0

In this answer a simple ad-hoc inner product for monomials is defined, since they are known to be orthogonal:

if $P = \sum_{n \ge 0} a_n X^n$ and $Q = \sum_{n \ge 0} b_n X^n$, then $$\left\langle P,Q \right\rangle = \sum_{n \ge 0} a_n b_n$$

My question is, is it legitimate to generalize the above to cases where the leading terms in the polynomials are not integer? For example:

if $P = X^\alpha\sum_{n \ge 0} a_n X^n$ and $Q = X^\beta\sum_{n \ge 0} b_n X^n$ with $\alpha,\beta\in\mathbb{C}$ , then $$\left\langle P,Q \right\rangle = \delta(\alpha-\beta)\sum_{n \ge 0} a_n b_n$$ Does this hold true? And if yes, how to prove that it does / where to read up on the proof?

Kagaratsch
  • 2,239
  • 1
    Your definition does not seem to be well-defined to me. What's the inner product of $P = X^{\pi}$ and $Q = X^{\pi - 1} (X)$? In any case, I imagine the thing you intended to write down is to make $X^{\alpha}$ an orthonormal basis for some arbitrary collection of exponents $\alpha$, which works fine but I don't know of any applications of doing this. – Qiaochu Yuan May 11 '18 at 01:10
  • @QiaochuYuan What if we define $X>0$, then on the principal branch $X^{\pi-1}X=X^\pi$, so that $P=Q$ in your example. However, I would like to have $\langle X^{2\pi},X^{\pi}\rangle=0$. You are saying that for a collection of exponents $\alpha$ one could define such an orthonormal basis. How would one make this rigorous? – Kagaratsch May 11 '18 at 01:20
  • 1
    my point is that according to your definition we would have $\langle P, Q \rangle = 0$ because $\pi \neq \pi-1$, despite the fact that $P = Q$; the problem is that the representation you use is not unique and the "inner product" depends on the choice of representation, so it's not well-defined. In general, if $V$ is any real or complex vector space and $B$ any basis of it (in your example, $V$ is polynomials with whatever exponents you want, and $B$ is polynomials with those exponents), then there's a unique inner product on $V$ with respect to which $B$ is orthonormal. – Qiaochu Yuan May 11 '18 at 02:40
  • @QiaochuYuan I see, thank you for pointing this out! – Kagaratsch May 11 '18 at 12:53

0 Answers0