Does the sequence $\{e_i\}$ converges in $l_2$? does the sequence $\{f(e_i)\}$ converges for any continuous linear functional on $l_2$?
I know that in $l_2$ norm $\{e_i\}$ does not converges as $d(l_i,l_j)=\sqrt{2}$ always! But I dont know the other! could any one help me?
The latter one is true since $\{e_i\}$ converges weakly in $l^2$. You can argue as follows:
Since $(l^2)^*=l^2$, for any given $f \in \mathcal{L}(l^2)$ we can find some $x=(x_i) \in l^2$ such that $f(y)=<x,y>_{l^2}=\sum_{i=1}^{\infty}x_i y_i$ holds for $\forall y \in l^2$. Notice $x_n \to 0$ as $n \to \infty$ we find $f(e_n)=x_n \to 0$.
As you point out the answer to the first is no. The answer to the second is yes. This is called weak convergence in $l^2$.
To see this note that $l^2$ is a Hilbert space and is thus (anti-) isomorphic to it's dual, via the inner product. So give $f\in (l^2)^*$ there is a $\xi=\sum_{i=1}^\infty a_ie_i\in l^2$ such that $f(\eta)=\langle\xi, \eta\rangle$ for all $\eta\in l^2$.
Thus we get that $f(e_i)=\overline{a_i}\rightarrow0$ since $\xi\in l^2$
