A sequence $(x_n)_{n \in \mathbb{N}}$ is said to be Frame, if there exist constants $A,B > 0$ such that $$ A \lVert x \rVert^2 \leq \sum_{n=0}^{\infty} \lvert (x,x_n) \rvert^2 \leq B \lVert x \rVert^2$$ for all $x \in H$.
Consider the Hilbertspace $L^2[0,1]$ endowed with the Lebesgue measure and the sequence of monomials, i.e. the functions $x_n(t) = t^n, t \in [0,1], n \in \mathbb{N}$. Using Hilbert's inequality one can verify that $(x_n)_{n \in \mathbb{N}}$ fulfills $$\sum_{n=0}^{\infty} \lvert (f,x_n)\rvert^2 \leq B \lVert f \rVert^2$$ for all $f \in L^2[0,1]$. Furthermore it is a consequence of Lusin's theorem that the span of $(x_n)_{n \in \mathbb{N}}$ is dense in $L^2[0,1]$. My question is if $(x_n)_{n \in \mathbb{N}}$ is a frame, that is it fulfills the lower frame bound. Since I have no idea from what I could infer a lower bound, my first thought was that it is not a frame. Since a frame provides expansions of the form $$ f = \sum_{n=0}^{\infty}a_n x^n$$ with complex coefficients $a_n$ where the series converges unconditionally and hence also in $L^2$-norm, I'm trying to find a $f \in L^2[0,1]$ which does not have a power series expansion of this form. Is this true or is $(x_n)_{n \in \mathbb{N}}$ a frame after all (or something inbetween, since the latter statement, if false, does not imply that $(x_n)_{n \in \mathbb{N}}$ is a frame).
I expect the calculations below are correct, although that I did not check all details. I am going do this later.
For each $\varepsilon\in (0,1/2)$ and $x\in [0,1]$ put $$f_\varepsilon(x)= \cases{ 1, & when $x\in [0,\varepsilon)$,\\ -1, & when $x\in [1,2\varepsilon)$,\\ 0, & otherwise. } $$
Then $(f,f)=2\varepsilon$. For each nonnegative integer $n$, we have $$(f,x_n)=\int_{0}^{\varepsilon} t^n dt-\int_{\varepsilon}^{2\varepsilon} t^n dt=\frac{t^{n+1}}{n+1}{\Huge|}^\varepsilon_0-\frac{t^{n+1}}{n+1}{\Huge|}^{2\varepsilon}_\varepsilon=$$ $$ \frac{1}{n+1}\left(2\varepsilon^{n+1}-(2\varepsilon)^{n+1}\right)=\frac{2\varepsilon^{n+1}(1-2^n)}{n+1}.$$ Thus $|(f,x_0)|=0$ and $$|(f,x_n)|\le\frac{(2\varepsilon)^{n+1}}{n+1}\le\frac{4\varepsilon^2}{n+1} $$ for each positive integer $n$. Then $$\sum_{n=0}^{\infty} |(f,x_n)|^2\le \sum_{n=1}^{\infty} \frac{16\varepsilon^4}{(n+1)^2}=16\varepsilon^4\left(\frac{\pi^2}6-1\right).$$
So there is no $A>0$ such that $ A \lVert f_\varepsilon \rVert^2 \leq \sum_{n=0}^{\infty} \lvert (f_\varepsilon,x_n)\rvert$ for all $\varepsilon>0$.