sos. For positive definiteness, simply note that as $p(x)>p(A_jx)$ for any $x \neq 0$ and $p$ is nonnegative, we get $p^k(x)>p^k(A_jx)$ for any $k\geq 1$ and $x\neq 0.$ To show that $q(x)-q(A_jx)$ is sos, we make use of the following identity: $$\begin{aligned}
\label{eq:identity}
    a^k-b^k=(a-b)\sum_{l=0}^{k-1} a^{k-1-l}b^l.
    \end{aligned}$$ Applying (\[eq:identity\]), we have $$\begin{aligned}
    &p^k(x)-\left(p(A_jx)+\alpha_j (\sum_i x_i^2)^d\right)^k \nonumber \\ &=\left(p(x)-p(A_jx)-\alpha_j (\sum_i x_i^2)^d\right)\cdot \sum_{l=0}^{k-1}p(x)^{k-1-l}\left(p(A_jx)+\alpha_j (\sum_i x_i^2)^d\right)^l\nonumber\\
    &=\sum_{l=0}^{k-1}\left(p(x)-p(A_jx)-\alpha_j (\sum_i x_i^2)^d\right)\cdot p(x)^{k-1-l}\left(p(A_jx)+\alpha_j (\sum_i x_i^2)^d\right)^l. \label{eq:sum}
    \end{aligned}$$ For any $l \in \{0,\ldots,k-1\}$, either $k-1-l\geq \frac{k-1}{2}$ or $l \geq \frac{k-1}{2}$. Suppose that the index $l$ is such that $k-1-l \geq \frac{k-1}{2}$: by definition of $k$, this implies that $k-1-l\geq K$. Since the polynomial in (\[eq:claus.p\]) sos and since $(p(A_jx)+\alpha_j (\sum_i x_i^2)^d)^l$ is sos, we get that the term $$\left(p(x)-p(A_jx)-\alpha_j (\sum_i x_i^2)^d\right)\cdot p(x)^{k-1-l}\left(p(A_jx)+\alpha_j (\sum_i x_i^2)^d\right)^l$$ in the sum (\[eq:sum\]) is sos. Similarly, if the index $l$ is such that $l \geq \frac{k-1}{2}$, we have that $l \geq K'$ by definition of $k.$ Sine the polynomial in (\[eq:claus.pAj\]) is sos, we come to the conclusion that the term $$\left(p(x)-p(A_jx)-\alpha_j (\sum_i x_i^2)^d\right)\cdot p(x)^{k-1-l}\left(p(A_jx)+\alpha_j (\sum_i x_i^2)^d\right)^l$$ in the sum (\[eq:sum\]) is sos. When summing over all possible $l \in \{0,\ldots,k-1\}$, as each term in the sum is sos, we conclude that the 