Weak convergence in $l^2$ spaces We say that $x_n$ converges weakly to $x$ in normed space $X$ when for any linear and continuous functional $f$ we have $f(x_n) \rightarrow f(x)$.
Let's define sequence $(e_n)$ as $e_n=1$ and $e_j = 0$ for $j \neq n$. e.g.
$$e_2 = (0, 1, 0, 0, 0...)$$
I want to judge on convergence/divergence of $\sqrt{n}e_n$ in $l^2$ space.
My work so far
We know that every linear and continuous functional in $l^2$ has to be in form $f(x_n) = \sum_{n=1}^\infty x_na_n$ where $a_n \in l^2$ and $x_n \in l^2$.
We are asking is there is a sequence $b_n \in l^2$ that $\sum_{j=1}^\infty a_j\sqrt{j}e_j$ converges to.
But when I take $a_j = e_j$ I have that $a_j \in l^2$ and
$$\sum_{j =1}^\infty a_j\sqrt{j}e_j = \sum_{j=1}^\infty \sqrt{j}e_j^2 = 0 + 0 +... +\sqrt{n} + 0 +.. = \sqrt{n} \rightarrow \infty$$
So I found linear, continuous functional that doesn't converge. Is this sufficient argument to say that $\sqrt{n}e_n$ diverges in $l^2$?
 A: It looks like you are mixing numbers and vectors: $e_j$ is a vector, while $a_j$ is a number. The way you wrote $f$ makes no  sense, as you wrote a product of elements on $\ell^2$. The bounded functionals, due to the Riesz Representation Theorem, are of the form
$$
f(x)=\sum_nx_na_n,
$$
with $a\in\ell^2$; that is, $\sum_n|a_n|^2<\infty$.
To test weak convergence, we need to consider, for each $y\in\ell^2$,
$$
\langle \sqrt n\,e_n,y\rangle=\sum_k\sqrt n\,\delta_{n}(k)\overline{y_k}=\sqrt n\,\overline{y_n}. 
$$
Let
$$
y=\sum_k k^{-1}\,e_{k^2}.
$$
Then, when $n=m^2$,
$$
\langle \sqrt n\,e_n,y\rangle=\langle m\,e_{m^2}\,y\rangle=1.
$$
This shows that $\sqrt n\,e_n$ does not converge weakly to zero.
Pushing this idea, we can take
$$
y=\sum_k k^{-1}\,e_{k^4}.
$$
Then, when $n=m^4$,
$$
\langle \sqrt n\,e_n,y\rangle=\langle m^2\,e_{m^4}\,y\rangle=m.
$$
This shows that $\sqrt n\,e_n$ does not converge weakly.
A: What do you mean by taking $a_j = e_j$ when $e_j$ is an element of $\ell^2$ and $a_j$ is just the $j$-th term of an element $a$ of $\ell^2$? I think that using subscripts both for coordinate indices (such as in $a_j$) and sequence indies (such as in $\sqrt{n} e_n$) is causing confusion in your attempt.
For that reason, let me use parenthesized superscripts for sequence indices and boldface font for elements of $\ell^2$. For instance, $\mathbf{e}^{(n)} = (e^{(n)}_j)_{j=1}^{\infty}$ will denote the element of $\ell^2$ defined by
$$ e^{(n)}_j = \begin{cases} 1, & j = n, \\ 0, & j \neq n. \end{cases} $$
Then we want to check whether the sequence $\mathbf{x}^{(n)} = \sqrt{n} \mathbf{e}^{(n)} = (\sqrt{n}e^{(n)}_j)_{j=1}^{\infty}$ converges weakly in $\ell^2$ or not. For that let $\mathbf{a} = (a_j)_{j=1}^{\infty}$ be the element of $\ell^2$ defined by
$$ a_j = \begin{cases} 2^{-k}, & \text{if $j = 4^k$ for some $k$}, \\ 0, & \text{otherwise}. \end{cases} $$
It is clear that $\mathbf{a}$ is in $\ell^2$, since
$$ \sum_{j=1}^{\infty} a_j^2 = \sum_{k=0}^{\infty} \left(\frac{1}{2^k}\right)^2 < \infty. $$
On the other hand, if $f$ is a continuous linear functional on $\ell^2$ represented by $\mathbf{a}$, then
$$ f(\mathbf{x}^{(n)})
= \sum_{j=1}^{\infty} a_j \sqrt{n} e^{(n)}_j
= \sqrt{n} a_n
= \begin{cases}
\dfrac{\sqrt{4^k}}{2^k} = 1 & \text{if $n = 4^k$ for some $k$}, \\
0, & \text{otherwise}.
\end{cases} $$
So $f(\mathbf{x}^{(n)})$ does not converge as $n\to\infty$.
