I think I have the right idea how to solve this, but I can't find a specific sequence that does what I want it to.

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 $\|\varphi_a\|=\sup\{|\varphi_a(x)|:\|x\|_\infty=1\}$.

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.

To show $\|\varphi_a\|\leq\|a\|_1$ is easy. I would like to find some $x\in c$ with $\|x\|_\infty=1$ such that $|\varphi_a(x)|\geq\|a\|_1$; this would show equality since it is a lower bound on $\|\varphi_a\|$.

Also, any idea how to show it is surjective, i.e. taking any $x\in c$ and expressing it as some $\varphi_a(x)$?

  • $\begingroup$ It is not always true that $\psi_a$ attains its norm. Show instead that $\forall d>0\;\exists x\;(\|x\|=1\land |\psi_a(x)\geq \|a\|_1(1-d)).$ $\endgroup$ Dec 17, 2015 at 23:52

1 Answer 1



Indeed, the inequality $$\left| a_1\lim_{n\to\infty}x_n+\sum_{n=1}^\infty a_{n+1}x_n \right|\le \|x\|_{\infty}\|a\|_1$$ is immediate from the triangle inequality.

To attain equality, we want everything inside the absolute value to have the same sign, that is $x_n=\operatorname{sign} a_{n+1}$. A problem is, this doesn't necessarily have a limit. A compromise is to set $x_n=\operatorname{sign} a_{n+1}$ for $n\le N$ and $x_n=\operatorname{sign} a_{1}$ for $n>N$, thus ensuring $\lim_{n\to\infty}x_n=\operatorname{sign} a_{1}$. This allows the terms $a_{n+1}x_n $ with $n>N$ to potentially be negative, so the estimate becomes $$ \left| a_1\lim_{n\to\infty}x_n+\sum_{n=1}^\infty a_{n+1}x_n \right|\ge \|x\|_{\infty}\|a\|_1 - 2 \sum_{n=N+1}^\infty |a_{n+1}| $$ Since $a\in\ell_1$, the subtracted term can be made arbitrarily small.


The linear combinations of the standard basis $e_n$ and the constant sequence $\sum_{n\in\mathbb{N}} e_n$ are dense in $c$. Therefore, every bounded functional $f$ on $c$ is determined by its values on this set; call them $\beta_n=f(e_n)$ and $\beta_\infty = f(\sum e_n)$. The boundedness of $f$ implies $\beta\in \ell_1$. But for any such $\beta $ there is a corresponding $a\in \ell_1$, namely $$ a= (\beta_\infty,\beta_1,\beta_2,\dots) $$

  • $\begingroup$ For Banach space $B$ with dual $B^*$, from def'n of $\|f\|$ we have $|f(x)|=\|f\|\cdot \inf \{\|x-y\|:f(y)=0\}=\|f\}\cdot d(x,f^{-1}0),$ and $B=\{a x+y: a\in S\land f(y)=0\}$ where $ S$ is the scalars (real or complex). From which : For $0\ne f\in B^*,$ we have $\forall x\in B\;\exists y \in f^{-1}0\;(\|x-y\|=d(x,f^{-1}0)$ $\iff \exists x\in B\backslash f^{-1}0\;\exists y\in f^{-1}0\;( \|x-y\|=d(x,f^{-1}0)$ $ \iff \exists x\in B\;(|f(x)|=\|f\|\cdot \|x\|\ne 0.)$ $\endgroup$ Feb 15, 2016 at 1:02
  • $\begingroup$ Be careful with the expression $\sum_{n\in\mathbb{N}}e_n$ as this series does not converge. I think you want to set $\beta_\infty=f(\mathbb{1})-\sum_{n\in\mathbb{N}}f(e_n)$ where $\mathbb{1}$ is the constant 1 sequence which is in c . It is quite easy to check that the sum involved here converges. $\endgroup$ Feb 3, 2021 at 2:09

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