Orthogonal polynomials

The monomials lie in $L^{2}_{w}$ and are independent. $\lVert x^{k}\rVert_{w}^{2}=\mu_{2k}<\infty$ by the moment hypothesis, so each monomial is in the space. A nontrivial linear combination of monomials is a nonzero polynomial $q$, and

because $\lvert q\rvert^{2}w$ is continuous, non-negative, strictly positive off the finitely many zeros of $q$ — this is where $w>0$ enters. So $q\neq 0$ in $L^{2}_{w}$ and the monomials are linearly independent.

Gram-Schmidt. Run the process of Orthonormal bases and Gram-Schmidt on $1,x,x^{2},\dots$. At step $n$ it subtracts from $x^{n}$ its projection onto the span of the previous outputs and divides by a norm that we have just shown is nonzero. Each output is a linear combination of $1,\dots,x^{n}$, hence a polynomial of degree $\le n$; and the coefficient of $x^{n}$ is the (nonzero) reciprocal of that norm times $1$, because the subtracted projection involves only $1,\dots,x^{n-1}$. So $\deg p_{n}=n$ exactly. This is the construction Kreyszig carries out on $[-1,1]$ with $w\equiv1$ to produce the Legendre polynomials, and Hassani's remark that orthogonalising the monomials must give polynomials.

Orthogonality to lower degrees. By construction $\operatorname{span}\{p_{0},\dots,p_{n-1}\}=\operatorname{span}\{1,\dots,x^{n-1}\}=:\mathcal{P}_{n-1}$, the polynomials of degree $\le n-1$. Since $p_{n}\perp p_{m}$ for $m<n$, $p_{n}\perp\mathcal{P}_{n-1}$. This is part 3.

Uniqueness. Let $\tilde p_{n}$ be any polynomial of degree $n$ orthogonal to $\mathcal{P}_{n-1}$. Write $\tilde p_{n}=\sum_{j=0}^{n}c_{j}p_{j}$, possible because $\deg\tilde p_{n}\le n$. For $m<n$, orthogonality to $p_{m}\in\mathcal{P}_{n-1}$ gives $0=(\tilde p_{n},p_{m})_{w}=c_{m}h_{m}$, and $h_{m}>0$, so $c_{m}=0$. Hence $\tilde p_{n}=c_{n}p_{n}$, with $c_{n}\neq0$ since $\deg\tilde p_{n}=n$. Any second orthogonal family satisfies the same hypothesis at each $n$, which proves part 2. $\square$

Step 1: kill the leading term. The polynomial

has no $x^{n+1}$: the coefficient of $x^{n+1}$ in $xp_{n}$ is $k_{n}^{(n)}$, and $\alpha_{n}$ was chosen to match. So $\deg r\le n$ and $r=\sum_{j=0}^{n}a_{j}p_{j}$ for some scalars $a_{j}$.

Step 2: test against $p_{m}$ for $m\le n-2$. Take $(\,\cdot\,,p_{m})_{w}$ of $r=p_{n+1}-\alpha_{n}xp_{n}$:

The first term vanishes by orthogonality since $m\le n-2<n+1$. In the second we moved the real factor $x$ across the inner product, which is legitimate because $x$ is real and the inner product has $\overline{x\,p_{m}}=x\,\overline{p_{m}}$. Now $\deg(xp_{m})=m+1\le n-1<n$, so $(p_{n},xp_{m})_{w}=0$ by part 3 of the previous statement. Hence $(r,p_{m})_{w}=0$, while also $(r,p_{m})_{w}=a_{m}h_{m}$ from the expansion. Since $h_{m}>0$,

which is exactly where the hypothesis "orthogonal to all lower degrees" does the work. Only $a_{n}$ and $a_{n-1}$ survive, giving $(\ast)$ with $\beta_{n}=a_{n}$ and $\gamma_{n}=a_{n-1}$.

Step 3: identify $\beta_{n}$. Compare coefficients of $x^{n}$ in $p_{n+1}=\alpha_{n}xp_{n}+\beta_{n}p_{n}+\gamma_{n}p_{n-1}$. The last term contributes nothing ($\deg p_{n-1}=n-1$), so

and solving, with $\alpha_{n}k_{n}^{(n)}=k_{n+1}^{(n+1)}$,

Hassani's formula.

Step 4: identify $\gamma_{n}$. Take $(\,\cdot\,,p_{n-1})_{w}$ of $(\ast)$. The left side gives $0$; on the right, $(p_{n},p_{n-1})_{w}=0$, so

To evaluate $(p_{n},xp_{n-1})_{w}$, apply $(\ast)$ at index $n-1$: $xp_{n-1}=\alpha_{n-1}^{-1}\bigl(p_{n}-\beta_{n-1}p_{n-1}-\gamma_{n-1}p_{n-2}\bigr)$, and only the $p_{n}$ term survives against $p_{n}$:

Therefore $\gamma_{n}=-\alpha_{n}h_{n}/(\alpha_{n-1}h_{n-1})$, as claimed. Since $h_{n},h_{n-1}>0$, the sign of $\gamma_{n}$ is opposite to that of $\alpha_{n}/\alpha_{n-1}$.

Step 5: the monic form. Put $\alpha_{n}=1$ for all $n$, write $B_{n}=-\beta_{n}$ and $C_{n}=-\gamma_{n}=h_{n}/h_{n-1}>0$; then $(\ast)$ becomes $(\ast\ast)$. For $B_{n}$ the coefficient formula is awkward to use directly, so take $(\,\cdot\,,p_{n})_{w}$ of $(\ast\ast)$ instead:

giving $B_{n}=(xp_{n},p_{n})_{w}/h_{n}$. For $n=0$ the same computation with $p_{0}=1$ gives $B_{0}=(x,1)_{w}/(1,1)_{w}=\mu_{1}/\mu_{0}$. $\square$

We prove Hassani's route; his Lemma 8.1.2 is left to a problem in the book, so we prove it too. Write $p_{\le k}$ for an unspecified polynomial of degree at most $k$, as Hassani does, and let $F_{n}=(K_{n}w)^{-1}\mathrm{d}^{n}(ws^{n})/\mathrm{d}x^{n}$. The hypotheses are: $\deg s\le2$ with real roots only; $w>0$ and integrable on $(a,b)$ with $w(a)s(a)=0=w(b)s(b)$; and $F_{1}$ a polynomial of degree $1$. The last one says precisely that

Lemma A (Hassani's 8.1.2). For $1\le m\le n$,

Proof, by induction on $m$. For $m=1$, expand and use $w'=w(t-s')/s$, which is (H) rearranged (valid where $s\neq0$, and the conclusion is an identity between continuous functions, so it extends to the whole interval):

In the bracket, $t$ and $s'$ have degree $\le1$, so the first product has degree $\le k+1$; and $s\,p_{\le k}'$ has degree $\le 2+(k-1)=k+1$. The bracket is therefore $p_{\le k+1}$, which is the claim for $m=1$. If the claim holds for $m<n$, apply the case $m=1$ once more with $n-m$ in place of $n$ and $k+m$ in place of $k$; the power of $s$ is still at least $1$, so the step is legitimate, and we get $ws^{\,n-m-1}p_{\le k+m+1}$. $\square$

Lemma B (Hassani's 8.1.3). For $0\le m<n$, $\mathrm{d}^{m}(ws^{n})/\mathrm{d}x^{m}$ vanishes at $x=a$ and at $x=b$. Proof. Take $k=0$, $p_{\le0}=1$ in Lemma A: the derivative equals $w\,s^{\,n-m}\,p_{\le m}$. Since $m<n$ we have $n-m\ge1$, so this is $(ws)\,s^{\,n-m-1}p_{\le m}$, and $ws$ vanishes at both endpoints by hypothesis. $\square$

$F_{n}$ is a polynomial of degree at most $n$. Lemma A with $m=n$, $k=0$ gives $\mathrm{d}^{n}(ws^{n})/\mathrm{d}x^{n}=w\,s^{0}\,p_{\le n}=w\,p_{\le n}$, so $F_{n}=p_{\le n}/K_{n}$ is a polynomial of degree at most $n$.

Orthogonality to lower degrees. Let $q$ be a polynomial with $\deg q=k<n$. Then

Integrate by parts, moving one derivative at a time onto $q$. After $j$ steps the boundary term is

which vanishes by Lemma B because $n-j<n$ for $j\ge1$. Carrying this out $k$ times leaves

where $C=q^{(k)}$ is the constant $k$th derivative of a degree-$k$ polynomial, and where $n-k-1\ge0$ because $k<n$. The last integral is $\bigl[\mathrm{d}^{\,n-k-1}(ws^{n})/\mathrm{d}x^{\,n-k-1}\bigr]_{a}^{b}=0$, again by Lemma B. So $\int_{a}^{b}qF_{n}w=0$.

The degree is exactly $n$. Write $F_{n}=p_{\le n-1}+k_{n}^{(n)}x^{n}$, multiply by $wF_{n}$ and integrate:

The left side is strictly positive, because $w>0$ and $F_{n}$ is a nonzero polynomial (it is nonzero: if $F_{n}\equiv0$ then $\mathrm{d}^{n}(ws^{n})/\mathrm{d}x^{n}\equiv0$, making $ws^{n}$ a polynomial of degree $<n$, which contradicts Lemma A at $m=n-1$ unless $w$ vanishes somewhere). Hence the right side is nonzero, so $k_{n}^{(n)}\neq0$ and $\deg F_{n}=n$.

Combining the last two steps: $(F_{n})$ has $\deg F_{n}=n$ and is orthogonal to all lower degrees, so by the uniqueness statement proved above it is the orthogonal family for $w$, standardised by the choice of $K_{n}$. The differential equation of Proposition 8.1.4, the equivalence with Arfken's route and the classification into three cases are reported from Hassani and from Arfken, Weber and Harris as cited; we have not reproved them. $\square$