12
$\begingroup$

There are plenty of examples of inner products on special sequences of polynomials such that they are orthogonal. I can't quite wrap my head around the inner product s.t. monomials are orthogonal. Say we have polynomials defined on the unit interval $[0, 1]$. I can define an inner product by stating: $$\langle x^m, x^n\rangle = \delta_{mn}$$ This then extends via linearity to a full inner product on the set of all polynomials on $[0,1]$.

I can't see how this inner product can be represented with respect to Lesbesgue measure however. If there was an $h$ s.t. $$\langle f, g\rangle = \int_0^1 f(x)g(x)h(x)dx$$ then $$\langle x^m, x^n\rangle = \int_0^1 x^mx^nh(x)dx = \langle x^{m+n}, 1\rangle$$ which can't satisfy the orthogonality requirements.

My question then is, does there exist a measure (maybe a discrete one) where this inner product has a representation wrt? (or even just a formula of some kind to make it less abstract).

$\endgroup$
2
  • 3
    $\begingroup$ I think that such conclusions are independent on the measure: don't you think? If you substitute $dx$ with any other measure $d \mu$, the same argument applies, so there is no hope in finding a suitable measure. $\endgroup$
    – Crostul
    Jun 4, 2015 at 14:01
  • $\begingroup$ Have you noticed that this inner product is just the dot product relative to the monomial basis? $\endgroup$
    – user14972
    Jun 4, 2015 at 21:08

3 Answers 3

22
$\begingroup$

If such a measure existed, then by your argument it follows that $$0 = \langle x^3, x^1 \rangle = \langle x^4,1 \rangle = \langle x^2 , x^2 \rangle = 1$$ which is a contradiction. Therefore such a measure can not exist.

$\endgroup$
10
$\begingroup$

Such a measure does not exist. Polynomials that are orthogonal with regards to a positive measure (you need a positive measure to get an inner product), must have simple roots inside the support of the measure (see for example this thread).

However, there's a simple formula for your inner product : if $P = \sum_{n \ge 0} a_n X^n$ and $Q = \sum_{n \ge 0} b_n X^x$, then

$$\left\langle P,Q \right\rangle = \sum_{n \ge 0} a_n b_n$$

$\endgroup$
2
$\begingroup$

This is not of the requested form, but you can come up with a simple integral expression which takes the dot product between coefficents in the power series. If you consider $e^{2\pi i x}$, it is clear that: $$\int_{0}^1 (e^{-2\pi i x})^n (e^{2\pi i x})^m dx = \delta_{n,m}$$ By a simple u substitution, $$\frac{1}{2\pi i}\int_{C} (1/x)^n x^m \frac{dx}{x} = \frac{1}{2\pi i}\int_{C} (x^n)^* x^m \frac{dx}{x} = \delta_{n,m}$$ where the contour C is the unit circle in the complex plane. Therefore, for $f(x) = \sum_n a_n x^{n}$ and $g(x) = \sum_n b_n x^{m}$ where $a_n$ and $b_n$ are $l^2$ sequences: $$\frac{1}{2\pi i}\int_{C} (f(x))^* g(x) \frac{dx}{x} = \sum_{n=0}^{\infty} a_n^* b_n$$ Thus giving the usual inner product for $l^2$ sequences.

$\endgroup$
1
  • $\begingroup$ btw, if f has real coefficients this integral can still be evaluated using the Residue Theorem and replacing $(f(x))^*$ with $f(1/x)$. $\endgroup$
    – Henry Hunt
    May 12, 2023 at 7:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .