how to prove isometric vector space isomorphism Let $(L^1)^*$ be the dual space to $L^1$, or bounded linear functional over $L^1$, i.e.,
$f:L^1\to \mathbb{R}$, $f(cx+y)=cf(x)+f(y)$, $|f(x)| \leq M|x|^1$
Define the norm for $f \in (L^1)^∗$ by $||f|| := \sup\{|f(x)| : \forall x \in L^1 \,$ such that $\, |x|^1 = 1\}$.
Let $L^{\infty}$ be the vector space of all bounded (real) sequences.
For each $(b_n) \in L^\infty$ define its norm by
$|(b_n)|^\infty := \sup\{|b_n| : \forall n\}$.
My question is how to show that there is an isometric vector space isomorphism $T : L^\infty \to (L^1)^*$ in the sense that $||T(x)|| = |x|^\infty$.
I know I need to start from here: a bounded sequence $b := (b_n)$ defines the linear functional $f_b : (a_1,a_2,a_3,···) \in L^1  \to  \sum_{n=1}^\infty (a_n \cdot b_n) \in \mathbb{R}$.
But where to move on? Could someone prove in details? Thanks!
 A: For $n\in N $ let $x_n=(x_{n,j})_{j\in N}\in l_1 $, where $x_{n,n}=1$ and $x_{n,j}=0$ for $j\ne n.$
For $f\in (l_1)^*$ let $f_n=f(x_n)$ and let $\psi( f)=(f_n)_{n\in N}.$
Show that $\psi (f)\in l_{\infty},$  that $\psi$ is a linear bijection from $(l_1)^*$ to $l_{\infty},$ and that $\|f\|=\|\psi (f)\|_{\infty}.$
Use the continuity of each $f\in l_1^*$ and use the  fact that if $y=(y_j)_{j\in N}\in l_1$ then $\lim_{n\to \infty}\|y-\sum_{j=1}^ny_jx_j\|_1=0.$
For $y=(y_j)_j\in l_1$ and $f\in l_1^*$ we have $0=f(y-\lim_{n\to \infty}\sum_{j=1}^n y_jx_j)=$ $f(y)-\lim_{n\to \infty}\sum_{j=1}^n y_j f(x_j)=$  $f(y)-\sum_{j=1}^{\infty}y_j f_j.$ So if $\psi (f)=\psi (g)$ then for all $y$ we have $f(y)=\sum_{j\in N}y_j f_j=\sum_{j\in N}y_jg_j=g(y).$ Therefore $$\psi \text { is injective}.$$
We have $\sup_{n\in N}|f_n|=$ $\sup_{n\in N}|f(x_n)|\leq$ $ \sup_{n\in N}\|x_n\|_1\cdot \|f\|=\|f\|$ because each $\|x_n\|_1=1.$ Therefore $\psi (f)\in l_{\infty}$ and $$\|\psi (f)\|_{\infty}\leq \|f\|.$$
We have $|f(y)|=|\sum_{j\in N}y_jf_j|\leq $ $\sum_{j\in N}|y_j|\cdot |f_j|\leq$ $ \sum_{j\in N}|y_j|\cdot \|\psi (f)\|=$ $\|y\|_1\cdot \|\psi (f)\|_{\infty}.$ So $\|f\|=\sup_{\|y\|_1=1}|f(y)|\leq$  $\sup_{\|y\|_1=1}\|y\|_1\cdot \|\psi (f)\|=\|\psi (f)\|_{\infty}$. Therefore $$\|f\|\leq \|\psi (f)\|_{\infty}.$$
From the previous two paragraphs we have $\|f\|=\|\psi (f)\|_{\infty}.$
For $g=(g_n)_{n\in N}\in l_{\infty}$ and for $y=(y_n)_{n\in N}\in l_1,$ let $g'(y)=\sum_{n\in N}y_n g_n.$ The sum converges absolutely because  $|g'(y)|\leq \sum_{n\in N}|y_n|\cdot |g_n|\leq$ $ \sum_{n\in N}|y_n|\cdot \|g\|_{\infty}=$ $\|y\|_1\cdot \|g\|_{\infty}$. So $g'(y)$ is defined, and $|g'(y)|\leq \|y\|_1\cdot \|g\|_{\infty}.$ And $g'$ is (obviously) linear. So $g'\in l_1^*.$ And we have $\psi (g')=g.$ Therefore $$\psi \text { is  surjective.}$$  
