Show that $F: X \to \ell^{2}$ is surjective

Question

Let $X$ be a infinite dimensional Hilbert Space that is separable and has $\{e_{n}\}_{n \in \mathbb N}$ a countable Orthonormal Basis.

I have been given the map:

$F: X \to \ell^{2}, x\mapsto (\langle e_{n},x\rangle)_{n \in\mathbb N}$

I need to show that: $2.$ $F$ is surjective

Idea on $2$:

I am supposed to use the following hint:

if for $(c_{n})_{n\in \mathbb N}\subset\mathbb R$ we know that $\sum\limits_{n \in \mathbb N}\vert c_{n}\vert^{2}<\infty$ then $\sum\limits_{n\in\mathbb N}c_{n}e_{n}$ is well defined, where $\{e_{n}\}_{n \in \mathbb N}$ is an orthnormal set.

How is this supposed to help me?

Edit, to show surjectivity:

Let $y \in \ell^{2}$ arbitrary: We know that therefore $\sum\limits_{n \in \mathbb N}\vert y_{n}\vert^{2}<\infty$ from the tip this then implies that $\sum\limits_{n \in \mathbb N}y_{n}e_{n}$ is well-defined in $X$, in other words, there exists a $x\in X$ so that $x=\sum\limits_{n \in \mathbb N}y_{n}e_{n}$.

Now by definition of $F$, it follows that $Fx=(\langle e_{n},x\rangle)_{n \in\mathbb N}=(\langle e_{n},\sum\limits_{n \in \mathbb N}y_{n}e_{n}\rangle)_{n \in\mathbb N}=(y_{n})_{n \in \mathbb N}=y$. But now I have a problem: what there if there is another $z \in \ell^{2}$ which is unequal to $y$ so that $x =\sum\limits_{n \in \mathbb N}z_{n}e_{n}$, then I'd have $Fx=y\neq z =Fx$. How does well-definedness alleviate this problem? Because well-definedness of $\sum\limits_{n \in \mathbb N}y_{n}e_{n}$ according to my understanding leads only to the fact that there cannot be two different $x,w$ where $x \neq w$ so that $x = \sum\limits_{n \in \mathbb N}y_{n}e_{n}$ and $w = \sum\limits_{n \in \mathbb N}y_{n}e_{n}$. Is my understanding of well-definedness correct?

$1.$ Question: What does well-defined in $X$ mean in this regard, does it mean there can be only one $x\in X$ so that $x=\sum\limits_{n \in \mathbb N}y_{n}e_{n}$?

Answer 1

To answer the surjectivity, I would argue in the following way:

If we would assume that $\exists c:= (c_n)_n \in {\ell}^2: c \notin \operatorname{ran}(F)$, then $\sum_{n \in \Bbb N} c_n e_n$ is well-defined by your hint, but $\sum_{n \in \Bbb N} c_n e_n \notin \overline{\operatorname{span}\{e_n\}}$ because otherwise $\sum_{n \in \Bbb N} c_n e_n = \sum_{n \in \Bbb N} \langle x, e_n \rangle e_n $ for some $x \in X$ which of course leads to a contradiction because $\{e_n\}$ is a ONB.

Answer 2

Suppose $\sum |c_n|^{2} <\infty$. Let $s_n=\sum\limits_{k=1}^{n} c_ne_n$. Then $\|s_n-s_m\|^{2}=\|\sum\limits_{k=n}^{m} c_ne_n\|^{2}$ for $m >n$. Exapnd this using inner products and use orthonormality to see that $\|s_n-s_m\|^{2}=\sum\limits_{k=n}^{m} |c_n|^{2}$. This lst quantity tends to $0$ as $n,m \to \infty$. Hnece $\{s_n\}$ is a Cauchy sequence. By completeness the infinite sum $\sum c_ne_n =\lim s_n$ exists. Also $\langle e_k, \lim s_n \rangle =c_k$ for all $k$ so $F(lim s_n)=\{c_n\}$.

Answer 3

For $n,m\in\mathbb{N}$ clearly $\langle F(e_n), F(e_m)\rangle_{\ell^2}=\delta_{n,m}=\langle e_n, e_m \rangle$ By the linearity and continuity of the inner product $F$ is then inner-product preserving and in particular an isometry. Hence $F(X)$ is closed. To show that $F$ is surjective it thus suffices to see that $F(X)$ contains the standard basis in $\ell^2$ consisting of sequences that $1$ at the $n$-th place and $0$ elsewehere. But this is nothing but $F(e_n)$.