Let $\ell \in (\ell^{\infty})^{*}$ Show that $\ell$ is uniquely represented by the sum of the functionals $\ell_{1}+\ell_{2}$, where $\ell_{1}((s_{n}))=\sum\limits_{n}s_{n}t_{n}$ and $\ell_{2}\vert_{c_{0}}=0$. Show further that $\vert \vert \ell \vert \vert = \vert \vert \ell_{1} \vert \vert + \vert \vert \ell_{2} \vert \vert$.

I am given the following hints: i) Consider $\ell(e_{n})$ ii) Choose $x,y \in \ell^{\infty}$ where $\vert \vert x \vert \vert_{\infty}=\vert \vert y \vert \vert_{\infty}=1$ and further that $\ell_{1}(x)$ approximates $\vert\vert \ell_{1} \vert\vert$ and $\ell_{2}(x)$ approximates $\vert\vert \ell_{2} \vert\vert$ iii) Now choose a sequence $z$ where $z_{n}=x_{n}$ for some $N\in \mathbb N$ and $n \leq N$ and $z_{n}=y_{n}$ for all $n > N$.

I am confused, even when following the hints:

i) for any element from the standard basis $e_{n}$ we have $\ell(e_{n})=\ell_{1}(e_{n})+\ell_{2}(e_{n})=t_{n}+0=t_{n}$

for ii) surely $\vert \vert \ell_{2} \vert \vert=0$ and $\vert \vert \ell_{1} \vert \vert=\vert\vert t \vert \vert_{1}$ (no more information was given on $t:=(t_{n})_{n}$). So I can choose for example $x:=\operatorname{sign}(t)$ and $y:=(1,0,....)$ but I do not see how this can help me

  • $\begingroup$ What is $\ell_{1}(x) \cong \vert\vert \ell_{1} \vert\vert$ supposed to mean? I know $V \cong W$, meaning that $V$ and $W$ are (isometrically) isomorphic to each other, but here we have a functional and a scalar. Also note that you can use \| \cdot \| instead of \vert \vert \cdot \vert \vert. $\endgroup$ – Viktor Glombik Jul 25 at 12:15
  • $\begingroup$ An approximation $\endgroup$ – SABOY Jul 25 at 12:23
  • 1
    $\begingroup$ Are you sure it is $\ell_{1}|_{c_0} = 0$ and not $\ell_2$? $\endgroup$ – Viktor Glombik Jul 25 at 12:27
  • $\begingroup$ Let $f\in (\ell^\infty)^*$. The restriction of $f$ to $c_0\subset \ell^\infty$ gives you a map $r:(\ell^\infty)^*\to (c_0)^*=\ell^1$. This map is clearly linear and continuous. Denote with $i:\ell^1\to (\ell^1)^{**}=(\ell^\infty)^*$ the inclusion of $\ell^1$ into its bi-dual. Note that $r\circ i$ is the identity on $\ell^1$. Then define $$f_1 = (i\circ r)(f) \qquad f_2=f-f_1$$ note that $f_2\lvert_{c_0}=r(f_2)=r(f)-r(f_1)=0$. This defines your deccomposition, you can check the properties either directly or by applying the hints. $\endgroup$ – s.harp Jul 25 at 12:32

Let $t_n=\ell(e_n)$.

Note that if $s\in\ell^\infty$ then $$\left|\sum_{n=1}^Ns_nt_n\right|\le||\ell||\,||s||.$$So $\sum|t_n|\le||\ell||<\infty$, so we can define $\ell_1\in(\ell^\infty)^*$ by$$\ell_1(s)=\sum s_nt_n.$$

If $s\in c_0$ then the sum $s=\sum s_ne_n$ converges in $\ell^\infty$, so $\ell s=\ell_1s.$ Hence if $\ell_2=\ell-\ell_1$ then $ls=0$ for every $x\in c_0$.

For uniqueness: If $\sum s_nt_n=\sum x_nt_n'$ for every $s\in c_0$ then $s=e_n$ shows that $t_n=t_n'$.

Or, less elementary but maybe more obvious: Say $K$ is the maximal ideal space of the Banach algebra $\ell^\infty$. Then $\ell^\infty\approx C(K)$ and $\Bbb N\subset K$ (or more properly, there is a canonical embedding of $\Bbb N$ in $K$).

So an element of $(\ell^\infty)^*$ "is" a measure $\mu$ on $K$; now $\ell_1$ is just the restriction of $\mu$ to $\Bbb N\subset K$.


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