For some $a=(a_n)\in\ell^1$, we define $\varphi_a:c\to\mathbb{F}$ by $$\varphi_a(x)=a_1\lim_{n\to\infty}x_n+\sum_{n=1}^\infty a_{n+1}x_n,\quad x=(x_n)\in c,$$ where $c$ is the space of convergent sequences and $\|\varphi_a\|=\sup\{|\varphi_a(x)|:\|x\|_\infty\leq1\}$.

I'm trying to prove that the mapping $\Phi:\ell^1\to c^*$ defined by $\Phi(a)=\varphi_a$ for $a\in\ell^1$ is an isometric isomorphism.

I went through several worked examples too but still I have some doubts:

For the Isometry, To show $||\Phi(a)||\leq\|a\|_1$ can be done through Triangle inequality. But how about the other direction?

I know since $a\in l^1$ we have $N$ such that $\sum \limits_{n=N}a_n<\epsilon$

Thus do we need to define a sequence $x=(x_n)\in c$ such that $|\varphi_a(x)|\geq||a||-\epsilon?$ for any $\epsilon$ So that then we can say $||\Phi(a)||\geq||a||$

In this answer there is a method given for the sequence to be. Which is $x_n=\operatorname{sign} a_{n+1}$ for $n\le N$ and $x_n=\operatorname{sign} a_{1}$ for $n>N$
But I can't understand why this will lead to the result $$ \left| a_1\lim_{n\to\infty}x_n+\sum_{n=1}^\infty a_{n+1}x_n \right|\ge \|x\|_{\infty}\|a\|_1 - 2 \sum_{n=N+1}^\infty |a_{n+1}| $$

And for the surjection if we define $f\in c^*$ with the basis elements of $c$ that is:
$\beta_n=f(e_n)$ can we prove that $\beta=\sum\beta_n\in l^1$ Appreciate your help:

  • $\begingroup$ "Thus do we need to define [...]?" Yes, exactly. $\endgroup$
    – amsmath
    Oct 26, 2019 at 17:19
  • $\begingroup$ Thank you but can you please tell why that the proposed sequence $x$ will lead to the desired result... $\endgroup$
    – Charith
    Oct 26, 2019 at 17:21

1 Answer 1


With the chosen $(x_n)$ you have \begin{align} a_1\lim_{n\to\infty}x_n+\sum_{n=1}^\infty a_{n+1}x_n &= |a_1| + \sum_{n=1}^N|a_{n+1}| + \operatorname{sign}(a_1)\sum_{n>N}a_{n+1}\\ &= \sum_{n=1}^N|a_n| + \operatorname{sign}(a_1)\sum_{n>N}a_{n+1}\\ &= \sum_{n=1}^\infty|a_n| - \sum_{n>N}|a_n| + \operatorname{sign}(a_1)\sum_{n>N}a_{n+1}\\ &= \|a\|_1 - \sum_{n>N}|a_n| + \operatorname{sign}(a_1)\sum_{n>N}a_{n+1}. \end{align} If $N$ is large, the last two summands are very small.

  • $\begingroup$ In here what is meant by "sign" is the positive or negative 1 (depending on the coefficient of $a_n$) am I correct? $\endgroup$
    – Charith
    Oct 26, 2019 at 17:33
  • 1
    $\begingroup$ @gune Exactly. That's the definition. So always $a\operatorname{sign}(a) = |a|$. $\endgroup$
    – amsmath
    Oct 26, 2019 at 17:34
  • $\begingroup$ Thank you very much for helping out. $\endgroup$
    – Charith
    Oct 26, 2019 at 17:39
  • 1
    $\begingroup$ @gune If you are content with my answer, please upvote and check it. Thanks. $\endgroup$
    – amsmath
    Oct 26, 2019 at 17:40
  • $\begingroup$ Sure. And I'm sorry I still have a little confusion,.. In order to obtain your third line we need to have, $\sum\limits_{n=1}^{N}|a_n|+\sum\limits_{n\geq N}|a_n|+...$ But instead from the second line don't we only have the series without the absolute value? $\endgroup$
    – Charith
    Oct 26, 2019 at 17:55

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