Consecutive Gaussian orthogonal polynomials and Jacobi truncations strictly interlace

The reduced-order uses in [F1] are valid: the original support has at least $m+1\ge k+1$ points and the absolute moments through $2k-1$ are among those assumed. By [F2], every $p_k$ has the stated simple roots and monic factorization. If $m=1$, there is no pair to prove. For the remaining case $m\ge 2$, take $j=1$; if $x_1$ is the root of $p_1$, [F3] gives $p_2(x_1)=-{\beta }_1<0$. At both $a$ and $b$, the degree-two monic factorization [F2] gives $p_2(a)>0$ and $p_2(b)>0$. By [F4] there is one root on each side of $x_1$ inside $(a,b)$, proving the first pair when $m\ge 2$.

For $2\le j<m$, assume the roots of $p_{j-1}$ strictly interlace those $x_1<\cdots <x_j$ of $p_j$. At $x_i$, exactly $j-i$ roots of the monic polynomial $p_{j-1}$ lie to its right and none equals it. Its factorization [F2] therefore makes its sign $(-1{)}^{j-i}$. By [F3], the sign of $p_{j+1}(x_i)=-{\beta }_j{p}_{j-1}(x_i)$ is $(-1{)}^{j-i+1}$. These signs alternate, so [F4] supplies a root in each $(x_i,x_{i+1})$. Also [F2] gives sign $(-1{)}^{j+1}$ for $p_{j+1}(a)$ and positive sign for $p_{j+1}(b)$; these are opposite to the signs $(-1{)}^j$ at $x_1$ and $-1$ at $x_j$. Thus [F4] supplies an additional root in each of $(a,x_1)$ and $(x_j,b)$.

These $j+1$ disjoint intervals contain $j+1$ distinct roots, and [F2] says there are exactly that many. They therefore exhaust the roots of $p_{j+1}$ in the required order. Starting with step 1.1, finite induction through step 1.2 proves every pair through $p_m$. By [F6], the same ordered roots are the eigenvalues of the leading principal truncations, since their recurrence coefficients are identical prefixes. This proves the spectral assertion as well; when $m=1$ the quantified pair set is empty.