[proofplan] We show that $p_n$ has exactly $n$ sign changes in $(a,b)$. Orthogonality to constants forces at least one sign change. If there were fewer than $n$ sign changes, the polynomial $q = \prod(x - x_i)$ (formed from the sign-change points) would have degree less than $n$ and the product $p_n q$ would not change sign, contradicting $\langle p_n, q \rangle = 0$. Since $p_n$ has degree $n$ and hence at most $n$ roots, the number of sign changes is exactly $n$, and all roots are simple. [/proofplan] [step:Show $p_n$ must change sign in $(a,b)$ at least once] Since $p_n$ is a nonzero polynomial and $n \ge 1$, orthogonality gives $\langle p_n, 1 \rangle = 0$, i.e., $\int_a^b w(x)\,p_n(x) \, d\mathcal{L}^1(x) = 0$. Since $w > 0$ on $(a,b)$, $p_n$ must change sign in $(a,b)$, so the number of sign changes $r$ satisfies $r \ge 1$. [/step] [step:Prove $r \ge n$ by contradiction using orthogonality] Let $x_1, \dots, x_r$ be the points in $(a,b)$ where $p_n$ changes sign, counted without multiplicity. Form the polynomial $q(x) = \prod_{i=1}^r(x - x_i)$, which has degree $r$ and changes sign at exactly the same points as $p_n$ in $(a,b)$. Therefore $p_n(x)\,q(x)$ does not change sign on $(a,b)$ (both factors change sign at the same points). Since $w > 0$ and the integrand $w\,p_n\,q$ is continuous and not identically zero, $\int_a^b w(x)\,p_n(x)\,q(x) \, d\mathcal{L}^1(x) \neq 0$. If $r < n$, then $q \in P_{n-1}[x]$, and orthogonality gives $\langle p_n, q \rangle = 0$, a contradiction. Therefore $r \ge n$. [/step] [step:Conclude $r = n$ with all roots simple and in $(a,b)$] Since $p_n$ has degree $n$, it has at most $n$ real roots, so $r \le n$. Combined with $r \ge n$, we get $r = n$. All $n$ roots are simple (since they are sign changes, the polynomial crosses zero transversally) and lie in the open interval $(a,b)$. [/step]