k+1}(L)) \subseteq L^1(\Delta_{k+1}(L))$ since $\Delta_{k+1}(L)$ has finite Lebesgue measure. The set $\{(t_1, \dots, t_{k+1}) : 0 < t_1 < \cdots < t_k < t_{k+1} < L\}$ is exactly $\Delta_{k+1}(L)$. Hence \begin{align*} S(x^\tau)^{(k+1)} = \int_{\Delta_{k+1}(L)} (x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_{k+1}})' \, d\mathcal{L}^{k+1}(t_1, \dots, t_{k+1}), \end{align*} completing the induction. [guided] We claim that the level-$k$ signature has the closed-form representation \begin{align*} S(x^\tau)^{(k)} = \int_{\Delta_k(L)} (x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_k})' \, d\mathcal{L}^k(t_1, \dots, t_k), \end{align*} where $\Delta_k(L) := \{(t_1, \dots, t_k) \in [0, L]^k : 0 < t_1 < t_2 < \cdots < t_k < L\}$ is the open $k$-simplex. Two ingredients combine here: (a) **Stieltjes-to-Lebesgue conversion** — replace each $dx^\tau$ by $(x^\tau)' \, d\mathcal{L}^1$ — and (b) **Fubini's theorem** — fold the iterated integral into a single multidimensional integral. We prove the claim by induction on $k$, keeping careful track of the hypotheses being verified at each step. **Base case $k = 1$.** By definition of the level-$1$ signature, \begin{align*} S(x^\tau)^{(1)}_{[0, L]} = x^\tau_L - x^\tau_0. \end{align*} By Step 2.2 (the fundamental theorem of calculus for absolutely continuous functions), \begin{align*} x^\tau_L - x^\tau_0 = \int_0^L (x^\tau_t)' \, d\mathcal{L}^1(t) = \int_{(0, L)} (x^\tau_t)' \, d\mathcal{L}^1(t), \end{align*} where the second equality holds because removing the two endpoints — a $\mathcal{L}^1$-null set — does not change the integral. Since $\Delta_1(L) = (0, L)$, this is exactly the claim at level $k = 1$. **Inductive hypothesis.** Suppose at level $k$ that for every $t \in [0, L]$, \begin{align*} S(x^\tau)^{(k)}_{[0, t]} = \int_{\Delta_k(t)} (x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_k})' \, d\mathcal{L}^k(t_1, \dots, t_k). \end{align*} **Inductive step: $k \to k+1$.** By the recursive definition of iterated integrals, \begin{align*} S(x^\tau)^{(k+1)}_{[0, L]} = \int_0^L S(x^\tau)^{(k)}_{[0, t_{k+1}]} \otimes dx^\tau_{t_{k+1}}. \end{align*} The outer integral is a Stieltjes integral with respect to the path-measure $dx^\tau$. To convert it to a Lebesgue integral, we use that $x^\tau$ is absolutely continuous (Step 2.2): the Stieltjes measure $dx^\tau$ on $[0, L]$ is absolutely continuous with respect to $\mathcal{L}^1$ with Radon–Nikodym density $(x^\tau)'$. Concretely, \begin{align*} \int_0^L f(t) \, dx^\tau_t = \int_0^L f(t) (x^\tau_t)' \, d\mathcal{L}^1(t) \end{align*} for every Borel $f: [0, L] \to V^{\otimes k}$ that is integrable against $|dx^\tau|$. Applying this with $f(t_{k+1}) = S(x^\tau)^{(k)}_{[0, t_{k+1}]}$ — which is bounded (by the inductive expression and the bound $\|(x^\tau)'\| \le 1$ a.e.) and continuous in $t_{k+1}$, hence Borel measurable and $|dx^\tau|$-integrable — gives \begin{align*} S(x^\tau)^{(k+1)}_{[0, L]} = \int_0^L S(x^\tau)^{(k)}_{[0, t_{k+1}]} \otimes (x^\tau_{t_{k+1}})' \, d\mathcal{L}^1(t_{k+1}). \end{align*} Now substitute the inductive expression for the inner factor: \begin{align*} S(x^\tau)^{(k+1)} = \int_0^L \!\!\!\int_{\Delta_k(t_{k+1})} (x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_k})' \otimes (x^\tau_{t_{k+1}})' \, d\mathcal{L}^k(t_1, \dots, t_k) \, d\mathcal{L}^1(t_{k+1}). \end{align*} **Verification of Fubini's hypotheses.** We wish to apply [Fubini's Theorem](/theorems/???) on the product space $\mathbb{R}^k \times \mathbb{R}$ with the product Lebesgue measure $\mathcal{L}^k \otimes \mathcal{L}^1 = \mathcal{L}^{k+1}$, swapping the iterated integral into a single integral on $\mathbb{R}^{k+1}$. Fubini requires: - (F1) The integrand is jointly measurable on the product space. The integrand is a product of measurable factors $(x^\tau_{t_i})'$ projected onto coordinates $t_i$, so it is jointly measurable. (More precisely: $(x^\tau)' \in L^1([0, L], V; \mathcal{L}^1)$ has a Borel-measurable representative, and tensor products of Borel functions on different coordinates remain jointly Borel.) - (F2) The integrand is in $L^1$ of the product measure (or non-negative). The norm of the integrand satisfies \begin{align*} \|(x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_{k+1}})'\|_{V^{\otimes (k+1)}} \le \prod_{i=1}^{k+1} \|(x^\tau_{t_i})'\|_V \le 1 \end{align*} $\mathcal{L}^{k+1}$-a.e., by submultiplicativity of the projective tensor norm and the bound $\|(x^\tau)'\| \le 1$ a.e. from Step 2.3. The set on which the integrand is non-zero is contained in $[0, L]^{k+1}$, which has $\mathcal{L}^{k+1}$-measure $L^{k+1} < \infty$. Hence the integrand is bounded a.e. on a finite-measure set, so it lies in $L^\infty(\mathcal{L}^{k+1}) \subset L^1(\mathcal{L}^{k+1})$. Both hypotheses are verified. Fubini gives \begin{align*} S(x^\tau)^{(k+1)} = \int_E (x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_{k+1}})' \, d\mathcal{L}^{k+1}(t_1, \dots, t_{k+1}), \end{align*} where the combined region of integration is \begin{align*} E = \{(t_1, \dots, t_{k+1}) \in [0, L]^{k+1} : 0 < t_1 < \cdots < t_k < t_{k+1} \le L\}. \end{align*} The set $\{t_{k+1} = L\}$ has $\mathcal{L}^{k+1}$-measure zero, so we may equivalently integrate over the open simplex $\Delta_{k+1}(L) = \{0 < t_1 < \cdots < t_{k+1} < L\}$. This completes the induction. **Why is the Fubini step non-trivial?** The two iterated orders — first integrate over $\Delta_k(t_{k+1})$ then over $t_{k+1} \in [0, L]$, vs. integrate over $\Delta_{k+1}(L)$ in one go — agree only when an integrability hypothesis is satisfied. The pointwise bound $\|(x^\tau)'\| \le 1$ a.e. (Step 2.3) is doing real work here: without it, the integrand could be unbounded and Fubini would fail. The Lipschitz reduction in Steps 1–2 was precisely the vehicle for getting this bound. **Why convert Stieltjes to Lebesgue?** The recursive definition of the signature uses Stieltjes integrals against $dx^\tau$. Stieltjes integrals against an absolutely continuous path are Lebesgue integrals against $(x^\tau)' \, d\mathcal{L}^1$. The conversion is essential because Fubini's theorem applies to integrals against a product **measure** — and we need the same measure on each factor of the product. The Stieltjes measures $dx^\tau$ on each axis assemble into the product Stieltjes measure, but this product is opaque; replacing each with $(x^\tau)' \, d\mathcal{L}^1$ exposes a clean product measure $\mathcal{L}^{k+1}$ and pulls all the path data into the integrand. **The simplex $\Delta_k(L)$.** The chronological constraint $0 < t_1 < \cdots < t_k < L$ identifies the natural geometric object underlying the iterated integral: the $k$-dimensional standard simplex of side $L$. Its $\mathcal{L}^k$-measure is $L^k / k!$, computed in Step 4. [/guided] [/step] [step:Bound the integrand and compute the volume of the simplex] Apply the norm to the simplex representation from Step 3 and pass it through the integral via the [norm-integral inequality](/theorems/???) (valid for Bochner integrals: $\|\int f \, d\mu\| \le \int \|f\| \, d\mu$): \begin{align*} \|S(x^\tau)^{(k)}\|_{V^{\otimes k}} \le \int_{\Delta_k(L)} \|(x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_k})'\|_{V^{\otimes k}} \, d\mathcal{L}^k(t_1, \dots, t_k). \end{align*} The tensor norm on $V^{\otimes k}$ is **submultiplicative**: \begin{align*} \|v_1 \otimes \cdots \otimes v_k\|_{V^{\otimes k}} \le \|v_1\|_V \cdots \|v_k\|_V \qquad \forall v_1, \dots, v_k \in V. \end{align*} (This is part of the definition of the projective tensor norm on $V^{\otimes k}$ used throughout the rough-paths formalism, and any compatible cross-norm satisfies the same bound.) Applying this with $v_i = (x^\tau_{t_i})'$ and using $\|(x^\tau_{t_i})'\|_V \le 1$ a.e. from Step 2: \begin{align*} \|(x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_k})'\|_{V^{\otimes k}} \le \prod_{i=1}^k \|(x^\tau_{t_i})'\|_V \le 1 \qquad \mathcal{L}^k\text{-a.e. on } \Delta_k(L). \end{align*} Hence \begin{align*} \|S(x^\tau)^{(k)}\|_{V^{\otimes k}} \le \int_{\Delta_k(L)} 1 \, d\mathcal{L}^k(t_1, \dots, t_k) = \mathcal{L}^k(\Delta_k(L)). \end{align*} The Lebesgue volume of the open $k$-simplex $\Delta_k(L)$ in $\mathbb{R}^k$ is $L^k / k!$. (One way to see this: the cube $[0, L]^k$ has volume $L^k$ and decomposes into $k!$ congruent simplices, one for each permutation of the coordinates, where each simplex has the form $\{t_{\pi(1)} < t_{\pi(2)} < \cdots < t_{\pi(k)}\}$; the standard simplex $\Delta_k(L)$ corresponds to the identity permutation.) Therefore \begin{align*} \|S(x^\tau)^{(k)}\|_{V^{\otimes k}} \le \frac{L^k}{k!}. \end{align*} [/step] [step:Combine to obtain the factorial decay bound for $S(x)$] By Step 1, $S(x)^{(k)} = S(x^\tau)^{(k)}$ as elements of $V^{\otimes k}$, so \begin{align*} \|S(x)^{(k)}\|_{V^{\otimes k}} = \|S(x^\tau)^{(k)}\|_{V^{\otimes k}} \le \frac{L^k}{k!} = \frac{\|x\|_{1;[a,b]}^k}{k!}. \end{align*} This completes the proof. [/step]