Little q-Jacobi polynomials

Remarks

Orthogonality: if $0 < aq < 1$ and $bq < 1$, then for $m \ne n$ $$\sum_{k=0}^{\infty} \frac{(bq; q)_k}{(q; q)_k} (aq)^k \, p_m(q^k) \, p_n(q^k) = 0,$$ and the weights $\frac{(bq; q)_k}{(q; q)_k} (aq)^k$ are positive.

Example: for $n = 1$ the sum has two terms, and $(1 - q^{-1}) q / (1 - q) = -1$ gives $$p_1(x) = 1 - \frac{1 - abq^2}{1 - aq} \, x.$$ Check of orthogonality to $p_0 = 1$ when $b = 0$: the weights are $w_k = (aq)^k / (q; q)_k$, and the $q$-binomial theorem gives $\sum_k w_k = 1/(aq; q)_\infty$ and $\sum_k w_k q^k = 1/(aq^2; q)_\infty$. So $\sum_k w_k p_1(q^k) = \frac{1}{(aq; q)_\infty} - \frac{1}{(1 - aq)(aq^2; q)_\infty} = 0$, since $(aq; q)_\infty = (1 - aq)(aq^2; q)_\infty$.

Non-example: for $b = 0$ the monomial $x$ is not the little $q$-Jacobi polynomial of degree $1$, since $x$ is not orthogonal to $p_0 = 1$ for these positive weights: $\sum_k w_k q^k > 0$.

Limit: $\lim_{q \to 1} p_n(x; q^\alpha, q^\beta \mid q) = P_n^{(\alpha, \beta)}(1 - 2x) / P_n^{(\alpha, \beta)}(1)$, where $P_n^{(\alpha,\beta)}$ is the Jacobi polynomial.