Let $X$ be a set, $k \colon X \times X \to \mathbb R$ be a symmetric positive definite kernel and $H := \overline{\text{span}}(\{ k(x, \cdot): x \in X \})$ the induced real reproducing kernel Hilbert space.
Is the followingclaim and its proof correct?
Claim. Let $(h_n)_{n \in \mathbb N} \subset H$ and $h \in H$. We have $h_n \xrightarrow[w]{n \to \infty} h$ (weak convergence) if and only if $h_n(x) \xrightarrow{n \to \infty} h(x)$ for all $x \in X$.
Proof. "$\implies$": If $h_n \xrightarrow[w]{n \to \infty} h$, then $\langle h_n - h, f \rangle \xrightarrow{n \to \infty} 0$ for all $f \in H$. Since $k(x, \cdot) \in H$, this in particular implies (this is the reproducing property of the RKHS) $\langle h_n - h, k(x, \cdot) \rangle = h_n(x) - h(x) \xrightarrow{n \to \infty} 0$ for all $x \in X$.
"$\impliedby$": Conversely, if $h_n \to h$ pointwise, that is, $\langle h_n - h, k(x, \cdot) \rangle \xrightarrow{n \to \infty} 0$ for all $x \in X$, then also $\langle h_n - h, f \rangle \xrightarrow{n \to \infty} 0$ for all $f \in \text{span}(\{ k(x, \cdot): x \in X \})$ by the linearity of the inner product and by its continuity, even for all $f \in \overline{\text{span}}(\{ k(x, \cdot): x \in X \}) = H$, which means that $h_n \to h$ weakly.
Motiviation. A standard fact about reproducing kernel Hilbert spaces is that strong convergence implies pointwise convergence. A standard fact from Functional Analysis is that strong convergence implies weak convergence.