Projection of vector on subspaces in a Hilbert space This may be a vague title and I think that this question must have an easy answer.
 Let $\mathcal{H}$ be a  weighted $\ell^2$ space of complex
sequence $\{x(n)\}_{n \geqslant 1}$ such that $$\|x\|^2 = \sum
_{n=1}^\infty \frac{|x(n)|^2}{n^2} < \infty .$$ It has the inner
product defined by:
$$\langle x, y \rangle=\sum_{n=1}^\infty \frac{x(n) \overline{y(n)}}{n^2}.$$Let, $E=\{e_n:{n\ge 1}\}$ be a set of vectors and $\mathcal{M}=span(E)$. Also suppose that, $E_N=\{e_n:1\le n\le N\}$ and $\mathcal{M}_N=span(E_N)$.
Now let $P_N(x)$ and $P(x)$ be projection of $x\in \mathcal{H}$ on $\mathcal{M}_N$ and $\overline{\mathcal{M}}$ respectively. My questions are:


*

*$\lim\limits_{N\to\infty}||P_N(x)-P(x)||=0 \ \forall x\in\mathcal{H}$, this is a true statement (isn't it?).

*We can write $P_N(x)=\sum_{n=1}^Nf_N(n,x)e_n$, can we write $P(x)=\sum_{n=1}^\infty f(n,x)e_n \ \forall x\in\mathcal{H}$? If not. which stronger condition do we essentially need to have this property? (Note: If $\{e_n\}$ is a schauder basis then we will have this.)

*If $(2)$ holds, then does $f_N(n,x)$ converge to $f(n,x)$ uniformly in $n$?


All of the above seem to be true to me, but I am being unable to come up with a rigorous proof to see my intuition. Any help/reference will be highly appreciated.
 A: *

*Is correct. For proof, you need the basic fact about Hilbert spaces, that given a vector not in a closed subspace, there is a unique vector in the subspace closest to that other point. This property fails in general in Banach spaces, where there may fail to be any closest point, or there may be infinitely-many. From this, one manufactures orthogonal projections.


1b. Probably to prove your limit assertion you'd want to choose a family of orthonormal bases of your subspaces.
For 2. Yes, the finite part of your property here is just a restatement in coordinates of 1. For the this to be well-defined, probably you want $e_1,\ldots,e_n$ linearly independent for all $n$? But, even then, if these aren't a Hilbert-space basis, infinite linear combinations may be zero even if no finite one is, which indicates that the infinite expression only makes unambiguous sense for the $e_n$'s a Hilbert space basis of the closure of subspace they span algebraically.
For 3. Revision: (my earlier worried reaction seems too strong, although not entirely unfounded) Thus, assume $v_1, \ldots,v_n,\ldots$ are a Hilbert space basis, though not necessarily orthogonal. That is, the closure of the algebraic span is the whole Hilbert space, and no individual $v_i$ is in the closure of the span of the others. Then we can make a dual basis $\lambda_n$, that is, such that $\lambda_i(v_j)=\delta_{ij}$ (Kronecker delta). Namely, let $\lambda_j$ be the orthogonal projection of $v_j$ to the orthogonal complement of the span of all $v_i$'s except $v_j$. Adjust $\lambda_j$ by a scalar so that $\langle \lambda_j,v_j\rangle=1$. Then $v=\lim_N \sum_{n\le N} \lambda_n(v)\cdot v_n$.
These $\lambda_n$'s are continuous. There is no "revision" of coefficients, as the uniqueness of projections shows.
The potential hazard is in trying to estimate the size of the tails in the infinite sum, if one doesn't have orthogonality, but we seem to have skirted that issue here.
