[proofplan] The proof has three parts. First, we cast the equation in the standard Young-CDE form $dY_t = f(Y_t)(dx_t)$ with $f: E \to \mathcal{L}(V, E)$ given by left multiplication, and verify that $f$ is a bounded linear (hence Lipschitz) map using the Banach-algebra structure of $E$ and the boundedness of $i_1: V \to E$; the [Picard theorem for Young CDEs](/theorems/???) then yields a unique $C_p$-solution. Second, we show by induction on the level $k$ that any solution component $Y^k_t$ must equal $(A \cdot S(x)_{[a,t]})^{(k)}$: the inductive step writes the level-$k$ equation as a Young integral, applies the inductive hypothesis on $Y^{k-1}$, and recognises the resulting iterated integral as the $k$-th level of $A \cdot S(x)_{[a,t]}$. Third, we verify directly that the candidate $\tilde Y_t := A \cdot S(x)_{[a,t]}$ is a solution by computing its Riemann sums against $dx$ levelwise — the Banach-algebra structure of $E$ ensures the products $\tilde Y_s \cdot (x_t - x_s)$ stay in $E$, and the bound $\|S(x)_{[s,t]} - \mathbf{1} - i_1(x_t - x_s)\|_E \lesssim \|x\|_{p\text{-var};[s,t]}^2$ controls the second-order remainder, which has $p/2$-variation strictly less than $1$ for $p < 2$, hence vanishes in the Young-integral limit. [/proofplan] [step:Cast the equation as a Young CDE and verify the vector-field hypotheses] Define \begin{align*} f: E &\to \mathcal{L}(V, E) \\ Y &\mapsto \bigl( v \mapsto Y \cdot i_1(v) \bigr), \end{align*} where the product on the right is the multiplication in the Banach algebra $E$. **Linearity and boundedness of $f(Y) \in \mathcal{L}(V, E)$.** For fixed $Y \in E$, the map $v \mapsto Y \cdot i_1(v)$ is linear in $v$ because $i_1$ is linear and left-multiplication by $Y$ is linear. By sub-multiplicativity of the Banach-algebra norm and the boundedness of $i_1$, \begin{align*} \|Y \cdot i_1(v)\|_E \le \|Y\|_E \cdot \|i_1(v)\|_E \le K\|Y\|_E \cdot \|v\|_V, \end{align*} so $f(Y) \in \mathcal{L}(V, E)$ with $\|f(Y)\|_{\mathcal{L}(V, E)} \le K\|Y\|_E$. **Linearity and boundedness of $f: E \to \mathcal{L}(V, E)$.** For $Y_1, Y_2 \in E$ and $\lambda \in \mathbb{R}$, $f(Y_1 + \lambda Y_2)(v) = (Y_1 + \lambda Y_2) \cdot i_1(v) = f(Y_1)(v) + \lambda f(Y_2)(v)$, so $f$ is linear. From the bound above with $Y$ replaced by $Y_1 - Y_2$, \begin{align*} \|f(Y_1) - f(Y_2)\|_{\mathcal{L}(V, E)} = \|f(Y_1 - Y_2)\|_{\mathcal{L}(V, E)} \le K\|Y_1 - Y_2\|_E, \end{align*} so $f$ is globally Lipschitz with constant $K$. By the [Picard theorem for Young CDEs](/theorems/???) (Lyons-Caruana-Lévy Theorem 1.28), applied to the equation $dY_t = f(Y_t)(dx_t)$ on $[a, b]$ driven by $x \in C_p([a,b], V)$ with $p \in [1, 2)$ and initial datum $Y_a = A \in E$ — the hypotheses being (i) $E$ a Banach space, satisfied since $E$ is a Banach algebra; (ii) $f$ Lipschitz, just verified; (iii) $p < 2$ so that $1/p + 1/p > 1$, satisfied by hypothesis — there exists a unique solution $Y \in C_p([a, b], E)$ to the Young-integral equation \begin{align*} Y_t = A + \int_a^t Y_s \cdot dx_s, \qquad t \in [a, b], \end{align*} where the integral is the Young integral in $E$ of the integrand $s \mapsto f(Y_s)$ against $x$. [guided] We need to fit the equation $dY_t = Y_t \cdot dx_t$ into the standard Young CDE framework. The standard form of a Young CDE driven by $x \in C_p([a,b], V)$ with values in a Banach space $E$ is \begin{align*} dY_t = f(Y_t)(dx_t), \qquad Y_a = A \in E, \end{align*} where $f: E \to \mathcal{L}(V, E)$ is the **vector field**, mapping each state $Y \in E$ to a bounded linear map $V \to E$ that tells the equation how to convert an infinitesimal increment of $x$ (an element of $V$) into an infinitesimal increment of $Y$ (an element of $E$). Define \begin{align*} f: E &\to \mathcal{L}(V, E) \\ Y &\mapsto \bigl( v \mapsto Y \cdot i_1(v) \bigr). \end{align*} Here $i_1: V \to E$ is the canonical level-$1$ embedding given in the hypotheses, and the product $Y \cdot i_1(v)$ is multiplication in the Banach algebra $E$. **Verification 1: $f(Y) \in \mathcal{L}(V, E)$.** We need $f(Y)$ to be a bounded linear map $V \to E$. Linearity in $v$ is clear: $i_1$ is linear and left-multiplication by $Y$ in $E$ is linear, so their composition is linear. Boundedness: by sub-multiplicativity of the Banach-algebra norm, \begin{align*} \|Y \cdot i_1(v)\|_E \le \|Y\|_E \cdot \|i_1(v)\|_E. \end{align*} The hypothesis $\|i_1(v)\|_E \le K\|v\|_V$ gives \begin{align*} \|f(Y)\|_{\mathcal{L}(V, E)} \le K\|Y\|_E, \end{align*} so $f(Y)$ is bounded with operator norm at most $K\|Y\|_E$. **Verification 2: $f: E \to \mathcal{L}(V, E)$ is linear.** Direct from bilinearity of the algebra product: \begin{align*} f(Y_1 + \lambda Y_2)(v) = (Y_1 + \lambda Y_2) \cdot i_1(v) = Y_1 \cdot i_1(v) + \lambda Y_2 \cdot i_1(v) = f(Y_1)(v) + \lambda f(Y_2)(v). \end{align*} So $f$ is a linear map between Banach spaces. **Verification 3: $f$ is Lipschitz.** Taking the operator-norm bound from Verification 1 and using linearity in $Y$: \begin{align*} \|f(Y_1) - f(Y_2)\|_{\mathcal{L}(V, E)} = \|f(Y_1 - Y_2)\|_{\mathcal{L}(V, E)} \le K\|Y_1 - Y_2\|_E, \end{align*} so $f$ is globally Lipschitz with constant $K$. (We track the constant $K$ honestly here — it equals $\|i_1\|_{\mathcal{L}(V, E)}$, which depends only on the chosen norm on $E$ and the embedding.) The Picard theorem for Young CDEs admits a global Lipschitz vector field as a special case of its standard $C^1_b$ hypothesis, so $f$ qualifies. **Apply the Picard theorem.** The [Picard theorem for Young CDEs](/theorems/???) (Lyons-Caruana-Lévy Theorem 1.28) states: given a Banach space $E$, a Lipschitz vector field $f: E \to \mathcal{L}(V, E)$, an initial datum $A \in E$, and a driving path $x \in C_p([a,b], V)$ with $p < 2$, the Young-integral equation \begin{align*} Y_t = A + \int_a^t f(Y_s)(dx_s) \end{align*} has a unique solution $Y \in C_p([a, b], E)$. The hypotheses are (i) $E$ a Banach space — satisfied; (ii) $f$ Lipschitz — satisfied with constant $K$; (iii) $p < 2$, so the Young integrability condition $1/p + 1/p > 1$ holds — satisfied by hypothesis. The conclusion gives our unique solution $Y \in C_p([a,b], E)$ to \begin{align*} Y_t = A + \int_a^t Y_s \cdot dx_s. \end{align*} Why does the linear vector field admit such a clean global solution theory? Because the equation is **linear** in $Y$: the contraction-mapping argument inside the Picard proof produces no blow-up, and the solution exists for all of $[a, b]$, not just on a small interval. Linear CDEs are the well-behaved special case where local-to-global is automatic. [/guided] [/step] [step:Establish the level-$0$ and level-$1$ identifications] Write $Y_t = (Y^0_t, Y^1_t, Y^2_t, \dots) \in E$ with $Y^k_t := \pi_k \circ \iota_E(Y_t) \in V^{\otimes k}$. Each level projection $\pi_k \circ \iota_E: E \to V^{\otimes k}$ is continuous (composition of the continuous $\iota_E$ and the continuous $\pi_k$), so $Y^k \in C_p([a,b], V^{\otimes k})$. Projecting the Young integral equation $Y_t = A + \int_a^t Y_s \cdot dx_s$ onto level $k$ and using that the algebra product satisfies $\pi_k(Y \cdot i_1(v)) = Y^{k-1} \otimes v$ for $k \ge 1$ and $\pi_0(Y \cdot i_1(v)) = 0$ (since $i_1(v)$ is purely level-$1$ and the product is graded), \begin{align*} Y^k_t = A^k + \int_a^t Y^{k-1}_s \otimes dx_s \quad (k \ge 1), \qquad Y^0_t = A^0. \end{align*} For level $0$: $Y^0_t = A^0$ for all $t$, matching $(A \cdot S(x)_{[a,t]})^{(0)} = A^0 \cdot 1 = A^0$. For level $1$: $Y^1_t = A^1 + \int_a^t Y^0_s \otimes dx_s = A^1 + \int_a^t A^0 \otimes dx_s = A^1 + A^0 (x_t - x_a)$, since $A^0 \in \mathbb{R}$ is a scalar constant in $s$ and $\int_a^t dx_s = x_t - x_a$ by the Young-integral fundamental theorem. On the other hand, $S(x)_{[a, t]}^{(0)} = 1$ and $S(x)_{[a, t]}^{(1)} = x_t - x_a$, so by the graded product formula, \begin{align*} (A \cdot S(x)_{[a, t]})^{(1)} = A^0 \otimes S(x)_{[a, t]}^{(1)} + A^1 \otimes S(x)_{[a, t]}^{(0)} = A^0 (x_t - x_a) + A^1 = Y^1_t. \end{align*} Hence $Y^k_t = (A \cdot S(x)_{[a, t]})^{(k)}$ holds for $k = 0, 1$. [guided] The product in $T((V))$ is graded: for $Y \in T((V))$ written as $(Y^0, Y^1, Y^2, \dots)$ and $i_1(v) = (0, v, 0, \dots)$ purely level-$1$, the convolution structure of the tensor product gives \begin{align*} (Y \cdot i_1(v))^{(k)} = \sum_{k_1 + k_2 = k} Y^{k_1} \otimes (i_1(v))^{(k_2)} = Y^{k-1} \otimes v \quad (k \ge 1), \qquad (Y \cdot i_1(v))^{(0)} = 0, \end{align*} where the only contributing term is $k_1 = k-1$, $k_2 = 1$ (since $(i_1(v))^{(k_2)} = 0$ unless $k_2 = 1$). Replacing $i_1(v)$ by $dx_t$ via the Young integral, the integral equation $Y_t = A + \int_a^t Y_s \cdot dx_s$ projects onto level $k$ as \begin{align*} Y^k_t = A^k + \int_a^t Y^{k-1}_s \otimes dx_s \quad (k \ge 1), \qquad Y^0_t = A^0, \end{align*} where we use that level projection commutes with the Young integral: $\pi_k \circ \iota_E$ is continuous linear, and the Young integral is built as a norm-limit of Riemann sums, on each of which level projection commutes by linearity of $\pi_k$. **Level $0$.** $Y^0_t = A^0$ is constant in $t$. The level-$0$ component of the signature is $S(x)_{[a, t]}^{(0)} = 1$ (by definition: the level-$0$ component of any signature is the scalar $1$). So $(A \cdot S(x)_{[a, t]})^{(0)} = A^0 \cdot 1 = A^0 = Y^0_t$. The level-$0$ identification holds. **Level $1$.** The equation $Y^1_t = A^1 + \int_a^t A^0 \otimes dx_s$ has the integrand $s \mapsto A^0$ constant. Pulling the constant outside the Young integral (justified by linearity of the Young integral), \begin{align*} \int_a^t A^0 \otimes dx_s = A^0 \otimes \int_a^t dx_s = A^0 \otimes (x_t - x_a), \end{align*} where $\int_a^t dx_s = x_t - x_a$ is the Young integral of the constant function $1$ against $x$ (immediate from the Riemann-sum definition: $\sum (x_{s_{i+1}} - x_{s_i}) = x_t - x_a$ telescopes). Hence $Y^1_t = A^1 + A^0 (x_t - x_a)$. Compute the level-$1$ component of $A \cdot S(x)_{[a, t]}$ using the graded product: \begin{align*} (A \cdot S(x)_{[a, t]})^{(1)} = \sum_{k_1 + k_2 = 1} A^{k_1} \otimes S(x)_{[a, t]}^{(k_2)} = A^0 \otimes S(x)_{[a, t]}^{(1)} + A^1 \otimes S(x)_{[a, t]}^{(0)}. \end{align*} Substituting $S(x)_{[a, t]}^{(0)} = 1$ and $S(x)_{[a, t]}^{(1)} = x_t - x_a$, \begin{align*} (A \cdot S(x)_{[a, t]})^{(1)} = A^0 (x_t - x_a) + A^1 = Y^1_t. \end{align*} The base case of the induction matches. [/guided] [/step] [step:Run the induction at level $k$ using the Young-integral recursion] Suppose $Y^j_t = (A \cdot S(x)_{[a, t]})^{(j)}$ for all $j \le k - 1$ and $t \in [a, b]$. The level-$k$ projection of the integral equation reads \begin{align*} Y^k_t = A^k + \int_a^t Y^{k-1}_s \otimes dx_s = A^k + \int_a^t (A \cdot S(x)_{[a, s]})^{(k-1)} \otimes dx_s, \end{align*} where we used the inductive hypothesis. Expanding the level-$(k-1)$ product, \begin{align*} (A \cdot S(x)_{[a, s]})^{(k-1)} = \sum_{j=0}^{k-1} A^j \otimes S(x)_{[a, s]}^{(k-1-j)}, \end{align*} so by linearity of the Young integral (each $A^j$ is a constant in $s$), \begin{align*} Y^k_t = A^k + \sum_{j=0}^{k-1} A^j \otimes \int_a^t S(x)_{[a, s]}^{(k-1-j)} \otimes dx_s. \end{align*} By the [iterated-integral recursion for the signature](/theorems/???), \begin{align*} \int_a^t S(x)_{[a, s]}^{(m-1)} \otimes dx_s = S(x)_{[a, t]}^{(m)} \quad \text{for } m \ge 1, \end{align*} which is the defining recursion: the level-$m$ component of the signature is, by definition, the $m$-fold iterated Young integral, which splits along the last variable as the integral of the level-$(m-1)$ component against $dx$. Applying this with $m = k - j$, \begin{align*} Y^k_t = A^k + \sum_{j=0}^{k-1} A^j \otimes S(x)_{[a, t]}^{(k-j)} = \sum_{j=0}^{k} A^j \otimes S(x)_{[a, t]}^{(k-j)} = (A \cdot S(x)_{[a, t]})^{(k)}, \end{align*} where in the second equality we incorporated the $j = k$ term $A^k \otimes S(x)_{[a, t]}^{(0)} = A^k$. By induction, $Y^k_t = (A \cdot S(x)_{[a, t]})^{(k)}$ for all $k \ge 0$ and $t \in [a, b]$. [guided] We now show, by induction on $k$, that any solution $Y_t$ of the CDE in $E$ must satisfy \begin{align*} Y^k_t = (A \cdot S(x)_{[a, t]})^{(k)} \quad \text{for all } t \in [a, b]. \end{align*} The base cases $k = 0, 1$ are done. Assume the inductive hypothesis holds for all levels $\le k - 1$. The level-$k$ projection of the integral equation gives \begin{align*} Y^k_t = A^k + \int_a^t Y^{k-1}_s \otimes dx_s, \end{align*} where the Young integral is well-defined because $s \mapsto Y^{k-1}_s$ is in $C_p([a,b], V^{\otimes(k-1)})$ (continuous level projection of the $C_p$-solution $Y$) and the driving path $x \in C_p([a,b], V)$ — and $1/p + 1/p > 1$ since $p < 2$ — so the [Young integrability condition](/theorems/???) is satisfied. Apply the inductive hypothesis to expand $Y^{k-1}_s$: \begin{align*} Y^{k-1}_s = (A \cdot S(x)_{[a, s]})^{(k-1)} = \sum_{j=0}^{k-1} A^j \otimes S(x)_{[a, s]}^{(k-1-j)}. \end{align*} The constants $A^j$ are independent of $s$, so they pull outside the Young integral by linearity (the Young integral, being a limit of Riemann sums, is linear in the integrand): \begin{align*} \int_a^t Y^{k-1}_s \otimes dx_s = \sum_{j=0}^{k-1} A^j \otimes \int_a^t S(x)_{[a, s]}^{(k-1-j)} \otimes dx_s. \end{align*} The crucial identification is now: the Young integral \begin{align*} \int_a^t S(x)_{[a, s]}^{(m-1)} \otimes dx_s = S(x)_{[a, t]}^{(m)} \qquad (m \ge 1). \end{align*} This is the **defining recursion of the signature**: the iterated integral \begin{align*} S(x)_{[a, t]}^{(m)} = \int_{a \le t_1 \le \cdots \le t_m \le t} dx_{t_1} \otimes \cdots \otimes dx_{t_m} \end{align*} satisfies, by Fubini-style splitting on the last variable $t_m$, \begin{align*} S(x)_{[a, t]}^{(m)} = \int_a^t \left( \int_{a \le t_1 \le \cdots \le t_{m-1} \le s} dx_{t_1} \otimes \cdots \otimes dx_{t_{m-1}} \right) \otimes dx_s = \int_a^t S(x)_{[a, s]}^{(m-1)} \otimes dx_s. \end{align*} Substituting with $m = k - j$, \begin{align*} Y^k_t = A^k + \sum_{j=0}^{k-1} A^j \otimes S(x)_{[a, t]}^{(k-j)}. \end{align*} The missing $j = k$ term is $A^k \otimes S(x)_{[a, t]}^{(0)} = A^k \otimes 1 = A^k$, which is exactly the standalone $A^k$ on the right. Combining, \begin{align*} Y^k_t = \sum_{j=0}^{k} A^j \otimes S(x)_{[a, t]}^{(k-j)} = (A \cdot S(x)_{[a, t]})^{(k)}. \end{align*} The induction closes. [/guided] [/step] [step:Verify that $\tilde Y_t := A \cdot S(x)_{[a,t]}$ solves the integral equation by direct Riemann-sum computation] Define $\tilde Y_t := A \cdot S(x)_{[a, t]}$. We verify (i) $\tilde Y \in C_p([a,b], E)$, (ii) $\tilde Y_a = A$, and (iii) $\tilde Y$ satisfies the Young integral equation $\tilde Y_t = A + \int_a^t \tilde Y_s \cdot dx_s$. Since the Picard theorem in Step 1 guarantees the unique solution lies in $C_p([a,b], E)$, the verification (iii) closes the proof: $\tilde Y$ is a solution, hence equals the unique solution $Y$. **Membership in $E$ and regularity.** Since $\mathcal{S}_p \subset \iota_E(E)$, we have $S(x)_{[a,t]} \in \iota_E(E)$ for every $t \in [a,b]$, and we identify $S(x)_{[a,t]}$ with its preimage in $E$. Multiplication is closed in $E$, so $\tilde Y_t = A \cdot S(x)_{[a,t]} \in E$. By the [signature increment bound for $p$-variation paths](/theorems/???), \begin{align*} \|S(x)_{[s, t]} - \mathbf{1}\|_E \lesssim_p \|x\|_{p\text{-var};[s, t]} + \|x\|_{p\text{-var};[s, t]}^{\lfloor p\rfloor + 1} = O\bigl(\|x\|_{p\text{-var};[s,t]}\bigr) \quad \text{as } |t-s|\to 0, \end{align*} where the implicit constant depends on $p$ and $K$. Combined with Chen's identity $S(x)_{[a,t]} = S(x)_{[a,s]} \cdot S(x)_{[s,t]}$, this yields \begin{align*} \|\tilde Y_t - \tilde Y_s\|_E = \|A \cdot S(x)_{[a,s]} \cdot (S(x)_{[s,t]} - \mathbf{1})\|_E \le \|A\|_E \cdot \|\tilde Y_s\|_E^{\!\!*} \cdot \|S(x)_{[s,t]} - \mathbf{1}\|_E \lesssim \|x\|_{p\text{-var};[s, t]}, \end{align*} where in particular $\sup_{s \in [a,b]} \|S(x)_{[a,s]}\|_E < \infty$ (signatures of $C_p$ paths are uniformly bounded in $E$ on compact intervals — same standard bound). This shows $\tilde Y \in C_p([a,b], E)$. **Initial condition.** $S(x)_{[a, a]} = \mathbf{1}$ (the signature of a degenerate interval is the unit), so $\tilde Y_a = A \cdot \mathbf{1} = A$. **Direct Riemann-sum verification.** Fix $t \in [a, b]$ and let $\mathcal{P}_n = \{a = t_0 < t_1 < \cdots < t_n = t\}$ be partitions of $[a, t]$ with mesh $|\mathcal{P}_n| \to 0$. The Young integral $\int_a^t \tilde Y_s \cdot dx_s$ is, by definition, the limit in $E$ of the left-point Riemann sums \begin{align*} R(\mathcal{P}_n) := \sum_{i=0}^{n-1} \tilde Y_{t_i} \cdot i_1(x_{t_{i+1}} - x_{t_i}). \end{align*} By Chen's identity, $\tilde Y_{t_{i+1}} - \tilde Y_{t_i} = A \cdot S(x)_{[a,t_i]} \cdot (S(x)_{[t_i, t_{i+1}]} - \mathbf{1}) = \tilde Y_{t_i} \cdot (S(x)_{[t_i, t_{i+1}]} - \mathbf{1})$. Decompose \begin{align*} S(x)_{[t_i, t_{i+1}]} - \mathbf{1} = i_1(x_{t_{i+1}} - x_{t_i}) + \rho_{t_i, t_{i+1}}, \qquad \rho_{s, u} := S(x)_{[s, u]} - \mathbf{1} - i_1(x_u - x_s). \end{align*} The remainder $\rho_{s,u}$ encodes the levels $\ge 2$ of the signature increment. By the [signature increment bound](/theorems/???) at order $\ge 2$, \begin{align*} \|\rho_{s, u}\|_E \lesssim_p \|x\|_{p\text{-var};[s, u]}^2, \end{align*} since the level-$2$ component scales like $\|x\|_{p\text{-var}}^2$ and higher levels are even smaller for $|u - s|$ small. Telescoping, \begin{align*} \tilde Y_t - \tilde Y_a &= \sum_{i=0}^{n-1} (\tilde Y_{t_{i+1}} - \tilde Y_{t_i}) = \sum_{i=0}^{n-1} \tilde Y_{t_i} \cdot \bigl( i_1(x_{t_{i+1}} - x_{t_i}) + \rho_{t_i, t_{i+1}} \bigr) \\ &= R(\mathcal{P}_n) + \sum_{i=0}^{n-1} \tilde Y_{t_i} \cdot \rho_{t_i, t_{i+1}}. \end{align*} We bound the remainder sum using sub-multiplicativity in $E$ and the $p$-variation bound on $\rho$: \begin{align*} \Bigl\| \sum_{i=0}^{n-1} \tilde Y_{t_i} \cdot \rho_{t_i, t_{i+1}} \Bigr\|_E \le \sup_{s \in [a,t]} \|\tilde Y_s\|_E \cdot \sum_{i=0}^{n-1} \|\rho_{t_i, t_{i+1}}\|_E \lesssim_p \sup_{s} \|\tilde Y_s\|_E \cdot \sum_{i=0}^{n-1} \|x\|_{p\text{-var};[t_i, t_{i+1}]}^2. \end{align*} Since $\|x\|_{p\text{-var}}^p$ is super-additive across partitions, $\sum_i \|x\|_{p\text{-var};[t_i, t_{i+1}]}^p \le \|x\|_{p\text{-var};[a,t]}^p < \infty$. With $q := p/2 \in [1/2, 1)$ — strictly less than $1$ since $p < 2$ — and writing $\|x\|_{p\text{-var};[t_i, t_{i+1}]}^2 = (\|x\|_{p\text{-var};[t_i, t_{i+1}]}^p)^{2/p}$, we apply Hölder/the $p$-variation refinement: \begin{align*} \sum_{i=0}^{n-1} \|x\|_{p\text{-var};[t_i, t_{i+1}]}^2 \le \Bigl( \max_i \|x\|_{p\text{-var};[t_i, t_{i+1}]} \Bigr)^{2-p} \cdot \sum_{i=0}^{n-1} \|x\|_{p\text{-var};[t_i, t_{i+1}]}^p \le \omega(|\mathcal{P}_n|)^{2-p} \cdot \|x\|_{p\text{-var};[a,t]}^p, \end{align*} where $\omega(\delta) := \sup_{|s-u| \le \delta} \|x\|_{p\text{-var};[s, u]} \to 0$ as $\delta \to 0$ by uniform continuity of $\|x\|_{p\text{-var}; [s, \cdot]}$ on $[a, b]$ (a standard property of $p$-variation control functions). Since $2 - p > 0$, the right-hand side tends to $0$ as $|\mathcal{P}_n| \to 0$. Therefore $R(\mathcal{P}_n) \to \tilde Y_t - \tilde Y_a$ in $E$ as $|\mathcal{P}_n| \to 0$. By definition of the Young integral, $\int_a^t \tilde Y_s \cdot dx_s = \tilde Y_t - \tilde Y_a$, i.e. \begin{align*} \tilde Y_t = A + \int_a^t \tilde Y_s \cdot dx_s. \end{align*} **Combining with uniqueness.** By Step 1, the integral equation has a unique solution $Y \in C_p([a,b], E)$. The candidate $\tilde Y \in C_p([a,b], E)$ satisfies the same equation. Hence $\tilde Y = Y$, and $Y_t = A \cdot S(x)_{[a, t]}$ for all $t \in [a, b]$. **Special case $A = \mathbf{1}$.** $\mathbf{1} \cdot S(x)_{[a, t]} = S(x)_{[a, t]}$, so the unique solution started at $\mathbf{1}$ is the signature itself. [guided] We have two ingredients in hand: (a) the Picard theorem (Step 1) gives a unique solution $Y \in C_p([a,b], E)$ to the integral equation, and (b) the level-by-level induction (Steps 2-3) shows that *any* solution must equal $A \cdot S(x)_{[a, \cdot]}$. To close the loop and obtain *existence* of the explicit form, we now verify that the candidate $\tilde Y_t := A \cdot S(x)_{[a, t]}$ actually satisfies the integral equation. The verification is direct, via Riemann sums in $E$ — no Sewing Lemma is invoked. **Step a: Regularity in $E$.** The hypothesis $\mathcal{S}_p \subset \iota_E(E)$ tells us $S(x)_{[a,t]}$ lies in $E$ for every $t$. Since $E$ is a Banach algebra, products of elements of $E$ stay in $E$, so $\tilde Y_t = A \cdot S(x)_{[a, t]} \in E$. For the regularity, write \begin{align*} \tilde Y_t - \tilde Y_s = A \cdot \bigl( S(x)_{[a,t]} - S(x)_{[a,s]} \bigr). \end{align*} Apply Chen's identity $S(x)_{[a,t]} = S(x)_{[a,s]} \cdot S(x)_{[s,t]}$ to factor: \begin{align*} S(x)_{[a,t]} - S(x)_{[a,s]} = S(x)_{[a,s]} \cdot \bigl( S(x)_{[s,t]} - \mathbf{1} \bigr). \end{align*} The signature-increment bound $\|S(x)_{[s,t]} - \mathbf{1}\|_E \lesssim_p \|x\|_{p\text{-var};[s,t]}$ (a standard estimate for $p$-variation signatures) and sub-multiplicativity of the Banach-algebra norm give \begin{align*} \|\tilde Y_t - \tilde Y_s\|_E \le \|A\|_E \cdot \|S(x)_{[a,s]}\|_E \cdot \|S(x)_{[s,t]} - \mathbf{1}\|_E \lesssim \|x\|_{p\text{-var};[s,t]}, \end{align*} where $\sup_{s \in [a,b]}\|S(x)_{[a,s]}\|_E < \infty$ by the same signature bound. So $\tilde Y$ has the same $p$-variation regularity as $x$, i.e. $\tilde Y \in C_p([a,b], E)$. **Step b: Initial condition.** At $t = a$, the signature is $S(x)_{[a, a]} = \mathbf{1}$ (the signature over a degenerate interval $[a,a]$ has all iterated integrals vanishing except the level-$0$ scalar $1$), so $\tilde Y_a = A \cdot \mathbf{1} = A$. **Step c: The integral equation, by direct Riemann sums.** We must show \begin{align*} \tilde Y_t = A + \int_a^t \tilde Y_s \cdot dx_s, \end{align*} where the integral is the Young integral in $E$. By definition, the Young integral is the limit (in $E$) of the left-point Riemann sums \begin{align*} R(\mathcal{P}_n) = \sum_{i=0}^{n-1} \tilde Y_{t_i} \cdot i_1(x_{t_{i+1}} - x_{t_i}) \end{align*} along any sequence of partitions $\mathcal{P}_n = \{t_0 < \cdots < t_n\}$ with $|\mathcal{P}_n| \to 0$. The challenge is to compute $\lim R(\mathcal{P}_n)$ explicitly and recognise it as $\tilde Y_t - \tilde Y_a$. **Decompose each signature increment into level-$1$ + remainder.** For each interval $[t_i, t_{i+1}]$, Chen's identity gives \begin{align*} \tilde Y_{t_{i+1}} - \tilde Y_{t_i} = \tilde Y_{t_i} \cdot \bigl( S(x)_{[t_i, t_{i+1}]} - \mathbf{1} \bigr). \end{align*} Now write the signature increment as its level-$1$ component plus a higher-order remainder: \begin{align*} S(x)_{[t_i, t_{i+1}]} - \mathbf{1} = i_1(x_{t_{i+1}} - x_{t_i}) + \rho_{t_i, t_{i+1}}, \end{align*} where $\rho_{s,u} := S(x)_{[s,u]} - \mathbf{1} - i_1(x_u - x_s)$ collects the levels $\ge 2$ of the signature increment. The signature increment bound at order $\ge 2$ — i.e. the bound on the levels $\ge 2$ of $S(x)_{[s,u]}$ — gives \begin{align*} \|\rho_{s,u}\|_E \lesssim_p \|x\|_{p\text{-var};[s,u]}^2, \end{align*} with the implicit constant depending on $p$, the Banach-algebra constants, and $K$. (The level-$2$ component is the iterated integral $\int_{s \le t_1 \le t_2 \le u} dx_{t_1} \otimes dx_{t_2}$, whose $p/2$-variation is bounded by $\|x\|_{p\text{-var}}^2$; higher levels scale even faster.) **Telescope and split.** Telescoping over the partition, \begin{align*} \tilde Y_t - \tilde Y_a = \sum_{i=0}^{n-1}(\tilde Y_{t_{i+1}} - \tilde Y_{t_i}) = \sum_{i=0}^{n-1} \tilde Y_{t_i} \cdot i_1(x_{t_{i+1}} - x_{t_i}) + \sum_{i=0}^{n-1} \tilde Y_{t_i} \cdot \rho_{t_i, t_{i+1}} = R(\mathcal{P}_n) + \mathrm{Rem}(\mathcal{P}_n). \end{align*} So the question is: does $\mathrm{Rem}(\mathcal{P}_n) \to 0$ as $|\mathcal{P}_n| \to 0$? If yes, then $R(\mathcal{P}_n) \to \tilde Y_t - \tilde Y_a$, which by definition of the Young integral means $\int_a^t \tilde Y_s \cdot dx_s = \tilde Y_t - \tilde Y_a$, the equation we want. **Bound the remainder sum.** By sub-multiplicativity in $E$, \begin{align*} \|\mathrm{Rem}(\mathcal{P}_n)\|_E \le \Bigl( \sup_{s \in [a,t]} \|\tilde Y_s\|_E \Bigr) \cdot \sum_{i=0}^{n-1} \|\rho_{t_i, t_{i+1}}\|_E \lesssim M \cdot \sum_{i=0}^{n-1} \|x\|_{p\text{-var};[t_i, t_{i+1}]}^2, \end{align*} where $M := \sup_{s \in [a,t]} \|\tilde Y_s\|_E < \infty$. Now we bound the $p$-variation sum. The key tool is the **super-additivity** of $\|x\|_{p\text{-var}}^p$: for any partition $\{t_i\}$ of $[a,t]$, \begin{align*} \sum_{i=0}^{n-1} \|x\|_{p\text{-var};[t_i, t_{i+1}]}^p \le \|x\|_{p\text{-var};[a,t]}^p. \end{align*} (This is immediate: each side is a sum over partitions; the LHS uses a coarser refinement than the RHS sup.) Now factor each $\|x\|_{p\text{-var}}^2$ as $\|x\|_{p\text{-var}}^p \cdot \|x\|_{p\text{-var}}^{2-p}$: \begin{align*} \sum_{i=0}^{n-1} \|x\|_{p\text{-var};[t_i, t_{i+1}]}^2 = \sum_{i=0}^{n-1} \|x\|_{p\text{-var};[t_i, t_{i+1}]}^p \cdot \|x\|_{p\text{-var};[t_i, t_{i+1}]}^{2-p} \le \Bigl(\max_i \|x\|_{p\text{-var};[t_i, t_{i+1}]}\Bigr)^{2-p} \cdot \|x\|_{p\text{-var};[a,t]}^p. \end{align*} Why does the maximum vanish as $|\mathcal{P}_n| \to 0$? The function $\omega: \delta \mapsto \sup_{u-s \le \delta} \|x\|_{p\text{-var};[s,u]}$ is monotone in $\delta$, and $\omega(\delta) \to 0$ as $\delta \to 0$ — this is a standard property of the $p$-variation control function on a compact interval, equivalent to uniform continuity of $s \mapsto \|x\|_{p\text{-var};[a,s]}$. Hence $\max_i \|x\|_{p\text{-var};[t_i, t_{i+1}]} \le \omega(|\mathcal{P}_n|) \to 0$. Since $2 - p > 0$ (recall $p < 2$), \begin{align*} \|\mathrm{Rem}(\mathcal{P}_n)\|_E \lesssim \omega(|\mathcal{P}_n|)^{2-p} \cdot \|x\|_{p\text{-var};[a,t]}^p \to 0 \quad \text{as } |\mathcal{P}_n| \to 0. \end{align*} This is the **key place where $p < 2$ is consumed**: we needed $2 - p > 0$ to make the maximum vanish in the limit; if $p \ge 2$, the second-order remainder would not be negligible and the rough-path machinery would be required. **Conclude.** From $\tilde Y_t - \tilde Y_a = R(\mathcal{P}_n) + \mathrm{Rem}(\mathcal{P}_n)$ and $\mathrm{Rem}(\mathcal{P}_n) \to 0$, we conclude $R(\mathcal{P}_n) \to \tilde Y_t - \tilde Y_a$. By the definition of the Young integral, \begin{align*} \int_a^t \tilde Y_s \cdot dx_s = \lim_{|\mathcal{P}_n| \to 0} R(\mathcal{P}_n) = \tilde Y_t - \tilde Y_a, \end{align*} so $\tilde Y_t = A + \int_a^t \tilde Y_s \cdot dx_s$, proving that $\tilde Y$ solves the integral equation. **Step d: Combine with uniqueness.** Step 1 gave a unique solution $Y \in C_p([a,b], E)$ to the integral equation. We have just shown $\tilde Y \in C_p([a,b], E)$ solves the same equation. By uniqueness, $Y = \tilde Y$, i.e. $Y_t = A \cdot S(x)_{[a, t]}$ for all $t$. (The level-by-level induction in Step 3 was an independent confirmation; here uniqueness alone closes the argument.) **Special case $A = \mathbf{1}$.** $\mathbf{1} \cdot S(x)_{[a, t]} = S(x)_{[a, t]}$, so the unique solution started at $\mathbf{1}$ is the signature itself. This is the canonical realisation of the signature as the solution of a linear CDE — a fundamental property used throughout rough path theory. [/guided] [/step]