Is there a representation of an inner product where monomials are orthogonal?

1.9k Views Asked by At

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).

2

There are 2 best solutions below

0
On BEST ANSWER

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

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

0
On

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.