I was wondering if, given a natural number $i\in \mathbb N$, there exists an orthonormal basis (w.r.t. the standard scalar product) $(p_n)_{n \in \mathbb N}$ of $L^2([0,1])$ such that $p_n$ is a polynomial of degree at least $i$ for all $n\in \mathbb N$.
I know this is true if $i=0$ using kind of Legendre polynomials.
A way to answer this question could be to determine if it is possible to find an orthonormal basis of the $i+1$-dimensional vector space of polynomials of degree less than or equal to $i$, consisting of $i+1$ polynomials of degree $i$. In deed we could then complete this finite family with $\{ x^{i+1}, x^{i+2}, \dots \} $ and then use a Gram–Schmidt procedure to orthogonalize the added monomials.
Does anyone have an answer to those questions?
Many thanks for your help.
There is a theorem called Muntz-Szasz theorem I believe, which says that polynomials in ${1, t^{n_1}, t^{n_2}, ...}$ form a dense set in $C[0,1]$ (so in particular in $L^2[0,1])$ if $\sum \frac{1}{t^{n_k}}$ diverges.
Clearly, if one removes finite many natural numbers (i.e. powers of $t$) then the remaining series of reciprocals diverges.
In words, Gram-Schmidt produces linear combinations of these polynomials and we already know that any polynomial can be approximated by (different but still linear) combination of these. This would show that if a polynomial is orthogonal to Gram-Schmidt combinations it is zero.