So, I have the following exercise I need to solve:
"For $x, y \in \mathbb{R}, \: K_1(x,y) = (xy + 1)^2 \:$ and $\: K_2(x,y) = (xy - 1)^2$. Describe the RKHS of $K_1$, $K_2$ and $K_1 + K_2$."
By "describe", I'm pretty sure it's asking for the general function form of the corresponding RKHS, the inner product, and the norm.
Now, beginning with $K_1$, what I tried was to do $f(x) = \sum_i a_i K_1(x,y_i)$, keeping $y$ fixed, and developping the square in the kernel, which gives me: $$f(x) = \sum_i a_i (y_i^2x^2 + 2y_ix + 1) = \alpha x^2 + \beta x + c$$ Where $\alpha = \sum_i a_i y_i^2$, $\beta = \sum_i 2a_i y_i$ and $c=\sum_i a_i$.
(i) Here, I have my first question. When I do $\sum_i$, I have to consider $x,y \in \mathbb{R}^d$, right? But since it's stated that $x,y \in \mathbb{R}$, is it possible/necessary to drop the $\sum_i$ when searching for a form of $f(x)$?
Now, using the results I got before, if the form of $f(x)$ is correct, I am able to describe any function as a vector: $$f = [\alpha \: \beta \: c]^T$$ With inner product: $$
(ii) I'm not sure about the inner product and norm though, since in this case $c$ would not have the same dimension as $\alpha$ and $\beta$, which seems awkward, and I believe is linked to question (i)?
Following the same logic for $K_2$, since all that changes between $K_1$ and $K_2$ is a sign, then only $\beta$ will change and become $\beta = -\sum_i 2a_i y_i$.
(iii) However, if I do that, and describe the functions of $K_2$ as $f = [\alpha \: \beta \: c]^T$, then $f(x)$ for $K_2$ will be exactly the same as for $K_1$, which seems wrong to me, since that would break the unicity between kernels and RKHSs. I could describe the functions for $K_2$ as $f = [\alpha \: -\beta \: c]^T$, but it stills seems off to me. I mean, I could just change the choice of $\beta$ to get a function that is supposed to be in the RKHS of $K_1$, no?
If I grasp how to do $K_1$ and $K_2$ separately, I imagine I'll have no problem doing $K_1 + K_2$ afterwards.
Any advice (or better ways to think about and solve this) are welcome.