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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.

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  • $\begingroup$ You mean $\color{red}{p_0},p_1,\dots,p_n$? $\endgroup$
    – azif00
    Aug 4, 2020 at 22:44
  • $\begingroup$ @Azif00 Thanks for pointing that out, fixed it. $\endgroup$ Aug 4, 2020 at 22:46

1 Answer 1

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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 $\deg(p^i) = i$.

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  • $\begingroup$ Interesting, thanks for pointing that out! $\endgroup$ Aug 4, 2020 at 22:53

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