Properties of semi inner product space by integration against weight function Given an interval $[a,b]$, and a Lebesgue integrable non-negative function $\omega \in L^1[a,b]$, I am considering the semi inner product on $C[a,b]$ by:
$$ \langle f,g \rangle_\omega:= \int_a^b f(t)\overline{g}(t)\cdot \omega(t)dt $$
And I want to consider the "Hilbert" space $L^2_\omega[a,b]$ with respect to this semi inner product, and obtain a Fourier series in that space using a complete orthonormal system for a continuous function on $[a,b]$. But as I saw in this thread, $L_\omega^2[a,b]$ is not necessarily contained in $L^2[a,b]$. But I was wondering whether these following properties that I concluded to myself are indeed true:


*

*If $\omega>0$ almost everywhere on $[a,b]$, then this is an inner product on 
$C[a,b]$.

*For every $f\in C[a,b]$, we have that $\langle f,f\rangle_\omega\leq \Vert f\Vert_\infty\cdot \Vert \omega \Vert_{L^1[a,b]}$. In particular $C[a,b]\subseteq L^2_\omega[a,b]$.

*Assume $f$ is analytic in $[a,b]$, i.e $f(x)=\sum\limits_{n=0}^\infty a_n x^n$ for all $x\in [a,b]$. If we assume that $\{T_n \}_{n=0}^\infty  $ is an orthonormal basis of $L^2_\omega[a,b]$ where $T_n$ is a polynomial of degree $n$, we can write $f(x)=\sum\limits_{n=0}^\infty \alpha_n T_n(x)$. Assume that for all $n\in \mathbb{N}\cup\{0\}$, we have:
$$ x^n=\sum\limits_{m=0}^n c_{n,m} \cdot T_m(x) $$
Then for all $m$, we have:
$$ \alpha_m=\sum\limits_{n=m}^\infty a_n\cdot c_{n,m} $$

I think (2) follows immediatly from the triangle inequality for integrals. The fact that this is an inner product in (1) follows that the measure of any non-degenerate interval  would be positive, and a continuous non-zero function would be bounded from below on an interval. I want to say that (3) follows from uniqueness of the coefficients for a power series, but have not been able to formalize it in my head. This follows from the formal manipulation:
$$ \sum\limits_{n=0}^\infty a_n x^n= \sum\limits_{n=0}^\infty a_n \Bigg( \sum\limits_{m=0}^n c_{m,n} T_m(x) \Bigg)= $$
$$ = \sum\limits_{m=0}^\infty \Bigg( \sum\limits_{n=m}^\infty a_n \cdot c_{n,m} \Bigg) T_m(x) $$
I would appreciate input on whether any of this reasoning is wrong, and how one can prove (3) formally.
 A: You should first understand what $f(x) = \sum_{n=0}^\infty \alpha_n T_n(x)$ means. It does not mean that for all $x$, $f(x) = \lim_{N \to \infty} \sum_{n=0}^N \alpha_n T_n(x)$. Rather, it means that the functions $\sum_{n=0}^N \alpha_n T_n$ converge in $L^2$ to $f$ (since that's what it means for $(T_n)_n$ to be a basis for $L^2$). On the other hand, when we say $f(x) = \sum_{n=0}^\infty a_n x^n$, we do mean that for each $x$, $f(x) = \lim_{N \to \infty} \sum_{n=0}^N a_n x^n$. But it's an easy exercise in power series to show that $\sum_{n=0}^N a_n x^n$, as a (sequence of) function(s), converges in $L^2$ to the function $f$. 
With that out of the way, (3) is not too bad. Simply do $$\alpha_m = \lim_{N \to \infty} \sum_{n=0}^N \langle \alpha_n T_n, T_m \rangle = \langle \sum_{n=0}^\infty \alpha_n T_n, T_m \rangle = \langle f, T_m \rangle = \langle \sum_{n=0}^\infty a_n x^n , T_m \rangle = \lim_{N \to \infty} \sum_{n=0}^N a_n \langle x^n, T_m \rangle = \lim_{N \to \infty} \sum_{n=0}^N a_n \langle \sum_{k=0}^n c_{n,k} T_k, T_m \rangle = \sum_{n=m}^\infty a_n c_{n,m},$$ where the second and fifth equalities follow from the fact that convergence in $L^2$ implies continuity in the inner product.
