Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer

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 $x\in \ell^1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.