Prove that $k(u,v) = f(k_1(u,v))$ is a valid kernel, where $f$ is a polynomial with positive coefficients I want to show that $k(u, v) = f(k1(u, v))$, is a valid kernel where $f$ is a polynomial with positive coefficients.
Proof.
Since each polynomial term is a product of kernels with a positive coefficient, the proof follows by applying (1) and (2).
(1) $k(u,v) = \alpha k_a(u,v) + \beta k_b(u,v)$, where for any choice of $\alpha,\beta \geq 0$, $k$is a valid kernel.
(2)$k(u,v) = k_1(u,v)k_2(u,v)$ is a valid kernel.
I think this is really simple, but I'm having a hard time understanding the proof. It isn't clear to me with each polynomial term is a product of kernel.
 A: A polinomial of degree $n$ can be written as follows:
$$f(x)=\sum_{i=0}^n a_i x^i$$

Claim: For ever $n$, and every $a_i\geq 0$, if $k$ is a valid kernel, so is 
      $f(k)$, provided the conditions $(1)$ and $(2)$ given in the question.

Applying the second condition iteratively, one can see that if $k$ is a valid kernel, $k^i$ is also a valid kernel for all $i>0$.
Next, consider the first condition assume that $\beta=0$. Hence, $y_i=a_i k^i$ is also a valid kernel for all $i$. As a result,
$$f(x)=\sum_{i=0}^n a_i x^i$$
is a sum of valid kernels. This sum can be written as $n$ pairwise sums of the terms $a_i x^i$. For example let $n=3$, then
$$f(x)=\sum_{i=0}^3 a_i x^i=a_0+a_1 x+ a_2x^2+a_3x^3=(((a_0+a_1 x)+ a_2x^2)+a_3x^3)$$
the results of the sum of every paranthesis is a valid sum by the first condition and since $a_i x^i$ is a valid kernel as shown above. Hence the whole sum is also a valid kernel.
Note that if $a_0\neq 0$ one needs an extra condition that $$k(u,v)=k_1(u,v)+a_0\quad a_0\geq 0\quad\mbox{is a valid kernel}$$
