$\ell \in (\ell^{\infty})^{*}$ can be uniquely represent by the two functionals

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

• 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. – Viktor Glombik Jul 25 at 12:15
• An approximation – SABOY Jul 25 at 12:23
• Are you sure it is $\ell_{1}|_{c_0} = 0$ and not $\ell_2$? – Viktor Glombik Jul 25 at 12:27
• 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. – 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 $$\Bbb N\subset K$$.