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I'm trying to prove that the map $T:x \in \ell^1 \mapsto Tx \in c'$ given by $$(Tx)(y)=x_1\lim_{n\to\infty}y_n + \sum_{k=1}^\infty x_{k+1}y_k$$ is an isometric isomorphism between $\ell^1$ and $c'$ (the dual of the subspace of $\ell^\infty$ consisting of all convergent sequences). I have already shown that $T$ is a bounded linear isometry. To prove the surjectivity, I found that given $y' \in c'$ the sequence $$x=\left(y'(1)-\sum_{n=1}^\infty y'(e_n),y'(e_1),y'(e_2),y'(e_3),\ldots\right)$$ satisfies $Tx=y$. I'm denoting by $1$ the sequence $(1)_{n \in \mathbb{N}}$. It is remaining to prove that $x$ indeed belongs to $\ell^1$, i.e., that $\Vert x \Vert_1 < \infty$. I see that $$\Vert x \Vert_1 = \vert y'(1) - \sum_{n=1}^\infty y'(e_n)\vert + \sum_{n=1}^\infty \vert y'(e_n)\vert$$ but I can't see why this should be finite. I feel I'm missing something that should be clear. I also would like to justify why $\sum_{n=1}^\infty y'(e_n)$ would be convergent since I'm using this to define $x$.

lfsm
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    Check out this https://math.stackexchange.com/questions/1942370/isometrical-isomorphism-t-ell-1-to-c?rq=1 – Salcio Jan 01 '23 at 01:05
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    Statement needs fixing. n is a varable $(y_n)$ for limit as well as a dummy in sum where $(x_{n+1}y_n)$ are terms. – herb steinberg Jan 01 '23 at 01:12
  • Yes @Salcio, they did almost the same as I did, but the given answer does not justify why $\sum_{n=1}^\infty y'(e_n)$ should converge and why $\Vert x \Vert_1$ should be finite. – lfsm Jan 01 '23 at 01:13
  • Sorry @herbsteinberg, I didn't get what you mean. I'm using $x=(x_n){n \in \mathbb{N}} \in \ell^1$ and $y=(y_n){n \in \mathbb{N}} \in c$ – lfsm Jan 01 '23 at 01:15
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    Looking at the bare statement. $y_n$ appears in two places: outside the sum where a limit is taken as $n\to \infty$ and inside the sum for $1\le n \lt infty$. You need to use a different index for the sum. – herb steinberg Jan 01 '23 at 01:26
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    I edited it. But to be honest I don't see why this was wrong. They are two distinct limits. – lfsm Jan 01 '23 at 01:33
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    You have to show that the series $\sum |y'(e_n)|$ is finite using the fact that $y'$ is a bounded linear functional. To achieve that let $y'(e_n)=r_nu_n,$ where $r_n=|y'(e_n)|.$ Let $x_n=\overline{u_n}$ for $n\le N$ and $x_n= 0$ for $n>N.$ The sequence $x$ tends to $0$ and is bounded by $1.$ Thus $\sum_{n=1}^N|y'(e_n)|=y'(x)\le |y'|.$ As $N$ is arbitrary you get convergence of the sum. – Ryszard Szwarc Jan 01 '23 at 02:06
  • @RyszardSzwarc Thanks!! This makes sense and solves my problem. – lfsm Jan 01 '23 at 12:49

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