# RKHS for polynomial kernel

Say we have a polynomial kernel of degree two: $k(x,x')=\langle x,x' \rangle^2$ for $X=\mathbb{R}^2$. I know that a feature map $\phi(x)=(x_1^2,\sqrt{2}x_1x_2,x_2^2)$ exist. What I want to know is what is the corresponding RKHS?

So as you say, $$\phi(x)=(x_1^2,\sqrt{2}x_1x_2,x_2^2)\in\mathbb{R}^3$$ is a feature map (since $$k(x,x')=\langle\phi(x),\phi(x')\rangle_{\mathbb{R}^3}$$), hence the functions in the RKHS are of the form $$f(x)=\langle\phi(x),\beta\rangle_{\mathbb{R}^3}=x_1^2\beta_1+\sqrt{2}x_1x_2\beta_2+x_2^2\beta_3$$ That is, the RKHS is $$\{f:\mathbb{R}\rightarrow\mathbb{R}|f(x)=x_1^2\beta_1+\sqrt{2}x_1x_2\beta_2+x_2^2\beta_3, \beta\in\mathbb{R}^3\}$$ and the RKHS norm of such an $$f$$ is $$||f||^2=\beta_1^2+\beta_2^2+\beta_3^2$$.

You can find the answer here page 42

Otherwise, if this is self-study here are the steps:

1) Look for an inner-product:

$K (x, y) = trace(x^Tyx^Ty)$ ...

2) Propose a candidate RKHS

$f(x) = \sum_i a_i K(x_i,x)...$

3) Check that the candidate is a Hilbert space

4) Check that $\mathcal{H}$ is the RKHS