TITLE: Is there a representation of an inner product where monomials are orthogonal?
QUESTION [12 upvotes]: 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).

REPLY [9 votes]: 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
$$\left\langle  P,Q  \right\rangle = \sum_{n \ge 0} a_n b_n$$