# Inner product of orthonormal basis vectors and their derivatives.

Let $$w$$ be a positive continuous function and let $$n$$ be a nonnegative integer. Equip $$\mathcal{P_n}(\mathbb{R})$$ with the inner product $$\langle p, q \rangle = \int_{0}^{1}p(x)q(x)w(x)dx.$$ Let $$p_0, p_1, ..., p_n$$ be an orthonormal basis for $$\mathcal{P}_n(\mathbb{R})$$ where each $$deg(p_k) = k$$. Show that $$\langle p_k, p_k' \rangle = 0$$ for each $$k$$, where $$p_k'$$ is the derivative.

I don't know where to begin with this. I was thinking of proving it arithmetically using the general formula of $$p_k$$ and $$p_k'$$ from Gram-Schmidt, but I was hoping that there is a more elegant solution.

EDIT: Added bottom explanation.

• You mean $\color{red}{p_0},p_1,\dots,p_n$? Aug 4, 2020 at 22:44
• @Azif00 Thanks for pointing that out, fixed it. Aug 4, 2020 at 22:46

Hint: It suffices to observe that $$\deg (p'_k) = k-1$$. Thus $$p'_k$$ is a linear combination of $$p_0,\ldots,p_{k-1}$$.
Note that the special definition of the inner product is completely irrelevant. If $$\langle -, - \rangle$$ is any inner product on $$\mathcal{P_n}(\mathbb{R})$$, you can apply the Gram–Schmidt process to the basis $$\{1, x, x^2,\ldots, x^n\}$$ and obtain an orthonormal basis $$\{p_0, p_1, p_2, \ldots , p_n\}$$ such that $$\operatorname{span} (p_0,\dots,p_i) = \operatorname{span} (1,\dots,x^i)$$ for $$i = 0,\ldots, n$$. This implies that $$\deg(p^i) = i$$.