[proofplan] Symmetry follows immediately from the symmetry of the underlying $\phi$-weighted inner product on $T_\phi((V))$. For positive semidefiniteness we exploit the kernel's defining representation $k_\phi(x,y) = \langle S(x), S(y)\rangle_\phi$: any quadratic form $\sum_{i,j} c_i c_j k_\phi(x_i, x_j)$ rewrites as the squared $\phi$-norm of the linear combination $\sum_i c_i S(x_i)$ in $T_\phi((V))$, which is non-negative because $\phi \geq 0$ makes $\langle \cdot, \cdot\rangle_\phi$ a genuine semi-inner product. [/proofplan]

[step:Verify symmetry from the symmetry of the weighted inner product] Fix $x, y \in \mathcal{C}_p$. By definition of the kernel, \begin{align*} k_\phi(x, y) = \langle S(x), S(y) \rangle_\phi = \sum_{k=0}^\infty \phi(k)\, \langle S(x)^{(k)}, S(y)^{(k)} \rangle_{V^{\otimes k}}. \end{align*} Each Hilbert--Schmidt inner product on $V^{\otimes k}$ is symmetric, $\langle S(x)^{(k)}, S(y)^{(k)} \rangle_{V^{\otimes k}} = \langle S(y)^{(k)}, S(x)^{(k)} \rangle_{V^{\otimes k}}$. Since $\phi(k) \in \mathbb{R}$, the weighted sum is symmetric term by term, hence \begin{align*} k_\phi(x, y) = \langle S(y), S(x) \rangle_\phi = k_\phi(y, x). \end{align*} The series converges absolutely under the summability condition (hence symmetry of the limit follows from symmetry of every partial sum), but for symmetry we do not need convergence: we need only that the operation defining the limit treats $x$ and $y$ symmetrically, which it does. [/step]

[step:Reduce positive semidefiniteness to a squared norm via bilinearity]Let $n \in \mathbb{N}$, $x_1, \ldots, x_n \in \mathcal{C}_p$, and $c_1, \ldots, c_n \in \mathbb{R}$. We must show \begin{align*} \sum_{i,j=1}^n c_i c_j\, k_\phi(x_i, x_j) \geq 0. \end{align*} The summability condition on $\phi \geq 0$ together with the factorial bound on signatures guarantees that each $S(x_i)$ lies in $T_\phi((V))$, the Hilbert space underlying the weighted inner product (see the [Sufficient Condition for Signature Membership](/theorems/???)). Hence the linear combination $h := \sum_{i=1}^n c_i S(x_i)$ also lies in $T_\phi((V))$. Substituting the kernel's defining identity $k_\phi(x_i, x_j) = \langle S(x_i), S(x_j)\rangle_\phi$ and using the bilinearity of $\langle \cdot, \cdot \rangle_\phi$, \begin{align*} \sum_{i,j=1}^n c_i c_j\, k_\phi(x_i, x_j) &= \sum_{i,j=1}^n c_i c_j\, \langle S(x_i), S(x_j) \rangle_\phi \\ &= \Bigl\langle \sum_{i=1}^n c_i S(x_i),\; \sum_{j=1}^n c_j S(x_j) \Bigr\rangle_\phi \\ &= \langle h, h \rangle_\phi. \end{align*}[/step]

[guided]The strategy is to recognise the quadratic form as a squared norm in a Hilbert space. The defining identity $k_\phi(x,y) = \langle S(x), S(y)\rangle_\phi$ writes the kernel as an inner product evaluated on the images of $x$ and $y$ under the signature map $S : \mathcal{C}_p \to T_\phi((V))$. This realises $k_\phi$ as a "feature-space" kernel, and positive semidefiniteness of any feature-space kernel is automatic — provided the feature space is genuinely an inner-product space. We need each $S(x_i)$ to lie in the ambient Hilbert space $T_\phi((V))$ so that we can take linear combinations and inner products. This is exactly the content of the [Sufficient Condition for Signature Membership](/theorems/???): under the summability condition on $\phi$, the signature map lands inside $T_\phi((V))$. With this verified, set $h := \sum_{i=1}^n c_i S(x_i) \in T_\phi((V))$. Bilinearity of $\langle \cdot, \cdot\rangle_\phi$ — both arguments are linear, with real scalars — distributes the double sum over the inner product: \begin{align*} \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*} Reading the chain backwards gives $\sum_{i,j} c_i c_j\, k_\phi(x_i, x_j) = \langle h, h\rangle_\phi$.[/guided]

[step:Conclude using non-negativity of the squared norm]The hypothesis $\phi : \mathbb{N} \cup \{0\} \to \mathbb{R}_+$ ensures every weight $\phi(k) \geq 0$, so the form \begin{align*} \langle u, v \rangle_\phi = \sum_{k=0}^\infty \phi(k)\, \langle u_k, v_k \rangle_{V^{\otimes k}} \end{align*} is a non-negative semi-inner product on $T_\phi((V))$. In particular, \begin{align*} \langle h, h \rangle_\phi = \sum_{k=0}^\infty \phi(k)\, \|h_k\|_{V^{\otimes k}}^2 \geq 0, \end{align*} because each summand is the product of a non-negative weight and a non-negative squared Hilbert--Schmidt norm. Combined with the previous step, \begin{align*} \sum_{i,j=1}^n c_i c_j\, k_\phi(x_i, x_j) = \langle h, h \rangle_\phi \geq 0. \end{align*} Since $n$, the points $x_1, \ldots, x_n$, and the coefficients $c_1, \ldots, c_n$ were arbitrary, $k_\phi$ is positive semidefinite, completing the proof together with the symmetry established in the first step.[/step]

[guided]We finish by showing $\langle h, h\rangle_\phi \geq 0$ for the element $h = \sum_i c_i S(x_i) \in T_\phi((V))$ constructed in the previous step. Why is this immediate? Write $h = (h_k)_{k \geq 0}$ with each $h_k \in V^{\otimes k}$. By definition of the $\phi$-weighted inner product, \begin{align*} \langle h, h\rangle_\phi = \sum_{k=0}^\infty \phi(k)\, \langle h_k, h_k\rangle_{V^{\otimes k}} = \sum_{k=0}^\infty \phi(k)\, \|h_k\|_{V^{\otimes k}}^2. \end{align*} Each $\|h_k\|^2_{V^{\otimes k}} \geq 0$ because the Hilbert--Schmidt inner product on $V^{\otimes k}$ is a genuine inner product; each weight $\phi(k) \geq 0$ by the hypothesis $\phi : \mathbb{N} \cup \{0\} \to \mathbb{R}_+$. Hence the sum has all non-negative terms and is bounded below by zero. A subtlety worth flagging: $\langle \cdot, \cdot\rangle_\phi$ is in general only a *semi*-inner product on $T_\phi((V))$ (not a full inner product) — it can fail to be positive definite if $\phi$ has zeros, since then any element supported on the levels where $\phi$ vanishes has $\phi$-norm zero without being itself zero. The genuine Hilbert-space structure of $T_\phi((V))$ (used in the next theorem on the RKHS isomorphism) requires $\phi(k) > 0$ for every $k$, but here we need only positive *semi*-definiteness, and positive semidefiniteness of the kernel is equivalent to non-negativity of the semi-inner product, which is exactly what we have. Combining with the identity $\sum_{i,j} c_i c_j\, k_\phi(x_i, x_j) = \langle h, h\rangle_\phi$ from Step 2, \begin{align*} \sum_{i,j=1}^n c_i c_j\, k_\phi(x_i, x_j) = \langle h, h\rangle_\phi \geq 0. \end{align*} The choice of $n$, points $x_1, \ldots, x_n \in \mathcal{C}_p$, and coefficients $c_1, \ldots, c_n \in \mathbb{R}$ was arbitrary, so $k_\phi$ is positive semidefinite. Combined with the symmetry $k_\phi(x,y) = k_\phi(y,x)$ from Step 1, this finishes the proof.[/guided]