[proofplan] Define the linear evaluation map $\Psi : T_\phi((V)) \to \mathcal{H}_\phi$, $\Psi(\ell)(x) := \langle \ell, S(x)\rangle_\phi$, and let $W := \overline{\operatorname{span}\, \mathcal{S}} \subseteq T_\phi((V))$ be the closed linear span of the image of the signature transform $\mathcal{S} := S(\mathcal{C}_p)$. The argument has two stages: (i) prove $W^\perp = \{0\}$ inside $T_\phi((V))$ by an analytic continuation argument — for any $u \perp \mathcal{S}$, the entire-function $\lambda \mapsto \langle u, S(\lambda x)\rangle_\phi$ vanishes identically, so all its Taylor coefficients vanish, which combined with the [Driver lemma](/theorems/???) on linear independence of signature components forces $u = 0$; (ii) verify that $\Psi$ is an isometry whose image is dense in $\mathcal{H}_\phi$, hence extends to an isomorphism by abstract Hilbert-space theory. Step (ii) provides existence and uniqueness of $\ell_f$ for each $f$. [/proofplan] [step:Set up the closed linear span of signatures and identify its orthogonal complement as the obstruction] Let $\mathcal{S} := \{S(x) : x \in \mathcal{C}_p\} \subseteq T_\phi((V))$ and define \begin{align*} W := \overline{\operatorname{span} \mathcal{S}}^{\,\|\cdot\|_\phi}, \end{align*} the closure of the linear span of $\mathcal{S}$ in the $\phi$-norm of $T_\phi((V))$. Since $T_\phi((V))$ is a Hilbert space and $W$ is a closed subspace, the [orthogonal decomposition theorem](/theorems/???) gives \begin{align*} T_\phi((V)) = W \oplus W^\perp. \end{align*} The strategy is to show $W^\perp = \{0\}$, hence $W = T_\phi((V))$. The remainder of this step and the next two address this. [/step] [step:Construct the entire test function $\lambda \mapsto \langle u, S(\lambda x)\rangle_\phi$ for $u \in W^\perp$] Fix $u = (u_k)_{k \geq 0} \in W^\perp$ and a path $x \in \mathcal{C}_p$. Define \begin{align*} f_{x,u}: \mathbb{R} &\to \mathbb{R} \\ \lambda &\mapsto \langle u, S(\lambda x)\rangle_\phi. \end{align*} By the homogeneity of the signature, $S(\lambda x)^{(k)} = \lambda^k S(x)^{(k)}$ for every $k \geq 0$. Hence \begin{align*} f_{x,u}(\lambda) = \sum_{k=0}^\infty \phi(k)\, \langle u_k, S(\lambda x)^{(k)}\rangle_{V^{\otimes k}} = \sum_{k=0}^\infty \phi(k)\, \lambda^k\, \langle u_k, S(x)^{(k)}\rangle_{V^{\otimes k}}. \end{align*} We claim this series defines a real-analytic function on $\mathbb{R}$. By Cauchy--Schwarz on each $V^{\otimes k}$ and the [factorial decay $\|S(x)^{(k)}\| \leq L^k/k!$](/theorems/???) (with $L = \|x\|_{p\text{-var}}$), the $k$-th coefficient $a_k := \phi(k) \langle u_k, S(x)^{(k)}\rangle_{V^{\otimes k}}$ satisfies \begin{align*} |a_k| \leq \phi(k)\, \|u_k\|_{V^{\otimes k}} \cdot \frac{L^k}{k!}. \end{align*} Cauchy--Schwarz on the weighted $\ell^2$ space then gives, for any $R > 0$, \begin{align*} \sum_{k=0}^\infty |a_k| R^k &\leq \sum_{k=0}^\infty \phi(k) \|u_k\|\, \frac{(LR)^k}{k!} \\ &\leq \Bigl(\sum_{k=0}^\infty \phi(k)\|u_k\|^2\Bigr)^{1/2} \Bigl(\sum_{k=0}^\infty \phi(k) \frac{(LR)^{2k}}{(k!)^2}\Bigr)^{1/2} \\ &= \|u\|_\phi \cdot \Bigl(\sum_{k=0}^\infty \phi(k) \frac{(LR)^{2k}}{(k!)^2}\Bigr)^{1/2}. \end{align*} The hypothesis "$\phi$ satisfies the summability condition" gives $\sum_k \phi(k) C^k/(k!)^2 < \infty$ for every $C > 0$; applied with $C = (LR)^2$, the right-hand factor is finite. Hence the power series for $f_{x,u}$ has infinite radius of convergence, so $f_{x,u}$ is entire (when extended to $\mathbb{C}$). [guided] Why introduce the function $f_{x,u}(\lambda) = \langle u, S(\lambda x)\rangle_\phi$ at all? The orthogonality $u \in W^\perp$ already says $\langle u, S(z)\rangle_\phi = 0$ for *every* $z \in \mathcal{C}_p$ — including $z = \lambda x$ for every $\lambda$. So why parametrise by $\lambda$? The point is that homogeneity of the signature, $S(\lambda x)^{(k)} = \lambda^k S(x)^{(k)}$, secretly turns the orthogonality statement into a statement about an *analytic function of $\lambda$*. By varying $\lambda$ continuously over $\mathbb{R}$ we obtain not just a single equation $\langle u, S(x)\rangle_\phi = 0$ but a *function* $f_{x,u}(\lambda)$ that vanishes everywhere. Power-series rigidity will then let us extract level-by-level information about $u_n$. We compute $f_{x,u}$ explicitly. Fix $u = (u_k)_{k \geq 0} \in W^\perp$ and $x \in \mathcal{C}_p$. By definition of $\langle \cdot, \cdot\rangle_\phi$ and the homogeneity of $S$, \begin{align*} f_{x,u}(\lambda) = \langle u, S(\lambda x)\rangle_\phi = \sum_{k=0}^\infty \phi(k)\, \langle u_k, S(\lambda x)^{(k)}\rangle_{V^{\otimes k}} = \sum_{k=0}^\infty \phi(k)\, \lambda^k\, \langle u_k, S(x)^{(k)}\rangle_{V^{\otimes k}}. \end{align*} This is a power series in $\lambda$ with coefficients $a_k := \phi(k)\, \langle u_k, S(x)^{(k)}\rangle_{V^{\otimes k}}$. We must verify that this series has infinite radius of convergence — without that, the function is only formal, and we cannot use analyticity to deduce things from "$f_{x,u} \equiv 0$". **Bound on $|a_k|$.** Cauchy--Schwarz on $V^{\otimes k}$ gives \begin{align*} |\langle u_k, S(x)^{(k)}\rangle_{V^{\otimes k}}| \leq \|u_k\|_{V^{\otimes k}}\, \|S(x)^{(k)}\|_{V^{\otimes k}}. \end{align*} The [factorial decay $\|S(x)^{(k)}\| \leq L^k/k!$](/theorems/???), valid with $L = \|x\|_{p\text{-var}}$, then yields \begin{align*} |a_k| = \phi(k)\, |\langle u_k, S(x)^{(k)}\rangle| \leq \phi(k)\, \|u_k\|_{V^{\otimes k}}\, \frac{L^k}{k!}. \end{align*} **Bound on $\sum_k |a_k| R^k$ for arbitrary $R > 0$.** We need to control this for every $R$ to get an entire function. Using the bound on $|a_k|$ and Cauchy--Schwarz on the *weighted $\ell^2$ space* (i.e., the inner product space underlying $T_\phi((V))$), \begin{align*} \sum_{k=0}^\infty |a_k| R^k &\leq \sum_{k=0}^\infty \phi(k)\, \|u_k\|\, \frac{(LR)^k}{k!} \\ &= \sum_{k=0}^\infty \bigl[\sqrt{\phi(k)}\, \|u_k\|\bigr] \cdot \bigl[\sqrt{\phi(k)}\, \tfrac{(LR)^k}{k!}\bigr] \\ &\leq \Bigl(\sum_{k=0}^\infty \phi(k)\, \|u_k\|^2\Bigr)^{1/2} \Bigl(\sum_{k=0}^\infty \phi(k)\, \tfrac{(LR)^{2k}}{(k!)^2}\Bigr)^{1/2} \\ &= \|u\|_\phi\, \Bigl(\sum_{k=0}^\infty \phi(k)\, \tfrac{(LR)^{2k}}{(k!)^2}\Bigr)^{1/2}. \end{align*} The first factor $\|u\|_\phi$ is finite because $u \in T_\phi((V))$. The second factor is finite by the summability condition on $\phi$ — applied with $C = (LR)^2$ — which states $\sum_k \phi(k)\, C^k/(k!)^2 < \infty$ for every $C > 0$. Since $R > 0$ was arbitrary, $\sum_k |a_k| R^k < \infty$ for every $R$, so the power series for $f_{x,u}$ has infinite radius of convergence. Extending $\lambda$ to $\mathbb{C}$, the function $f_{x,u}$ is entire. **Why we needed the summability condition here.** The hypothesis is exactly the right strength: weak enough to be a usable assumption on $\phi$, strong enough to make the test function entire. Without summability, the radius of convergence could collapse, and the analytic-rigidity argument in the next step would fail. [/guided] [/step] [step:Use $u \perp \mathcal{S}$ to force the entire function to vanish identically] For each $\lambda \in \mathbb{R}$, the path $\lambda x$ lies in $\mathcal{C}_p$ (the space is closed under positive scalar reparametrisation; for $\lambda < 0$ one uses the reversal symmetry of the signature, or one observes that $\lambda x \in \mathcal{C}_p$ directly because scaling preserves $p$-variation). Hence $S(\lambda x) \in \mathcal{S} \subseteq W$, and by hypothesis $u \in W^\perp$, \begin{align*} f_{x,u}(\lambda) = \langle u, S(\lambda x) \rangle_\phi = 0 \quad \text{for all } \lambda \in \mathbb{R}. \end{align*} Since $f_{x,u}$ is real-analytic on $\mathbb{R}$ and vanishes on all of $\mathbb{R}$, all its Taylor coefficients at $\lambda = 0$ vanish. Differentiating the power series term by term (justified inside its radius of convergence) and evaluating at $\lambda = 0$, \begin{align*} \frac{d^n f_{x,u}}{d\lambda^n}(0) = n!\, a_n = n!\, \phi(n)\, \langle u_n, S(x)^{(n)} \rangle_{V^{\otimes n}} = 0. \end{align*} Hence \begin{align*} \phi(n)\, \langle u_n, S(x)^{(n)} \rangle_{V^{\otimes n}} = 0 \quad \text{for every } n \in \mathbb{N} \cup \{0\} \text{ and every } x \in \mathcal{C}_p. \end{align*} [guided] The point of constructing the entire function $f_{x,u}(\lambda) = \sum_k a_k \lambda^k$ is that orthogonality to a *single* family $\{S(\lambda x) : \lambda \in \mathbb{R}\}$ secretly carries information about *every* signature level $u_n$ separately. How? By the homogeneity $S(\lambda x)^{(k)} = \lambda^k S(x)^{(k)}$, the parameter $\lambda$ enters as a weight $\lambda^k$ on the level-$k$ signature in the inner product, which is exactly the structure that lets us extract the level-$k$ contribution by differentiating $k$ times. Concretely: $u \in W^\perp$ means $\langle u, w\rangle_\phi = 0$ for every $w \in W$, in particular for every signature $S(\lambda x)$. So $f_{x,u} \equiv 0$ on $\mathbb{R}$. An entire function (or even just a real-analytic function on $\mathbb{R}$) that vanishes everywhere has all its Taylor coefficients zero — this is the [identity theorem for analytic functions](/theorems/???) at the simplest case. Reading off the $n$-th Taylor coefficient gives $\phi(n) \langle u_n, S(x)^{(n)}\rangle_{V^{\otimes n}} = 0$. The crucial role of the summability condition: it guarantees the series for $f_{x,u}$ converges on all of $\mathbb{C}$, so the analytic-vanishing argument applies. Without it we would have only a formal power series, and "all coefficients vanish" would not follow from "the function vanishes on $\mathbb{R}$." [/guided] [/step] [step:Conclude $u = 0$ using strict positivity of $\phi$ and the Driver lemma on signature components] The summability condition on $\phi$ together with the standing assumption that $\phi$ defines an inner product on $T_\phi((V))$ implies $\phi(n) > 0$ for every $n$. Hence the previous step simplifies to \begin{align*} \langle u_n, S(x)^{(n)} \rangle_{V^{\otimes n}} = 0 \quad \text{for every } n \geq 0 \text{ and every } x \in \mathcal{C}_p. \end{align*} Now we invoke the [Driver lemma on linear independence of signature components](/theorems/???), which asserts that the family $\{S(x)^{(n)} : x \in \mathcal{C}_p\}$ has dense linear span in $V^{\otimes n}$ for every $n \geq 0$. Concretely, this lemma says: any tensor $w \in V^{\otimes n}$ orthogonal in $V^{\otimes n}$ to every $S(x)^{(n)}$ must be zero. Applied to $w = u_n$, the chain above forces $u_n = 0 \in V^{\otimes n}$ for every $n$. Hence \begin{align*} u = (u_0, u_1, u_2, \ldots) = (0, 0, 0, \ldots) = 0 \in T_\phi((V)). \end{align*} This shows $W^\perp = \{0\}$, hence $W = T_\phi((V))$. [/step] [step:Define $\Psi$ as an isometry from $T_\phi((V))$ to $\mathcal{H}_\phi$] Define \begin{align*} \Psi: T_\phi((V)) &\to \mathcal{H}_\phi \\ \ell &\mapsto \bigl(x \mapsto \langle \ell, S(x)\rangle_\phi\bigr). \end{align*} Linearity in $\ell$ is immediate from linearity of $\langle \cdot, \cdot\rangle_\phi$ in its first argument. We verify that $\Psi$ is an isometry on the dense subspace $\operatorname{span} \mathcal{S} \subseteq T_\phi((V))$, and use this to extend. For $\ell = S(x)$ a single signature, $\Psi(S(x))(y) = \langle S(x), S(y)\rangle_\phi = k_\phi(x,y)$, which is exactly the kernel section $k_\phi(x, \cdot)$. By the [reproducing property of $\mathcal{H}_\phi$](/theorems/???), $k_\phi(x, \cdot) \in \mathcal{H}_\phi$ with $\|k_\phi(x,\cdot)\|_{\mathcal{H}_\phi}^2 = k_\phi(x,x) = \|S(x)\|_\phi^2$. By bilinearity, for any finite linear combination $\ell = \sum_i c_i S(x_i)$, \begin{align*} \|\Psi(\ell)\|_{\mathcal{H}_\phi}^2 = \Bigl\|\sum_i c_i k_\phi(x_i, \cdot)\Bigr\|_{\mathcal{H}_\phi}^2 = \sum_{i,j} c_i c_j k_\phi(x_i, x_j) = \Bigl\|\sum_i c_i S(x_i)\Bigr\|_\phi^2 = \|\ell\|_\phi^2, \end{align*} where the second equality is the standard RKHS computation $\langle k_\phi(x_i,\cdot), k_\phi(x_j,\cdot)\rangle_{\mathcal{H}_\phi} = k_\phi(x_i, x_j)$. Hence $\Psi$ is an isometry on $\operatorname{span}\mathcal{S}$. By the previous steps, $\operatorname{span}\mathcal{S}$ is dense in $T_\phi((V))$. By construction of $\mathcal{H}_\phi$ (the closure of the span of kernel sections in the RKHS norm), the image $\Psi(\operatorname{span}\mathcal{S}) = \operatorname{span}\{k_\phi(x,\cdot) : x \in \mathcal{C}_p\}$ is dense in $\mathcal{H}_\phi$. The [extension theorem for isometries](/theorems/???) — a densely defined isometry between Hilbert spaces with dense image extends uniquely to an isometric isomorphism between the two completions — applied to the densely defined isometry $\Psi$ gives a unique extension to a Hilbert space isomorphism \begin{align*} \Psi : T_\phi((V)) \to \mathcal{H}_\phi. \end{align*} [guided] The map $\Psi$ sends a "feature vector" $\ell \in T_\phi((V))$ to the function $x \mapsto \langle \ell, S(x)\rangle_\phi$ obtained by pairing $\ell$ against the signature feature map. We want to show $\Psi$ is a Hilbert-space isomorphism. The standard recipe is "isometry on a dense subspace, with dense image, extends uniquely". Let us execute it. **Define the candidate dense subspace.** From Steps 1--4 we have $W = T_\phi((V))$, where $W = \overline{\operatorname{span}\mathcal{S}}^{\|\cdot\|_\phi}$. So $\operatorname{span}\mathcal{S}$ is dense in $T_\phi((V))$. **Linearity of $\Psi$.** For $\ell, \ell' \in T_\phi((V))$ and $\alpha, \beta \in \mathbb{R}$, linearity of $\langle \cdot, \cdot\rangle_\phi$ in the first argument gives \begin{align*} \Psi(\alpha\ell + \beta\ell')(x) = \langle \alpha\ell + \beta\ell', S(x)\rangle_\phi = \alpha\langle \ell, S(x)\rangle_\phi + \beta\langle \ell', S(x)\rangle_\phi = (\alpha\Psi(\ell) + \beta\Psi(\ell'))(x). \end{align*} **Compute $\Psi$ on a single signature.** For $\ell = S(x)$ with $x \in \mathcal{C}_p$ and any $y \in \mathcal{C}_p$, \begin{align*} \Psi(S(x))(y) = \langle S(x), S(y)\rangle_\phi = k_\phi(x,y) = k_\phi(x, \cdot)(y). \end{align*} Hence $\Psi(S(x)) = k_\phi(x, \cdot)$, the kernel section at $x$. By the [reproducing property of $\mathcal{H}_\phi$](/theorems/???), the kernel section lies in $\mathcal{H}_\phi$ with \begin{align*} \langle k_\phi(x,\cdot), k_\phi(y, \cdot)\rangle_{\mathcal{H}_\phi} = k_\phi(x,y). \end{align*} In particular, $\|k_\phi(x,\cdot)\|^2_{\mathcal{H}_\phi} = k_\phi(x,x) = \langle S(x), S(x)\rangle_\phi = \|S(x)\|_\phi^2$. So $\Psi$ preserves squared norms on the generators $S(x)$. **Extend to finite linear combinations.** For $\ell = \sum_{i=1}^n c_i S(x_i) \in \operatorname{span}\mathcal{S}$, linearity of $\Psi$ gives $\Psi(\ell) = \sum_i c_i k_\phi(x_i, \cdot) \in \mathcal{H}_\phi$. We compute the squared norm in $\mathcal{H}_\phi$ using the inner product on kernel sections: \begin{align*} \|\Psi(\ell)\|^2_{\mathcal{H}_\phi} = \Bigl\|\sum_i c_i k_\phi(x_i, \cdot)\Bigr\|^2_{\mathcal{H}_\phi} = \sum_{i,j} c_i c_j\, \langle k_\phi(x_i, \cdot), k_\phi(x_j, \cdot)\rangle_{\mathcal{H}_\phi} = \sum_{i,j} c_i c_j\, k_\phi(x_i, x_j). \end{align*} On the source side, the same bilinear expansion of $\langle \ell, \ell\rangle_\phi$ gives \begin{align*} \|\ell\|_\phi^2 = \Bigl\langle \sum_i c_i S(x_i),\, \sum_j c_j S(x_j)\Bigr\rangle_\phi = \sum_{i,j} c_i c_j\, \langle S(x_i), S(x_j)\rangle_\phi = \sum_{i,j} c_i c_j\, k_\phi(x_i, x_j). \end{align*} Comparing, $\|\Psi(\ell)\|_{\mathcal{H}_\phi}^2 = \|\ell\|_\phi^2$. Hence $\Psi$ is an isometry on $\operatorname{span}\mathcal{S}$. **Density of the image.** The image $\Psi(\operatorname{span}\mathcal{S}) = \operatorname{span}\{k_\phi(x, \cdot) : x \in \mathcal{C}_p\}$ is dense in $\mathcal{H}_\phi$ by construction of $\mathcal{H}_\phi$ — the RKHS is by definition the closure of the span of kernel sections in the RKHS norm. **Extend by continuity.** The [extension theorem for isometries](/theorems/???) applies: a linear isometry from a dense subspace $D$ of a Hilbert space $H_1$ into a Hilbert space $H_2$ extends uniquely to a linear isometry $\tilde\Psi : H_1 \to H_2$. The proof is the usual one — for $h \in H_1$ pick $\ell_n \in D$ with $\ell_n \to h$, then $\Psi(\ell_n)$ is Cauchy in $H_2$ (because $\Psi$ is an isometry, so $\|\Psi(\ell_n) - \Psi(\ell_m)\|_{H_2} = \|\ell_n - \ell_m\|_{H_1} \to 0$) and one defines $\tilde\Psi(h) := \lim_n \Psi(\ell_n)$ in $H_2$. Density of the image of $D$ in $H_2$ then forces $\tilde\Psi$ to be surjective. Applied to $D = \operatorname{span}\mathcal{S}$, $H_1 = T_\phi((V))$, $H_2 = \mathcal{H}_\phi$, this produces the unique isometric isomorphism \begin{align*} \Psi : T_\phi((V)) \to \mathcal{H}_\phi, \qquad \ell \mapsto \bigl(x \mapsto \langle \ell, S(x)\rangle_\phi\bigr). \end{align*} The fact that the formula $\Psi(\ell)(x) = \langle \ell, S(x)\rangle_\phi$ continues to hold for general $\ell \in T_\phi((V))$ (not just those in $\operatorname{span}\mathcal{S}$) follows by approximation: for $\ell_n \to \ell$ in $T_\phi((V))$, both sides converge — the left side by continuity of the extension, the right side because $\langle \ell_n, S(x)\rangle_\phi \to \langle \ell, S(x)\rangle_\phi$ by Cauchy--Schwarz with $\|S(x)\|_\phi$ fixed. [/guided] [/step] [step:Read off existence and uniqueness of $\ell_f$ for every $f \in \mathcal{H}_\phi$] Since $\Psi$ is a Hilbert-space isomorphism, in particular a bijection, every $f \in \mathcal{H}_\phi$ has a unique preimage $\ell_f := \Psi^{-1}(f) \in T_\phi((V))$. By definition of $\Psi$, \begin{align*} f(x) = \Psi(\ell_f)(x) = \langle \ell_f, S(x)\rangle_\phi \quad \text{for all } x \in \mathcal{C}_p, \end{align*} which is the existence claim. Uniqueness of $\ell_f$ follows directly from injectivity of $\Psi$ (as part of being an isomorphism). The map $\ell_f \mapsto f$ is the inverse of $\ell_f \leftrightarrow f$ under $\Psi$, hence an isomorphism between $T_\phi((V))$ and $\mathcal{H}_\phi$, completing the proof. [/step]