I don't understand the following proof that the reproducing kernel $k: \mathcal X \times \mathcal X \rightarrow \mathbb C$ is positive definite.

$\forall f \in H, \langle f,k_x\rangle_{H} = f(x)$ and $k(y,x) = k_x(y)$.

$\sum_{i,j} c_i \bar{c_j} k(x_i, x_j) = \sum_{i,j} c_i \bar{c_j} \langle k(\cdot,x_i), k(\cdot, x_j)\rangle = \langle \sum_i c_i k(\cdot,x_i), \sum_j c_j k(\cdot,x_j) \rangle \ge 0 $


Why $k(x_i,x_j) = \langle k(\cdot,x_i), k(\cdot, x_j) \rangle$ ?

I think $k(x_i,x_j) = \langle k(\cdot,x_j), k(\cdot, x_i) \rangle = \overline{\langle k(\cdot,x_i), k(\cdot, x_j) \rangle} \ne \langle k(\cdot,x_i), k(\cdot, x_j) \rangle$. Should inner product of RKHS be symmetric?

  • $\begingroup$ Indeed, it rather should be $\langle k_{x_j}, k_{x_i}\rangle$ applying the condition for $f=k_{x_j}$ and $x=x_i$. But, it causes a harmless change of the above.. $\endgroup$ – Berci Oct 10 '18 at 23:31

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