Definition: Weak convergence- If a sequence $x_n\in X$ converges weakly then if $x_n\to x_0$ then $\lim_{n\to\infty}f(x_n)=f(x_0)$ for $f\in X'$

Proposition: If a sequence converges by the norm then it converges weakly. Proof: $|f(x_n)-f(x_0)|=|f(x_n-x_0)|\leqslant ||f|||x_n-x_0||\to 0$ as $n\to\infty$.

Then my doubt arises in the following example: Consider the space $C_0$(space of all the sequence that converge to 0) with norm $max$. $e_k=(0,0,0,...,1,0,0,0,0...0)$ then $||e_i-e_j||=1,i\neq j$, where $e_1,e_2...$ is a basis of $C_0$

$f\in (C_0)'\simeq l_1$



Then $|f(e_k)|\to 0$ as $k\to\infty$


1) How does the author establishes $\xi_k=f(e_k)$?

2) Is correct to assume that since $|f(e_k)|\to 0$ the sequences converge weakly but not strongly?

Thanks in advance!


1) For $e_k=(x_n)\in C_0$, we have $x_n=1$ if $n=k$ and $x_n=0$ otherwise. Thus we have $f(e_k)=\sum_{n}x_n\xi_n=\xi_k$.

2) From $|f(e_k)|\to 0$ you can conclude directly that $\{e_k\}$ converges weakly. To show it does not converge strongly you need to do some more work, presumably already covered (or assumed) by the author.

  • $\begingroup$ I think that the sequence does not converge by the norm once $||e_k-e_i||=1$ for $i\neq k$. Is it right? $\endgroup$ – Pedro Gomes Jun 11 '18 at 14:05
  • $\begingroup$ That's correct. $\endgroup$ – Aweygan Jun 11 '18 at 14:07
  • $\begingroup$ I am having a hard time understanding point 1 in your explanation. Why is $e_k=(x_n)$? $\endgroup$ – Pedro Gomes Jun 11 '18 at 14:11
  • $\begingroup$ $e_k$ is just a sequence, I'm saying this sequence is $(x_n)$ $\endgroup$ – Aweygan Jun 11 '18 at 14:14

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