[proofplan] The full $\phi$-signature kernel is the sum $k_\phi^{x,y}(s,t) = \sum_{i=0}^\infty \phi(i)\,\langle S(x)^i_{a,s}, S(y)^i_{a,t}\rangle_{V^{\otimes i}}$ and its truncation $k_\phi^{n,x,y}(s,t)$ is the partial sum to order $n$, so the difference is exactly the tail $\sum_{i=n+1}^\infty$. We control each tail term by Cauchy-Schwarz on the inner product followed by the factorial decay estimate $\|S(x)^i_{a,s}\|_{V^{\otimes i}} \leq \|x\|_{1,[a,s]}^i / i!$. The resulting tail sum is finite by the assumed summability condition on $\phi$, and convergence to $0$ as $n \to \infty$ follows from this same condition because the tail of a convergent non-negative series tends to zero. [/proofplan] [step:Identify the difference as the tail of the kernel series] By the definition of the $\phi$-signature kernel and its truncation, \begin{align*} k_\phi^{x,y}(s,t) &= \sum_{i=0}^\infty \phi(i)\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}, \\ k_\phi^{n,x,y}(s,t) &= \sum_{i=0}^n \phi(i)\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}. \end{align*} Both series converge absolutely (the first by the *Sufficient Condition for Signature Membership* applied to $\phi$; the second is a finite sum), so subtraction is term-by-term: \begin{align*} k_\phi^{x,y}(s,t) - k_\phi^{n,x,y}(s,t) = \sum_{i=n+1}^\infty \phi(i)\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}. \end{align*} Taking absolute values and applying the triangle inequality termwise, \begin{align*} \left| k_\phi^{x,y}(s,t) - k_\phi^{n,x,y}(s,t) \right| \leq \sum_{i=n+1}^\infty \phi(i)\,\left|\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}\right|, \end{align*} where we used $\phi(i) \geq 0$ to drop the absolute value on $\phi$. [guided] The first task is to write the difference $k_\phi^{x,y}(s,t) - k_\phi^{n,x,y}(s,t)$ as something concrete that we can estimate. By the definition of the $\phi$-signature kernel, \begin{align*} k_\phi^{x,y}(s,t) = \sum_{i=0}^\infty \phi(i)\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}, \end{align*} and the level-$n$ truncation simply cuts the series off: \begin{align*} k_\phi^{n,x,y}(s,t) = \sum_{i=0}^n \phi(i)\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}. \end{align*} Why are we allowed to subtract these series term by term? The first series converges absolutely by the *Sufficient Condition for Signature Membership* (this is precisely the condition we have assumed on $\phi$); the second is a finite sum and so converges directly. Two absolutely convergent series can be subtracted term by term, giving \begin{align*} k_\phi^{x,y}(s,t) - k_\phi^{n,x,y}(s,t) = \sum_{i=n+1}^\infty \phi(i)\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}. \end{align*} The difference is exactly the tail of the kernel series — every order $\le n$ cancels. Now we apply the triangle inequality, which for an absolutely convergent series of real numbers gives $|\sum a_i| \leq \sum |a_i|$. Since $\phi(i) \geq 0$ by hypothesis (we have $\phi : \mathbb{N} \cup \{0\} \to \mathbb{R}_+$), we may pull $\phi(i)$ out of the absolute value: \begin{align*} \left| k_\phi^{x,y}(s,t) - k_\phi^{n,x,y}(s,t) \right| \leq \sum_{i=n+1}^\infty \phi(i)\,\left|\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}\right|. \end{align*} The remaining task is to bound each absolute inner product. [/guided] [/step] [step:Bound each inner product via Cauchy-Schwarz on $V^{\otimes i}$] For each $i \geq n+1$, the tensor product space $V^{\otimes i}$ is a Hilbert space with the induced inner product $\langle \cdot, \cdot \rangle_{V^{\otimes i}}$ and associated norm $\|\cdot\|_{V^{\otimes i}}$. By the Cauchy-Schwarz inequality on $V^{\otimes i}$ applied to the vectors $S(x)^i_{a,s}, S(y)^i_{a,t} \in V^{\otimes i}$, \begin{align*} \left|\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}\right| \leq \|S(x)^i_{a,s}\|_{V^{\otimes i}}\,\|S(y)^i_{a,t}\|_{V^{\otimes i}}. \end{align*} [/step] [step:Apply the factorial decay estimate $\|S(x)^i_{a,s}\|_{V^{\otimes i}} \leq \|x\|_{1,[a,s]}^i / i!$] By the factorial decay estimate for iterated integrals of bounded variation paths (Chapter 1, Proposition on factorial decay), since $x \in C_1([a,b], V)$ has finite $1$-variation, \begin{align*} \|S(x)^i_{a,s}\|_{V^{\otimes i}} \leq \frac{\|x\|_{1,[a,s]}^i}{i!}. \end{align*} The same estimate applies to $y$ on the interval $[a,t]$: \begin{align*} \|S(y)^i_{a,t}\|_{V^{\otimes i}} \leq \frac{\|y\|_{1,[a,t]}^i}{i!}. \end{align*} Multiplying, \begin{align*} \|S(x)^i_{a,s}\|_{V^{\otimes i}}\,\|S(y)^i_{a,t}\|_{V^{\otimes i}} \leq \frac{(\|x\|_{1,[a,s]}\,\|y\|_{1,[a,t]})^i}{(i!)^2}. \end{align*} [guided] We invoke the factorial decay estimate for iterated integrals: if $x \in C_1([a,b], V)$ is a bounded variation path with $1$-variation $\|x\|_{1,[a,s]} := \sup_\pi\sum_k |x_{t_{k+1}} - x_{t_k}|$ over partitions $\pi$ of $[a,s]$, then the level-$i$ signature term obeys \begin{align*} \|S(x)^i_{a,s}\|_{V^{\otimes i}} \leq \frac{\|x\|_{1,[a,s]}^i}{i!}. \end{align*} Why does this hold? The level-$i$ signature is an iterated Riemann-Stieltjes integral $S(x)^i_{a,s} = \int_{a < t_1 < \dots < t_i < s} dx_{t_1} \otimes \cdots \otimes dx_{t_i}$. Bounding the integrand by $|dx_{t_1}|\cdots|dx_{t_i}|$ in the projective tensor norm and integrating over the simplex $\{a < t_1 < \dots < t_i < s\}$ — whose $1$-variation volume is $\|x\|_{1,[a,s]}^i / i!$ since the simplex has fraction $1/i!$ of the cube — yields the bound. The hypothesis $x \in C_1$ is consumed here: without finite $1$-variation, the iterated integral is not even defined and the bound has no content. The estimate applies symmetrically to $y$ on $[a,t]$, giving $\|S(y)^i_{a,t}\|_{V^{\otimes i}} \leq \|y\|_{1,[a,t]}^i / i!$. Multiplying the two estimates produces \begin{align*} \|S(x)^i_{a,s}\|_{V^{\otimes i}}\,\|S(y)^i_{a,t}\|_{V^{\otimes i}} \leq \frac{\|x\|_{1,[a,s]}^i\,\|y\|_{1,[a,t]}^i}{i!\cdot i!} = \frac{(\|x\|_{1,[a,s]}\,\|y\|_{1,[a,t]})^i}{(i!)^2}. \end{align*} The factorial-squared denominator is what makes signature kernels behave so well analytically — the tail decays much faster than any geometric series. [/guided] [/step] [step:Combine the estimates to prove the tail bound] Substituting the bound from the previous step into the inequality from Step 1, \begin{align*} \left| k_\phi^{x,y}(s,t) - k_\phi^{n,x,y}(s,t) \right| \leq \sum_{i=n+1}^\infty \phi(i)\,\frac{(\|x\|_{1,[a,s]}\,\|y\|_{1,[a,t]})^i}{(i!)^2}, \end{align*} which is the claimed bound. [/step] [step:Deduce convergence as $n \to \infty$ from summability of $\phi$] The *Sufficient Condition for Signature Membership* asserts that the series \begin{align*} \sum_{i=0}^\infty \phi(i)\,\frac{(\|x\|_{1,[a,s]}\,\|y\|_{1,[a,t]})^i}{(i!)^2} \end{align*} converges. Since this is a non-negative convergent series, its tail starting from index $n+1$ tends to zero as $n \to \infty$: \begin{align*} \sum_{i=n+1}^\infty \phi(i)\,\frac{(\|x\|_{1,[a,s]}\,\|y\|_{1,[a,t]})^i}{(i!)^2} \xrightarrow{n \to \infty} 0. \end{align*} Combining with the bound of the previous step, \begin{align*} \left| k_\phi^{x,y}(s,t) - k_\phi^{n,x,y}(s,t) \right| \xrightarrow{n \to \infty} 0, \end{align*} which is exactly $k_\phi^{n,x,y}(s,t) \to k_\phi^{x,y}(s,t)$ as $n \to \infty$. This completes the proof. [guided] We have a non-negative bound on the error: \begin{align*} \left| k_\phi^{x,y}(s,t) - k_\phi^{n,x,y}(s,t) \right| \leq \sum_{i=n+1}^\infty \phi(i)\,\frac{(\|x\|_{1,[a,s]}\,\|y\|_{1,[a,t]})^i}{(i!)^2} =: T_n. \end{align*} We need to show $T_n \to 0$ as $n \to \infty$. The hypothesis on $\phi$ — that $\phi$ satisfies the *Sufficient Condition for Signature Membership* — supplies exactly the missing ingredient: the full series \begin{align*} S := \sum_{i=0}^\infty \phi(i)\,\frac{(\|x\|_{1,[a,s]}\,\|y\|_{1,[a,t]})^i}{(i!)^2} \end{align*} converges to a finite number $S \in \mathbb{R}_{\ge 0}$. Why does this force $T_n \to 0$? A general fact about convergent non-negative series: if $\sum_{i=0}^\infty a_i = S < \infty$ with $a_i \ge 0$, then the tail \begin{align*} T_n := \sum_{i=n+1}^\infty a_i = S - \sum_{i=0}^n a_i \to S - S = 0 \end{align*} as $n \to \infty$, since the partial sums $\sum_{i=0}^n a_i$ converge to $S$ by definition of convergence of the series. Applying this with $a_i = \phi(i)\,(\|x\|_{1,[a,s]}\,\|y\|_{1,[a,t]})^i / (i!)^2 \ge 0$ (non-negativity using $\phi(i) \ge 0$, the bounded-variation norms $\ge 0$, and $i! > 0$), \begin{align*} T_n \xrightarrow{n \to \infty} 0. \end{align*} Combining with the bound from Step 4, \begin{align*} 0 \le \left| k_\phi^{x,y}(s,t) - k_\phi^{n,x,y}(s,t) \right| \le T_n \to 0, \end{align*} the squeeze theorem gives $|k_\phi^{x,y}(s,t) - k_\phi^{n,x,y}(s,t)| \to 0$, i.e. $k_\phi^{n,x,y}(s,t) \to k_\phi^{x,y}(s,t)$ as $n \to \infty$. A note on what fails without the hypothesis. If $\phi$ did not satisfy the summability condition (for example $\phi(i) = (i!)^2$ which makes the series diverge), then the truncations $k_\phi^{n,x,y}$ would still be well-defined finite quantities for each $n$, but the limit $k_\phi^{x,y}$ would not exist as a real number, and the convergence statement would be vacuous. The summability condition is precisely what makes the limit object exist and the truncation a genuine approximation. [/guided] [/step]