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Let $C = \left\{ \{x_i\} \in l_\infty | \exists \lim_{i \rightarrow \infty} x_i \right\}$. Let $\phi \in C^{*}$ ($C^{*}$ is a dual space). Let $y=\{y_i\} \in l_{\infty}$ $\phi_y: \{x_i\} \rightarrow \sum_i x_i y_i$. Prove that $\exists \alpha \in \mathbb{R}, \exists y \in l_{\infty}$ such that $\phi = \phi_y + \alpha * \lim$

I understand all of the definitions and conditions of the problem, but i don't know enough tools that can help me for solving this.

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  • $\begingroup$ The space $\;\ell_\infty\;$ is the space of all (real, or complex) bounded sequences. Why would the series $\;\sum x_iy_i\;$ converge at all, to begin with? Perhaps you're mixing notation? Perhaps I am... Even if it were the set of converging sequences, why would that series converge? Otherwise the functional isn't well-defined... $\endgroup$ – DonAntonio Nov 23 '17 at 22:46
  • $\begingroup$ Hint: if $e_k = \{\delta_i^k\}$, then $\phi_y(e_k) = y_k$, so you already know $y$. It only remains to prove that $\phi_y$ is well defined and that $\phi - \phi_y$ is a multiple of $\lim$. $\endgroup$ – Gribouillis Nov 23 '17 at 22:47
  • $\begingroup$ Sorry, I meant $\phi(e_k) = y_k$. It is then easy to prove that if $\lim x_i = 0$, then $\phi(x) = \phi_y(x)$ by using the continuity of $\phi$. $\endgroup$ – Gribouillis Nov 23 '17 at 22:53
  • $\begingroup$ @Gribouillis, what are $\delta_i$ and why $\phi(e_k) = y_k$? $\endgroup$ – Khan Nov 23 '17 at 23:05
  • $\begingroup$ Oh, I used the Kronecker delta. I only mean that $e_k$ is the sequence which all terms are 0 but the k-th term which has value 1. If you suppose that $\phi = \phi_y + \alpha \lim$, then $\phi(e_k) = \phi_y(e_k) + \lim(e_k) = y_k$, so that $y$ is entirely given by the values of the linear form $\phi$. $\endgroup$ – Gribouillis Nov 23 '17 at 23:27
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For $k \in \mathbb{N}$, let ${e}^{k} = {\left\{{{\delta}}_{i}^{k}\right\}}_{i \in \mathbb{N}}$ where ${\delta}$ is the Kronecker delta. Let ${y}_{k} = {\phi} \left({e}^{k}\right)$. Let $M = {\left\|{\phi}\right\|}_{{C}^{\ast }}$ and let ${u}^{n} = \sum _{k = 0}^{n} \text{sgn} \left({y}_{k}\right) {e}^{k} \in C$. We have

$$\sum _{k = 0}^{n} \left|{y}_{k}\right| = \sum _{k = 0}^{n} \text{sgn} \left({y}_{k}\right) {\phi} \left({e}^{k}\right) = {\phi} \left({u}^{n}\right) \leqslant M {\left\|{u}^{n}\right\|}_{C} \leqslant M$$

Hence $y = {\left\{{y}_{i}\right\}}_{i \in \mathbb{N}} \in {\ell }_{1}$ and ${\left\|y\right\|}_{{\ell }_{1}} \leqslant {\left\|{\phi}\right\|}_{{C}^{\ast }}$. It follows that the expression

$${{\phi}}_{y} \left(x\right) = \sum _{i \in \mathbb{N}} {y}_{i} {x}_{i}$$

defines a bounded linear functional on $C$.

We claim that if $x \in C$ and $\lim x = 0$, then ${\phi} \left(x\right) = {{\phi}}_{y} \left(x\right)$. Indeed the condition imply that ${v}^{n} \rightarrow x$ in $C$ where ${v}^{n} = \sum _{k = 0}^{n} {x}_{k} {e}^{k}$. But one has ${\phi} \left({v}^{n}\right) = {{\phi}}_{y} \left({v}^{n}\right)$ and as ${\phi}$ and ${{\phi}}_{y}$ are continuous, the result follows.

Now let $x \in C$ be such that $\lim x = {\lambda}$ one can write $x = \left(x-{\lambda} w\right)+{\lambda} w$ where $w = {\left\{1\right\}}_{i \in \mathbb{N}}$ is the constant sequence equal to $1$. We have

$${\phi} \left(x\right) = {\phi} \left(x-{\lambda} w\right)+{\lambda} {\phi} \left(w\right) = {{\phi}}_{y} \left(x-{\lambda} w\right)+{\lambda} {\phi} \left(w\right) = {{\phi}}_{y} \left(x\right)+\left({\phi} \left(w\right)-\sum _{i \in \mathbb{N}} {y}_{i}\right) \lim x$$

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