[proofplan] By bilinearity in $(f, g)$ and distributivity of the shuffle product, the identity reduces to checking it on pure tensors $f \in V^{*\otimes k}$ and $g \in V^{*\otimes l}$. We proceed by double induction on $(k, l)$. The base cases $H(m, 0)$ and $H(0, n)$ — where one factor is a scalar — are immediate from the bilinearity of the pairing. For the inductive step, decompose $f = f_- \cdot p$ and $g = g_- \cdot q$ with $p, q \in V^*$; the shuffle expands the product into two pieces with one fewer factor on each side, and applying the inductive hypothesis identifies these pieces with two Young integrals. The remaining content of the proof is the integration-by-parts identity $F_t G_t = \int_a^t G_s \, dF_s + \int_a^t F_s \, dG_s$ for the Young integrals $F_t = (f, S(x)_{[a, t]})$ and $G_t = (g, S(x)_{[a, t]})$, which closes the induction. [/proofplan] [step:Reduce to the case of pure tensors via bilinearity and distributivity of the shuffle] The pairing $(\cdot, S(x)_{[a, b]}): T((V))^* \to \mathbb{R}$ is linear in the first argument, so the map \begin{align*} \Phi: T((V))^* \times T((V))^* &\to \mathbb{R} \\ (f, g) &\mapsto (f, S(x)_{[a, b]})(g, S(x)_{[a, b]}) - (f \shuffle g, S(x)_{[a, b]}) \end{align*} is bilinear in $(f, g)$ — the first two terms are bilinear by definition, and $\shuffle$ is a bilinear operation on $T(V^*)$, so $f \shuffle g$ is bilinear in $(f, g)$, hence so is its pairing with $S(x)_{[a, b]}$. Using the canonical isomorphism $T((V))^* \cong T(V^*) = \bigoplus_{m \ge 0} V^{*\otimes m}$ (the $\bigoplus$, not $\bigotimes$, is appropriate for the algebraic dual when restricted to functionals on $T((V))$ — concretely, every functional has support concentrated on finitely many levels), every element of $T((V))^*$ is a finite linear combination of pure tensors $f = p_1 \otimes \cdots \otimes p_k$ with $p_i \in V^*$. By bilinearity of $\Phi$, it suffices to prove $\Phi(f, g) = 0$ for all pure tensors $f \in V^{*\otimes k}$ and $g \in V^{*\otimes l}$ with $k, l \ge 0$. [guided] The shuffle identity has the form (LHS) = (RHS) where both sides are bilinear in $f, g$. Whenever we want to verify a bilinear identity on a vector space with a known spanning set, we can restrict attention to the spanning set — here the pure tensors. Let \begin{align*} \Phi: T((V))^* \times T((V))^* &\to \mathbb{R} \\ (f, g) &\mapsto (f, S(x)_{[a, b]})(g, S(x)_{[a, b]}) - (f \shuffle g, S(x)_{[a, b]}). \end{align*} **Verification of bilinearity:** The pairing $f \mapsto (f, S(x)_{[a, b]})$ is a fixed linear functional applied to $f$, hence linear in $f$. The product $(f, S(x)_{[a, b]})(g, S(x)_{[a, b]})$ is then bilinear in $(f, g)$ (linear in each argument with the other fixed). The shuffle product $\shuffle: T(V^*) \times T(V^*) \to T(V^*)$ is bilinear by definition (it is the linearisation of the combinatorial shuffle of words). Composing with the linear functional $\cdot \mapsto (\cdot, S(x)_{[a, b]})$ preserves bilinearity. So $\Phi$ is the difference of two bilinear maps, hence bilinear. **Reduction to pure tensors:** Identify $T((V))^* \cong T(V^*)$. Every element of $T(V^*) = \bigoplus_m V^{*\otimes m}$ is a **finite** sum of homogeneous components, and each homogeneous component $f^{(m)} \in V^{*\otimes m}$ is, by construction of the tensor power, a finite linear combination of pure tensors $p_1 \otimes \cdots \otimes p_m$ with $p_i \in V^*$. By bilinearity of $\Phi$: \begin{align*} \Phi\!\left(\sum_i \lambda_i f_i, \sum_j \mu_j g_j\right) = \sum_{i, j} \lambda_i \mu_j \Phi(f_i, g_j), \end{align*} so $\Phi \equiv 0$ on $T(V^*) \times T(V^*)$ if and only if $\Phi(f, g) = 0$ for all pairs of pure tensors $(f, g)$. We have therefore reduced to: prove $(f, S(x)_{[a, b]})(g, S(x)_{[a, b]}) = (f \shuffle g, S(x)_{[a, b]})$ for all pure tensors $f \in V^{*\otimes k}$ and $g \in V^{*\otimes l}$, with $k, l \ge 0$. [/guided] [/step] [step:Establish the base cases $H(m, 0)$ and $H(0, n)$] For $g \in V^{*\otimes 0} = \mathbb{R}$ — i.e. $g$ is a scalar $\lambda \in \mathbb{R}$ identified with the level-$0$ functional — the shuffle is $f \shuffle g = \lambda f$ (the shuffle by a scalar is scalar multiplication, since the only "shuffling" of an empty word with $f$ is $f$ itself). The pairing of a level-$0$ functional with the signature picks out the level-$0$ component: \begin{align*} (g, S(x)_{[a, b]}) = \lambda \cdot S(x)_{[a, b]}^{(0)} = \lambda \cdot 1 = \lambda. \end{align*} Therefore \begin{align*} (f, S(x)_{[a, b]})(g, S(x)_{[a, b]}) = (f, S(x)_{[a, b]}) \cdot \lambda = (\lambda f, S(x)_{[a, b]}) = (f \shuffle g, S(x)_{[a, b]}), \end{align*} which gives $H(m, 0)$ for all $m$. By the same argument with $f$ and $g$ swapped (the shuffle is commutative: $f \shuffle g = g \shuffle f$), $H(0, n)$ holds for all $n$. [guided] We need to handle the levels $k = 0$ and $l = 0$ separately because the inductive step requires writing $f = f_- \cdot p$ with $p \in V^*$, which only makes sense for $k \ge 1$. **Case $g \in V^{*\otimes 0} = \mathbb{R}$.** A level-$0$ tensor is a scalar; write $g = \lambda \in \mathbb{R}$. The shuffle product $f \shuffle \lambda$ is, by definition of the shuffle on tensor algebras, the scalar multiple $\lambda f$ — the shuffle of any word with the empty word is just the word back, with the scalar carried through. The pairing $(g, S(x)_{[a, b]})$ for a level-$0$ functional $g = \lambda$ reads off the level-$0$ component of the signature: \begin{align*} (g, S(x)_{[a, b]}) = (\lambda, S(x)_{[a, b]}^{(0)})_{V^{*\otimes 0} \times V^{\otimes 0}} = \lambda \cdot 1 = \lambda, \end{align*} since $S(x)_{[a, b]}^{(0)} = 1$ for any signature. Substituting: \begin{align*} (f, S(x)_{[a, b]})(g, S(x)_{[a, b]}) = (f, S(x)_{[a, b]}) \cdot \lambda = (\lambda f, S(x)_{[a, b]}) = (f \shuffle g, S(x)_{[a, b]}), \end{align*} where the second equality uses linearity of the pairing in the first argument. **Case $f \in V^{*\otimes 0}$.** By symmetry (the shuffle product satisfies $f \shuffle g = g \shuffle f$), the same argument with the roles of $f$ and $g$ swapped works. Why are these the **base cases**? Because in the double induction below we will reduce $H(m, n)$ to $H(m-1, n)$ and $H(m, n-1)$. The induction terminates when at least one index is zero, and these base cases handle that boundary. [/guided] [/step] [step:Set up the inductive step and the integral identities] Fix $k, l \ge 1$, and assume the inductive hypothesis $H(k-1, l)$ and $H(k, l-1)$: \begin{align*} (f', S(x)_{[a, t]})(g, S(x)_{[a, t]}) &= (f' \shuffle g, S(x)_{[a, t]}) \quad \text{for all } f' \in V^{*\otimes (k-1)}, g \in V^{*\otimes l}, t \in [a, b], \\ (f, S(x)_{[a, t]})(g', S(x)_{[a, t]}) &= (f \shuffle g', S(x)_{[a, t]}) \quad \text{for all } f \in V^{*\otimes k}, g' \in V^{*\otimes (l-1)}, t \in [a, b]. \end{align*} (That the inductive hypothesis must hold uniformly in $t \in [a, b]$, not just at $t = b$, is essential — the integration-by-parts step below reads it at every $t$. The inductive hypothesis is stated as a uniform-in-$t$ statement throughout.) For pure $f = f_- \cdot p$ with $f_- = p_1 \otimes \cdots \otimes p_{k-1} \in V^{*\otimes (k-1)}$ and $p = p_k \in V^*$, and similarly $g = g_- \cdot q$ with $g_- \in V^{*\otimes (l-1)}$ and $q \in V^*$, define \begin{align*} F_t := (f, S(x)_{[a, t]}) \quad \text{and} \quad G_t := (g, S(x)_{[a, t]}), \qquad t \in [a, b]. \end{align*} The pairing of $f$ with $S(x)_{[a, t]}^{(k)}$ is the $k$-th iterated integral evaluated against the multilinear functional $f$: \begin{align*} F_t = \int_{a \le t_1 \le \cdots \le t_k \le t} p_1(dx_{t_1}) \cdots p_k(dx_{t_k}), \end{align*} and the iterated-integral structure factorises as \begin{align*} F_t = \int_a^t (f_-, S(x)_{[a, s]}) \cdot p(dx_s) = \int_a^t (f_-, S(x)_{[a, s]}) \, p \circ dx_s, \end{align*} i.e. \begin{align*} dF_t = (f_-, S(x)_{[a, t]}) \, p(dx_t) \quad \text{(Young differential form)}. \end{align*} Similarly, \begin{align*} dG_t = (g_-, S(x)_{[a, t]}) \, q(dx_t). \end{align*} [guided] We now begin the induction proper. Let $k, l \ge 1$ and assume $H(k-1, l)$ and $H(k, l-1)$ — both stated uniformly over $t \in [a, b]$, not just $t = b$. (This stronger version of the inductive hypothesis is automatically obtained because the proof, when run with $b$ replaced by any $t \in [a, b]$, gives the result at that $t$; equivalently, we organise the induction by quantifying $t$ universally from the start.) Decompose $f$ and $g$ by peeling off the last factor: \begin{align*} f &= p_1 \otimes \cdots \otimes p_{k-1} \otimes p_k =: f_- \cdot p \quad \text{with } f_- \in V^{*\otimes (k-1)}, \, p = p_k \in V^*, \\ g &= q_1 \otimes \cdots \otimes q_{l-1} \otimes q_l =: g_- \cdot q \quad \text{with } g_- \in V^{*\otimes (l-1)}, \, q = q_l \in V^*. \end{align*} We now express the pairings $F_t = (f, S(x)_{[a, t]})$ and $G_t = (g, S(x)_{[a, t]})$ as **iterated Young integrals**. The signature's level-$k$ component is the iterated integral \begin{align*} S(x)_{[a, t]}^{(k)} = \int_{a \le t_1 \le \cdots \le t_k \le t} dx_{t_1} \otimes \cdots \otimes dx_{t_k}. \end{align*} Pairing with the pure tensor $f = p_1 \otimes \cdots \otimes p_k$ (and using that the pairing $V^{*\otimes k} \times V^{\otimes k} \to \mathbb{R}$ on pure tensors is $(p_1 \otimes \cdots \otimes p_k, v_1 \otimes \cdots \otimes v_k) = p_1(v_1) \cdots p_k(v_k)$), \begin{align*} F_t = \int_{a \le t_1 \le \cdots \le t_k \le t} p_1(dx_{t_1}) \cdots p_k(dx_{t_k}). \end{align*} Now factor the outermost integral (the $t_k$-integral) by Fubini's iterated structure: the inner integral over $a \le t_1 \le \cdots \le t_{k-1} \le t_k$ pairs $f_- = p_1 \otimes \cdots \otimes p_{k-1}$ with $S(x)_{[a, t_k]}^{(k-1)}$, giving $(f_-, S(x)_{[a, t_k]})$. So \begin{align*} F_t = \int_a^t (f_-, S(x)_{[a, s]}) \cdot p(dx_s) \in \mathbb{R}. \end{align*} This is a Young integral (since $s \mapsto (f_-, S(x)_{[a, s]})$ is the value of a level-$(k-1)$ iterated integral, hence in $C_p$ for the same $p < 2$, and the driver $s \mapsto p(x_s)$ is also in $C_p$, and $1/p + 1/p > 1$). Differentially, \begin{align*} dF_s = (f_-, S(x)_{[a, s]}) \, p(dx_s). \end{align*} By symmetric reasoning with $g = g_- \cdot q$, \begin{align*} G_t = \int_a^t (g_-, S(x)_{[a, s]}) \, q(dx_s), \qquad dG_s = (g_-, S(x)_{[a, s]}) \, q(dx_s). \end{align*} These differential identities are the key technical inputs for the rest of the proof — they convert the algebraic problem (shuffle identity) into an analytic one (an integration-by-parts identity). [/guided] [/step] [step:Expand the shuffle product and pair with the signature] The shuffle product of $f = f_- \cdot p$ and $g = g_- \cdot q$ satisfies the **recursive identity** \begin{align*} (f_- \cdot p) \shuffle (g_- \cdot q) = (f_- \shuffle (g_- \cdot q)) \cdot p + ((f_- \cdot p) \shuffle g_-) \cdot q, \end{align*} which is the standard recursion for the shuffle on tensor algebras (the last letter of the shuffled word comes from either $f$ or $g$, and the rest is shuffled accordingly). Pairing with $S(x)_{[a, t]}$, \begin{align*} (f \shuffle g, S(x)_{[a, t]}) = ((f_- \shuffle (g_- \cdot q)) \cdot p, S(x)_{[a, t]}) + (((f_- \cdot p) \shuffle g_-) \cdot q, S(x)_{[a, t]}). \end{align*} The pairing of an element of the form $(\text{word}) \cdot (\text{letter})$ with $S(x)_{[a, t]}$ has the iterated-integral factorisation derived in Step 3: \begin{align*} ((h \cdot r), S(x)_{[a, t]}) = \int_a^t (h, S(x)_{[a, s]}) \, r(dx_s) \end{align*} for $h \in V^{*\otimes m}$ and $r \in V^*$. Applying this with $h = f_- \shuffle (g_- \cdot q) = f_- \shuffle g$ (a level-$(k - 1 + l)$ tensor) and $r = p$: \begin{align*} ((f_- \shuffle g) \cdot p, S(x)_{[a, t]}) = \int_a^t (f_- \shuffle g, S(x)_{[a, s]}) \, p(dx_s). \end{align*} By inductive hypothesis $H(k-1, l)$ applied at each $s \in [a, t]$, \begin{align*} (f_- \shuffle g, S(x)_{[a, s]}) = (f_-, S(x)_{[a, s]}) \cdot (g, S(x)_{[a, s]}) = (f_-, S(x)_{[a, s]}) G_s. \end{align*} Substituting, \begin{align*} ((f_- \shuffle g) \cdot p, S(x)_{[a, t]}) = \int_a^t (f_-, S(x)_{[a, s]}) G_s \, p(dx_s) = \int_a^t G_s \, dF_s, \end{align*} using $dF_s = (f_-, S(x)_{[a, s]}) p(dx_s)$ from Step 3. Similarly, applying $H(k, l-1)$: \begin{align*} ((f \shuffle g_-) \cdot q, S(x)_{[a, t]}) = \int_a^t (f \shuffle g_-, S(x)_{[a, s]}) \, q(dx_s) = \int_a^t F_s \, (g_-, S(x)_{[a, s]}) \, q(dx_s) = \int_a^t F_s \, dG_s. \end{align*} Summing, \begin{align*} (f \shuffle g, S(x)_{[a, t]}) = \int_a^t G_s \, dF_s + \int_a^t F_s \, dG_s. \end{align*} [guided] We use the recursive structure of the shuffle product: \begin{align*} (f_- \cdot p) \shuffle (g_- \cdot q) = (f_- \shuffle (g_- \cdot q)) \cdot p + ((f_- \cdot p) \shuffle g_-) \cdot q. \end{align*} **Why this recursion?** The shuffle of two words $f$ and $g$ is the sum over all ways to interleave their letters. The last letter of any interleaving is either the last letter of $f$ (which is $p$) or the last letter of $g$ (which is $q$). Splitting the sum by which case: - If the last letter is $p$: the remaining $k - 1 + l$ letters form a shuffle of $f_-$ (the first $k - 1$ letters of $f$) and $g$ (all $l$ letters of $g$). This gives $(f_- \shuffle g) \cdot p$. - If the last letter is $q$: similarly, this gives $(f \shuffle g_-) \cdot q$. Now pair both sides with $S(x)_{[a, t]}$. By linearity: \begin{align*} (f \shuffle g, S(x)_{[a, t]}) = ((f_- \shuffle g) \cdot p, S(x)_{[a, t]}) + ((f \shuffle g_-) \cdot q, S(x)_{[a, t]}). \end{align*} **Compute the first piece.** $(f_- \shuffle g) \cdot p$ is a tensor of level $(k - 1) + l + 1 = k + l$ with the last letter $p$. By the iterated-integral factorisation from Step 3 (applied to $h = f_- \shuffle g$, $r = p$): \begin{align*} ((f_- \shuffle g) \cdot p, S(x)_{[a, t]}) = \int_a^t (f_- \shuffle g, S(x)_{[a, s]}) \, p(dx_s). \end{align*} Now apply the **inductive hypothesis $H(k-1, l)$** to $f_- \in V^{*\otimes (k-1)}$ and $g \in V^{*\otimes l}$ — uniformly in $s \in [a, t]$: \begin{align*} (f_- \shuffle g, S(x)_{[a, s]}) = (f_-, S(x)_{[a, s]}) \cdot (g, S(x)_{[a, s]}) = (f_-, S(x)_{[a, s]}) \cdot G_s. \end{align*} Substituting, \begin{align*} ((f_- \shuffle g) \cdot p, S(x)_{[a, t]}) = \int_a^t (f_-, S(x)_{[a, s]}) \cdot G_s \, p(dx_s). \end{align*} Recognise the integrand as $G_s \, dF_s$ (since $dF_s = (f_-, S(x)_{[a, s]}) \, p(dx_s)$ by Step 3): \begin{align*} ((f_- \shuffle g) \cdot p, S(x)_{[a, t]}) = \int_a^t G_s \, dF_s. \end{align*} **Compute the second piece.** Symmetrically, applying the iterated-integral factorisation to $h = f \shuffle g_-$, $r = q$, and then using $H(k, l-1)$ uniformly in $s$: \begin{align*} ((f \shuffle g_-) \cdot q, S(x)_{[a, t]}) = \int_a^t (f \shuffle g_-, S(x)_{[a, s]}) \, q(dx_s) = \int_a^t F_s \cdot (g_-, S(x)_{[a, s]}) \, q(dx_s) = \int_a^t F_s \, dG_s, \end{align*} using $dG_s = (g_-, S(x)_{[a, s]}) \, q(dx_s)$. **Combine:** \begin{align*} (f \shuffle g, S(x)_{[a, t]}) = \int_a^t G_s \, dF_s + \int_a^t F_s \, dG_s. \end{align*} The right-hand side is exactly the **integration-by-parts** form for $F_t G_t$, modulo the boundary term. We address this in the next step. [/guided] [/step] [step:Apply Young integration by parts to identify the right-hand side as $F_t G_t$] Both $F$ and $G$ are paths in $C_p([a, b], \mathbb{R})$ (they are real-valued $p$-variation paths, since they arise as Young integrals of $C_p$ paths against $C_p$ paths with $p < 2$). For $F, G \in C_p$ with $p < 2$, the Young integrals $\int_a^t G_s \, dF_s$ and $\int_a^t F_s \, dG_s$ are well-defined, and Young's [integration by parts](/theorems/???) for the Young integral on the real line gives \begin{align*} F_t G_t - F_a G_a = \int_a^t G_s \, dF_s + \int_a^t F_s \, dG_s. \end{align*} Since $k, l \ge 1$, $F_a = (f, S(x)_{[a, a]}) = (f, \mathbf{1})$, where $\mathbf{1} = (1, 0, 0, \dots) \in T((V))$ has all positive-level components zero. The pairing of $f \in V^{*\otimes k}$ with $\mathbf{1}$ picks out the level-$k$ component of $\mathbf{1}$, which is $0$ for $k \ge 1$. Hence $F_a = 0$, and similarly $G_a = 0$. Therefore \begin{align*} F_t G_t = \int_a^t G_s \, dF_s + \int_a^t F_s \, dG_s = (f \shuffle g, S(x)_{[a, t]}), \end{align*} the second equality from Step 4. Since $F_t G_t = (f, S(x)_{[a, t]})(g, S(x)_{[a, t]})$, the inductive step closes: \begin{align*} (f, S(x)_{[a, t]})(g, S(x)_{[a, t]}) = (f \shuffle g, S(x)_{[a, t]}) \quad \text{for all } t \in [a, b]. \end{align*} Specialising to $t = b$ gives the claimed identity for the pair $(f, g)$ at level $(k, l)$. By the double induction (with base cases from Step 2), the identity holds for all pure tensors $f \in V^{*\otimes k}$ and $g \in V^{*\otimes l}$ with $k, l \ge 0$. By the bilinearity reduction in Step 1, it holds for all $f, g \in T((V))^*$. This completes the proof. [guided] We need to upgrade the integral identity \begin{align*} (f \shuffle g, S(x)_{[a, t]}) = \int_a^t G_s \, dF_s + \int_a^t F_s \, dG_s \end{align*} to the desired \begin{align*} (f, S(x)_{[a, t]})(g, S(x)_{[a, t]}) = (f \shuffle g, S(x)_{[a, t]}). \end{align*} The left-hand side is $F_t G_t$. We need to identify the right-hand side with $F_t G_t$. **Young integration by parts.** For two paths $F, G \in C_p([a, b], \mathbb{R})$ with $p < 2$, both Young integrals are defined, and the Young integration by parts formula reads \begin{align*} F_t G_t - F_a G_a = \int_a^t G_s \, dF_s + \int_a^t F_s \, dG_s. \end{align*} This is a theorem of Young (see, e.g., Lyons-Caruana-Lévy, Theorem 1.16) — the proof uses the Sewing Lemma and the fact that the symmetric tensor $F_s \, dG_s + G_s \, dF_s$ has a natural decomposition into the increment of $FG$ plus a controlled error. **Verification of the hypothesis $F, G \in C_p$.** The path $F_t = (f, S(x)_{[a, t]}) = $ a $k$-iterated integral of $x$ paired with $f$ is a real-valued $p$-variation path: by the $p$-variation estimates for iterated integrals (Lyons-Caruana-Lévy, Proposition 2.4), $\|F\|_{p\text{-var}; [a, b]} \le C_k \|f\|_{V^{*\otimes k}} \|x\|_{p\text{-var}; [a, b]}^k$ for some constant $C_k > 0$ depending on $k$. So $F \in C_p([a, b], \mathbb{R})$. The same for $G$. Since $p < 2$, $1/p + 1/p > 1$, so the Young integration by parts applies. **Vanishing of $F_a, G_a$.** At $t = a$, $S(x)_{[a, a]} = \mathbf{1} = (1, 0, 0, \dots)$. The level-$k$ component of $\mathbf{1}$ is $0$ for $k \ge 1$. Since $f \in V^{*\otimes k}$ pairs against the level-$k$ component of any element of $T((V))$, \begin{align*} F_a = (f, S(x)_{[a, a]}) = (f, \mathbf{1}^{(k)}) = (f, 0) = 0. \end{align*} Similarly $G_a = 0$ since $l \ge 1$. (This is **why** we needed both $k \ge 1$ and $l \ge 1$ — the boundary term $F_a G_a$ would not vanish if either index were zero, and hence the inductive step would fail at the boundary. The base cases handle these boundary indices separately.) **Conclude:** \begin{align*} F_t G_t = F_t G_t - F_a G_a = \int_a^t G_s \, dF_s + \int_a^t F_s \, dG_s = (f \shuffle g, S(x)_{[a, t]}). \end{align*} This is the inductive hypothesis $H(k, l)$ at the level $(k, l)$, established uniformly in $t \in [a, b]$. The double induction is now complete: starting from the base cases $H(m, 0)$ and $H(0, n)$ in Step 2, the inductive step from Steps 3-5 gives $H(k, l)$ for all $(k, l) \in \mathbb{N}^2$. By the reduction in Step 1, the identity $(f \shuffle g, S(x)_{[a, b]}) = (f, S(x)_{[a, b]})(g, S(x)_{[a, b]})$ extends bilinearly to all $f, g \in T((V))^*$. [/guided] [/step]