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 $l^2$?
Best Answer
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.