Let \(w\) be positive and continuous on \([0,1]\), let \(n\) be a positive integer, and equip the space \(P_n\) of real polynomials of degree at most \(n\) with \[ \langle p,q\rangle=\int_0^1 p(t)q(t)w(t)\,dt. \]
- Prove that \(P_n\) has an orthonormal basis \(p_0,p_1,\ldots,p_n\) with \(\deg p_k=k\).
- Prove that \[ \langle p_k,p_k'\rangle=0 \] for every \(k\).
Solution
1. The displayed bilinear form is an inner product on \(P_n\).
Symmetry and bilinearity are immediate. If \(p\neq0\), then \(p^2w\) is continuous and nonnegative on \([0,1]\). Since \(w(t)>0\) everywhere and a nonzero polynomial is nonzero at some point, \(p^2w\) is strictly positive on some interval. Hence \[ \langle p,p\rangle = \int_0^1p(t)^2w(t)\,dt >0. \] Thus the form is positive definite.
2. Applying Gram–Schmidt to\[ 1,t,t^2,\ldots,t^n \]produces orthonormal polynomials\[ p_0,p_1,\ldots,p_n \]with\[ \deg p_k=k \]for every \(k\).
At the \(k\)th Gram–Schmidt step, before normalization one obtains \[ q_k(t) = t^k -\sum_{j=0}^{k-1}\langle t^k,p_j\rangle p_j(t). \] Inductively, each \(p_j\) with \(j<k\) has degree \(j<k\). Therefore the subtracted sum has degree at most \(k-1\), so the coefficient of \(t^k\) in \(q_k\) remains \(1\). Hence \[ \deg q_k=k, \] in particular \(q_k\neq0\). Normalizing \(q_k\) multiplies it by a nonzero scalar and therefore preserves its degree. Gram–Schmidt gives orthonormality, so the resulting family is an orthonormal basis of the \((n+1)\)-dimensional space \(P_n\).
3. For every \(k\geq1\),\[ p_k'\in \operatorname{span}\{p_0,\ldots,p_{k-1}\}. \]
Step 2 gives \(\deg p_k=k\), so \[ \deg p_k'\leq k-1. \] Thus \(p_k'\in P_{k-1}\). Again by step 2, \[ p_0,\ldots,p_{k-1} \] are \(k\) linearly independent vectors in the \(k\)-dimensional space \(P_{k-1}\), hence form a basis of that space.
4. For every \(0\leq k\leq n\),\[ \langle p_k,p_k'\rangle=0. \]
For \(k=0\), \(p_0\) is constant, so \(p_0'=0\). For \(k\geq1\), step 3 writes \(p_k'\) as a linear combination of \(p_0,\ldots,p_{k-1}\). Since the family from step 2 is orthonormal, \(p_k\) is orthogonal to each of those polynomials. Hence it is orthogonal to their linear combination \(p_k'\).
5. Q.E.D.
Step 2 proves part 1, and step 4 proves part 2.