Let $\{x_n\}_{n=1}^{\infty}\subset \ell_1$ be a sequence in $\ell_1$ with $x_n = (x_n(1),x_n(2), x_n(3),\ldots )$

I want to show that $$\lim_{n\to\infty}\sum_{j=1}^{\infty} x_n(j)y(j) = 0 $$

for all $y\in c_0$ if and only if $\sup_n \left\|x_n\right\|_1<\infty$ and $\lim_{n\to\infty}x_n(j) = 0$ for $j=1,2,3,\ldots$.

Apparently we can use the fact that there's an isometric identification of $c_0^*$ and $\ell_1$ via the canonical pairing between $c_0$ and $\ell_1$.

So how does this identification help us? With this identification, do we interpret the $x_n$ as functionals, in the sense that $y\mapsto \sum_{j=1}^{\infty}x_n(j)y(j) $ ? . To me this seems like proving that $x_n$ converges to the $0$ - funtional iff those $2$ conditions hold. How can we show this?

Can someone shed some light over this?


For one of the directions, the identification makes things simple. If the functionals converge in a weak* sense to the 0 functional, what does it tell you about the norms of the functionals? Can you choose a simple $c_0$ sequence that can get you the limit of $x_n(j)$ for fixed $j$?

For the other direction, seek to bound the sum for a fixed sequence $y$ by a given $\epsilon > 0$. Try separating the sum into two parts, one where $y$ is small (you know $y \in c_0$) and the rest is just a finite sum where you can take advantage of linearity of limits.

  • $\begingroup$ Thanks! So for $\Leftarrow$, we can say: for fixed $y\in c_0$, and $\epsilon > 0$ we can find $M$ such that $y(j)<\epsilon$ for $j> M $ and then, split the sum into $\sum_{j=1}^M + \sum_{j=M+1}^{\infty}$ where the second part is then smaller then $\epsilon \sup_n |x_n|_1$. So the result follows by taking limits?. For $\Rightarrow$ I am little confused because if we take $y(j) = 1/j$ for example how do we see that $\lim_{n\to\infty}\sum x_n(j)/j = 0$ implies that $x_n(j)\to 0$ for all $j$? I see no reason why that should be true. $\endgroup$ – DinkyDoe Oct 7 '13 at 13:23
  • $\begingroup$ Ok so I was thinking about what you said: If $\Lambda_{x_n}(y) = \sum_j x_n(j)y(j)$ and if $\Lambda_{x_n}$ converges to the 0-functional ( and so $|\Lambda_{x_n}|\to 0$), doesn't this automatically mean, by the isometric identification, that $|x_n|_1= \sum_j |x_n(j)| \to 0$. And then it follows automatically. $\endgroup$ – DinkyDoe Oct 7 '13 at 13:58
  • 1
    $\begingroup$ @DinkyDoe careful it does not converge in operator norm, just weakly. you need a theorem about weak convergence. for the other point, why not use the standard basis as candidates of c0 and see what that tells you $\endgroup$ – Evan Oct 8 '13 at 12:18
  • $\begingroup$ Ha, ofcourse :p I'm stupid. Yes, $\Lambda_{x_n}(e_j) \to 0$ implies $x_n(j)\to 0$ for all $j$. $\endgroup$ – DinkyDoe Oct 8 '13 at 12:38
  • $\begingroup$ The theorem we need is ofcourse that every weakly convergent sequence is bounded to conclude $\sup_n|x_n|<\infty$. $\endgroup$ – DinkyDoe Oct 8 '13 at 13:27

Maybe i can you give some hints: you can sue for the $\Rightarrow$ direction the Banach-Steinhaus-Theorem, or Uniform Boundedness Principle. For the other direction construct a sequence which does not satisfied the asked property.

P.S. If you have find a solution please write this solution down here, not make clear that you have solved it, this is nicer also for other people which are interested in Mathematics.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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