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Could anyone help with the proof of the following: the standard basis of $l_1(\mathbb{N})$ does not have a limit in weak topology? I think it is the case in norm topology since that sequence is not a Cauchy sequence, so maybe one should use the fact that continuity is equivalent between these 2 topologies for liner functionals?

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up vote 4 down vote accepted

By standard basis, I guess you mean $e^n$ given by $e_k^n=\delta_{nk}$. Assume it converges to $x$ weakly in $\ell^1$. The map $L_j\colon \ell^1\to \Bbb R$, $L_j(y)=y_j$ is linear an continuous. Hence $\lim_{n\to +\infty}L_j(e^n)=L_j(x)=0$, so $e^n$ would converge weakly to $0$. Define now $L(y):=\sum_{k\in\Bbb N}y_k$, a linear functional on $\ell^1$. It's well-defined, linear and continuous, and $L(e^n)=1$ for all $n$, which proves we can't have weak convergence.

Actually, weak convergence in $\ell^1$ is the same as strong convergence for sequences (but the two topologies cannot be the same, as it's a infinite dimensional vector space). Using this result to solve the problem would be overkill, but I think it's interesting to know it.

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Here's an alternative proof which is just a tiny bit shorter: Consider $$L: \ell^1 \to \mathbb R, \quad L(y) = \textstyle\sum_{n=1}^\infty (-1)^n y_n.$$ Then $L$ clearly is a continuous linear functional on $\ell^1$ and $L(e_n) = (-1)^n$ does not converge. In particular it can't converge to $L(x)$ for any $x\in \ell^1$.

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