0
$\begingroup$

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:

$\endgroup$
2
  • $\begingroup$ "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

1
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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

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.