[proofplan] We expand the weighted signature kernel as the series $k_\phi(x,y)_{s,t} = \sum_{i \geq 0} \phi(i)\, \langle S(x)_{a,s}^{(i)}, S(y)_{c,t}^{(i)} \rangle_{V^{\otimes i}}$, isolate the $i = 0$ term as the constant $\phi(0)$, and apply the iterated-integral recursion $S(x)_{a,s}^{(i)} = \int_a^s S(x)_{a,u}^{(i-1)} \otimes dx_u$ to peel one tensor level off each remaining signature factor. Fubini's theorem, justified by a uniform $L^1$ dominating function on the simplex constructed from the factorial bound $\|S(x)^{(k)}\|_{V^{\otimes k}} \leq L^k/k!$ and the summability condition, lets us interchange the sum and the iterated integrals. The product structure $\langle A \otimes a, B \otimes b\rangle_{V^{\otimes j+1}} = \langle A, B\rangle_{V^{\otimes j}} \langle a, b\rangle_V$ of the Hilbert--Schmidt inner product on tensor products factors the integrand, after which re-indexing $j = i-1$ collapses the inner sum to $k_{\phi_+}(x,y)_{u,v}$. [/proofplan]

[step:Expand the weighted signature kernel as a series and isolate the order-zero term] By definition of the $\phi$-weighted inner product, \begin{align*} k_\phi(x, y)_{s,t} = \langle S(x)_{a,s}, S(y)_{c,t} \rangle_\phi = \sum_{i=0}^\infty \phi(i)\, \langle S(x)_{a,s}^{(i)}, S(y)_{c,t}^{(i)} \rangle_{V^{\otimes i}}. \end{align*} The order-zero components of the signature satisfy $S(x)_{a,s}^{(0)} = 1 \in \mathbb{R} \cong V^{\otimes 0}$ and $S(y)_{c,t}^{(0)} = 1$, so $\langle S(x)_{a,s}^{(0)}, S(y)_{c,t}^{(0)} \rangle_{V^{\otimes 0}} = 1$. Splitting off the $i = 0$ summand gives \begin{align*} k_\phi(x, y)_{s,t} = \phi(0) + \sum_{i=1}^\infty \phi(i)\, \langle S(x)_{a,s}^{(i)}, S(y)_{c,t}^{(i)} \rangle_{V^{\otimes i}}. \end{align*} [/step]

[step:Apply the iterated-integral recursion to peel off one tensor level from each signature factor]For each $i \geq 1$, the [iterated integral definition of the signature](/theorems/???) provides the recursion \begin{align*} S(x)_{a,s}^{(i)} &= \int_a^s S(x)_{a,u}^{(i-1)} \otimes dx_u \in V^{\otimes i}, \\ S(y)_{c,t}^{(i)} &= \int_c^t S(y)_{c,v}^{(i-1)} \otimes dy_v \in V^{\otimes i}. \end{align*} Both integrals are well defined because $x$ and $y$ are of bounded variation ($x \in C_1([a,b], V)$, $y \in C_1([c,d], V)$), so each of these is a [Riemann--Stieltjes integral](/theorems/???) of a continuous integrand against a path of bounded variation. Substituting into the previous step, \begin{align*} k_\phi(x, y)_{s,t} = \phi(0) + \sum_{i=1}^\infty \phi(i)\, \Bigl\langle \int_a^s S(x)_{a,u}^{(i-1)} \otimes dx_u,\; \int_c^t S(y)_{c,v}^{(i-1)} \otimes dy_v \Bigr\rangle_{V^{\otimes i}}. \end{align*}[/step]

[guided]The signature truncated at level $i$ has the recursive structure \begin{align*} S(x)_{a,s}^{(i)} = \int_a^s S(x)_{a,u}^{(i-1)} \otimes dx_u, \end{align*} and similarly for $y$ on $[c,t]$. Why is this useful here? The right-hand side of the identity we are trying to prove involves the kernel of the *shifted* weight $\phi_+$, which we recall is defined by $\phi_+(j) = \phi(j+1)$. Pairing the weight $\phi(i)$ with a tensor in $V^{\otimes i}$ corresponds, after shifting indices by $1$, to pairing $\phi_+(j) = \phi(j+1)$ with a tensor in $V^{\otimes j}$. The recursion above is exactly the tool that produces the index shift: it expresses an order-$i$ signature as an integral of order-$(i-1)$ signatures. We must verify that both Riemann--Stieltjes integrals are well defined. The hypothesis $x \in C_1([a,b], V)$ means $x$ has bounded $1$-variation, hence is of bounded variation in the classical sense; the integrand $u \mapsto S(x)_{a,u}^{(i-1)}$ is continuous on $[a,b]$ (by induction on $i$, since the level-$0$ signature is the constant $1$ and each integration produces a continuous function). The same applies on $[c,d]$ for $y$. By the [existence theorem for the Riemann--Stieltjes integral of a continuous function against a function of bounded variation](/theorems/???), both integrals exist. Substituting these representations into the inner product gives \begin{align*} \sum_{i=1}^\infty \phi(i)\, \Bigl\langle \int_a^s S(x)_{a,u}^{(i-1)} \otimes dx_u,\; \int_c^t S(y)_{c,v}^{(i-1)} \otimes dy_v \Bigr\rangle_{V^{\otimes i}}. \end{align*}[/guided]

[step:Use the product structure of the Hilbert--Schmidt inner product to factor the integrand]Recall that the Hilbert--Schmidt inner product on $V^{\otimes (j+1)} \cong V^{\otimes j} \otimes V$ satisfies \begin{align*} \langle A \otimes a,\, B \otimes b \rangle_{V^{\otimes (j+1)}} = \langle A, B \rangle_{V^{\otimes j}} \cdot \langle a, b \rangle_V \end{align*} for all $A, B \in V^{\otimes j}$ and $a, b \in V$. Applying this with $j = i - 1$, $A = S(x)_{a,u}^{(i-1)}$, $B = S(y)_{c,v}^{(i-1)}$, $a = dx_u$, $b = dy_v$, the integrand factorises pointwise. Treating the iterated integrals as limits of Riemann--Stieltjes sums and passing the bilinear pairing through, we obtain \begin{align*} \Bigl\langle \int_a^s S(x)_{a,u}^{(i-1)} \otimes dx_u,\; \int_c^t S(y)_{c,v}^{(i-1)} \otimes dy_v \Bigr\rangle_{V^{\otimes i}} = \int_a^s \int_c^t \langle S(x)_{a,u}^{(i-1)}, S(y)_{c,v}^{(i-1)} \rangle_{V^{\otimes (i-1)}} \, \langle dx_u, dy_v \rangle_V. \end{align*} The exchange of pairing and integration is justified because each of $\int_a^s S(x)_{a,u}^{(i-1)} \otimes dx_u$ and $\int_c^t S(y)_{c,v}^{(i-1)} \otimes dy_v$ is a strong limit of Riemann--Stieltjes sums in the finite-dimensional spaces $V^{\otimes i}$, and the inner product is bilinear and continuous in each argument. Substituting, \begin{align*} k_\phi(x,y)_{s,t} = \phi(0) + \sum_{i=1}^\infty \phi(i) \int_a^s \int_c^t \langle S(x)_{a,u}^{(i-1)}, S(y)_{c,v}^{(i-1)} \rangle_{V^{\otimes (i-1)}} \, \langle dx_u, dy_v \rangle_V. \end{align*}[/step]

[guided]The expression we currently have for the $i$-th summand is \begin{align*} \Bigl\langle \int_a^s S(x)_{a,u}^{(i-1)} \otimes dx_u,\; \int_c^t S(y)_{c,v}^{(i-1)} \otimes dy_v \Bigr\rangle_{V^{\otimes i}}. \end{align*} We want to rewrite this as a double integral over $[a,s] \times [c,t]$ with the inner product passed *inside* the integrals. Why is this the right move? Because the eventual goal is to recognise the integrand as $k_{\phi_+}(x,y)_{u,v}$, which is itself an inner product at the lower level $V^{\otimes (i-1)}$ — precisely the level that survives once one tensor factor has been peeled off. The tool is the multiplicative property of the Hilbert--Schmidt inner product on a tensor product: under the associativity isomorphism $V^{\otimes (j+1)} \cong V^{\otimes j} \otimes V$ given on simple tensors by $v_1 \otimes \cdots \otimes v_{j+1} \mapsto (v_1 \otimes \cdots \otimes v_j) \otimes v_{j+1}$ and extended bilinearly, the HS inner product satisfies \begin{align*} \langle A \otimes a,\, B \otimes b\rangle_{V^{\otimes (j+1)}} = \langle A, B\rangle_{V^{\otimes j}}\, \langle a, b\rangle_V \qquad \text{for } A, B \in V^{\otimes j},\ a, b \in V. \end{align*} This is the defining property of the HS inner product on a tensor product (and reduces to the standard Frobenius identity $\operatorname{tr}((A \otimes a)(B \otimes b)^*) = \operatorname{tr}(AB^*)\, a \cdot b$ in the matrix case). We apply this with $j = i-1$, $A = S(x)_{a,u}^{(i-1)}$, $B = S(y)_{c,v}^{(i-1)}$, $a = dx_u$, $b = dy_v$: \begin{align*} \langle S(x)_{a,u}^{(i-1)} \otimes dx_u,\, S(y)_{c,v}^{(i-1)} \otimes dy_v\rangle_{V^{\otimes i}} = \langle S(x)_{a,u}^{(i-1)}, S(y)_{c,v}^{(i-1)}\rangle_{V^{\otimes (i-1)}}\, \langle dx_u, dy_v\rangle_V. \end{align*} This factorisation is *pointwise* in $(u,v)$. To pass it through the integrals, recall that each Riemann--Stieltjes integral $\int_a^s S(x)_{a,u}^{(i-1)} \otimes dx_u$ is a *strong* limit of Riemann sums in the finite-dimensional space $V^{\otimes i}$. The HS inner product $\langle \cdot, \cdot\rangle_{V^{\otimes i}}$ is bilinear and continuous in each argument, so it commutes with the strong limit. Concretely, if $P_n$ is a sequence of Riemann sums tending to the integral on the $x$-side and $Q_n$ similarly on the $y$-side, then \begin{align*} \langle P_n, Q_n\rangle_{V^{\otimes i}} \to \Bigl\langle \int \cdots, \int \cdots \Bigr\rangle_{V^{\otimes i}}, \end{align*} and the left-hand side, by bilinearity of the inner product on Riemann sums and the pointwise factorisation above, equals \begin{align*} \sum_{j,k} \langle S(x)_{a,u_j}^{(i-1)}, S(y)_{c,v_k}^{(i-1)}\rangle_{V^{\otimes (i-1)}}\, \langle x_{u_{j+1}} - x_{u_j}, y_{v_{k+1}} - y_{v_k}\rangle_V, \end{align*} which is itself a Riemann sum approximating \begin{align*} \int_a^s \int_c^t \langle S(x)_{a,u}^{(i-1)}, S(y)_{c,v}^{(i-1)}\rangle_{V^{\otimes (i-1)}}\, \langle dx_u, dy_v\rangle_V. \end{align*} Taking the limit identifies the inner product of the iterated integrals with this double Riemann--Stieltjes integral, justifying the passage. Substituting back, we have \begin{align*} k_\phi(x,y)_{s,t} = \phi(0) + \sum_{i=1}^\infty \phi(i) \int_a^s \int_c^t \langle S(x)_{a,u}^{(i-1)}, S(y)_{c,v}^{(i-1)} \rangle_{V^{\otimes (i-1)}} \, \langle dx_u, dy_v \rangle_V. \end{align*} The sum and integrals are still in this order; swapping them is the next step's job, after we set up an integrable dominator.[/guided]

[step:Construct an integrable dominating function on the simplex via the factorial bound on signature levels]Before exchanging sum and integral we exhibit an $L^1$-dominator. By the [factorial decay of iterated integrals](/theorems/???), there exist constants $L_x \geq \|x\|_{1\text{-var};[a,b]}$ and $L_y \geq \|y\|_{1\text{-var};[c,d]}$ such that for all $u \in [a,s]$, $v \in [c,t]$ and all $k \geq 0$, \begin{align*} \|S(x)_{a,u}^{(k)}\|_{V^{\otimes k}} \leq \frac{L_x^k}{k!}, \qquad \|S(y)_{c,v}^{(k)}\|_{V^{\otimes k}} \leq \frac{L_y^k}{k!}. \end{align*} Cauchy--Schwarz on $V^{\otimes (i-1)}$ then gives \begin{align*} \bigl|\langle S(x)_{a,u}^{(i-1)}, S(y)_{c,v}^{(i-1)} \rangle_{V^{\otimes (i-1)}}\bigr| \leq \frac{(L_x L_y)^{i-1}}{((i-1)!)^2}. \end{align*} Setting $C := L_x L_y$ and using $\phi_+(i-1) = \phi(i)$, the absolute value of the $i$-th summand of the integrand is bounded by \begin{align*} |\phi(i)| \cdot \frac{C^{i-1}}{((i-1)!)^2} = |\phi_+(i-1)| \cdot \frac{C^{i-1}}{((i-1)!)^2}. \end{align*} Re-indexing $j = i-1$ and using the hypothesis that $|\phi_+|$ satisfies the summability condition $\sum_{j=0}^\infty |\phi_+(j)|\, C^j / (j!)^2 < \infty$ for every $C > 0$, we obtain \begin{align*} M := \sum_{j=0}^\infty |\phi_+(j)| \cdot \frac{C^j}{(j!)^2} < \infty. \end{align*} The function $(u,v) \mapsto M$ is constant on $[a,s] \times [c,t]$, hence integrable against the finite total-variation measure $|d\langle x,y\rangle|_V$ induced by the bilinear pairing of the bounded-variation paths $x, y$.[/step]

[guided]The exchange of sum and double integral that we are about to perform is the main analytic step of the proof, and it requires a dominating function. Where does the dominator come from? The factorial bound $\|S(x)^{(k)}\| \leq L^k/k!$ is the classical estimate for iterated integrals against a bounded-variation path: each integration introduces one factor of the total variation $L$ and one factorial in the denominator, by induction on $k$. We need $L_x \geq \|x\|_{1\text{-var};[a,b]}$ to control all of $u \in [a,s]$ uniformly (and similarly for $y$). Now the integrand of the $i$-th summand is, in absolute value, at most \begin{align*} |\phi(i)| \cdot \|S(x)_{a,u}^{(i-1)}\|_{V^{\otimes(i-1)}} \cdot \|S(y)_{c,v}^{(i-1)}\|_{V^{\otimes (i-1)}} \leq |\phi(i)|\, \frac{L_x^{i-1}}{(i-1)!}\, \frac{L_y^{i-1}}{(i-1)!} = |\phi_+(i-1)| \, \frac{(L_x L_y)^{i-1}}{((i-1)!)^2}, \end{align*} where in the last step we used $\phi(i) = \phi_+(i-1)$. Summing over $i \geq 1$ and re-indexing $j = i - 1$, \begin{align*} \sum_{i=1}^\infty |\phi(i)| \cdot \frac{(L_x L_y)^{i-1}}{((i-1)!)^2} = \sum_{j=0}^\infty |\phi_+(j)| \cdot \frac{(L_x L_y)^j}{(j!)^2} =: M. \end{align*} Why is $M$ finite? This is exactly where the hypothesis "$|\phi_+|$ satisfies the summability condition" is consumed: the summability condition says that for every $C > 0$, $\sum_j |\phi_+(j)|\, C^j / (j!)^2 < \infty$, and we apply it with $C = L_x L_y$. The dominator $M$ is constant on the rectangle $[a,s] \times [c,t]$, and the variation measure of the pairing $\langle dx, dy\rangle_V$ on this rectangle is finite (because both paths have bounded variation), so the constant $M$ is integrable against this measure.[/guided]

[step:Apply Fubini--Tonelli to interchange the infinite sum with the iterated integral]By the previous step, the integrand \begin{align*} F(u,v) := \sum_{i=1}^\infty \phi(i)\, \langle S(x)_{a,u}^{(i-1)}, S(y)_{c,v}^{(i-1)} \rangle_{V^{\otimes (i-1)}} \end{align*} is dominated in absolute value by the constant $M < \infty$ on $[a,s] \times [c,t]$, and the partial sums converge pointwise to $F$. Combining the pointwise series convergence with the uniform $L^1$ bound, [Fubini--Tonelli for series](/theorems/???) (equivalently, the [Dominated Convergence Theorem](/theorems/???) applied to the partial sums viewed as a sequence of integrable functions) lets us interchange the sum and the double integral: \begin{align*} \sum_{i=1}^\infty \phi(i) \int_a^s \int_c^t \langle S(x)_{a,u}^{(i-1)}, S(y)_{c,v}^{(i-1)} \rangle_{V^{\otimes (i-1)}}\, \langle dx_u, dy_v \rangle_V = \int_a^s \int_c^t F(u,v)\, \langle dx_u, dy_v \rangle_V. \end{align*}[/step]

[guided]We want to move the sum inside the double integral. The Fubini--Tonelli theorem for series-and-integrals says: if a sequence of integrable functions $f_i$ is dominated by an integrable $g$ in the sense that $\sum_i |f_i| \leq g$ pointwise with $g \in L^1$, then $\sum_i \int f_i = \int \sum_i f_i$. We verify the hypothesis. Set $f_i(u,v) := \phi(i) \langle S(x)_{a,u}^{(i-1)}, S(y)_{c,v}^{(i-1)} \rangle_{V^{\otimes(i-1)}}$. By Cauchy--Schwarz and the factorial bound from the previous step, \begin{align*} \sum_{i=1}^\infty |f_i(u,v)| \leq \sum_{i=1}^\infty |\phi_+(i-1)|\, \frac{(L_x L_y)^{i-1}}{((i-1)!)^2} = M. \end{align*} The constant $M$ is integrable against the finite total-variation measure on $[a,s] \times [c,t]$ induced by the bounded-variation paths $x,y$. So Fubini--Tonelli applies, and we may swap. What goes wrong without the summability condition on $|\phi_+|$? The dominator $M$ might be infinite, the partial sums could fail to converge in $L^1$, and the swap would be unjustified.[/guided]

[step:Re-index the inner sum and recognise the kernel of the shifted weight] Substitute $j = i - 1$ in the integrand $F(u,v)$ and use $\phi(i) = \phi(j+1) = \phi_+(j)$: \begin{align*} F(u,v) = \sum_{j=0}^\infty \phi_+(j)\, \langle S(x)_{a,u}^{(j)}, S(y)_{c,v}^{(j)} \rangle_{V^{\otimes j}} = \langle S(x)_{a,u}, S(y)_{c,v} \rangle_{\phi_+} = k_{\phi_+}(x,y)_{u,v}. \end{align*} Substituting this back into the expression from the previous step yields \begin{align*} k_\phi(x,y)_{s,t} = \phi(0) + \int_a^s \int_c^t k_{\phi_+}(x,y)_{u,v}\, \langle dx_u, dy_v \rangle_V, \end{align*} which is the claimed identity. [/step]