# Isometric identification of $c_0^*$ and $\ell^1$

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.

• 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. – DinkyDoe Oct 7 '13 at 13:23
• 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. – DinkyDoe Oct 7 '13 at 13:58
• @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 – Evan Oct 8 '13 at 12:18
• Ha, ofcourse :p I'm stupid. Yes, $\Lambda_{x_n}(e_j) \to 0$ implies $x_n(j)\to 0$ for all $j$. – DinkyDoe Oct 8 '13 at 12:38
• The theorem we need is ofcourse that every weakly convergent sequence is bounded to conclude $\sup_n|x_n|<\infty$. – 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.