# prove that the mapping $\Phi:\ell^1\to c^*$ is an isometric isomorphism

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:

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

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.
• In here what is meant by "sign" is the positive or negative 1 (depending on the coefficient of $a_n$) am I correct? Oct 26, 2019 at 17:33
• @gune Exactly. That's the definition. So always $a\operatorname{sign}(a) = |a|$. Oct 26, 2019 at 17:34
• 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? Oct 26, 2019 at 17:55