Feature maps and Reproducing Kernel Hilbert Spaces

Question

Let $K:E\times E\to \mathbb C$ be a positive definite function. I want to prove the complex case of what is stated on Wikipedia https://en.wikipedia.org/wiki/Reproducing_kernel_Hilbert_space#Feature_maps for the real case. So I want to prove that there is a Hilbert space $H$ and a feature map $\Phi:E\to H$ such that

$$ \forall x,y\in E\quad K(x,y)=\langle\Phi(x),\Phi(y)\rangle. $$ But when I use the Moore-Aronszajn theorem to get the reproducing kernel Hilbert space $H$ with reproducing kernel $K$, then I get by the reproducing property

$$ \forall x,y\in E\quad K(x,y)=\langle K(\cdot,y),K(\cdot,x)\rangle=\langle \Phi(y),\Phi(x)\rangle $$ if I define $\Phi:E \to H;t\mapsto K(\cdot,t)$. This means that it does not work the way I tried because $x$ and $y$ are in the wrong order. How do I prove it correctly?

Answer 1

The Wikipedia article you refer to defines $\Phi$ by $\Phi(t) = K(t, \cdot)$. Since $K(y,x)=\overline{K(x,y)}$, this is equivalent to $\Phi(t) = \overline{ K( \cdot, t)}$. The computation works:

$$K(x,y)=\langle K(\cdot,y),K(\cdot,x)\rangle=\langle \overline{\Phi(y)}, \overline{\Phi(x)}\rangle = \langle \Phi(x), \Phi(y) \rangle$$

A more concise way of saying the same is that when we're unhappy about the order of arguments in an inner product, the complex conjugate space offers an alternative: it's another Hilbert space in which the scalar multiplication is modified as $\alpha * v = \bar \alpha v$, and the inner product is accordingly $\langle x,y\rangle_* = \langle y, x\rangle$.