To prove: $T:l^1 \to c_0^*, \langle Tx, y\rangle = \sum x_ky_k$ is an isometric isomorphism
Is this proof correct? Particularly, I dont know if I proved injectivity correctly
Proof: (i) linearity: let $x, y \in l^{1}(\mathbb{K}), \eta \in c_0(\mathbb{K})$
$T(\alpha x+y)(\eta)=\sum(\alpha x+y)_k \eta_k=\sum \alpha x_k \eta_k+y_k \eta_k=\alpha \sum x_k \eta_k+\sum y_k \eta_k=\alpha Tx(\eta)+Ty(\eta)$
(ii) continuous: let $x \in l^1(\mathbb{K}), y \in c_0(\mathbb{K})$
$\lVert Tx\rVert_{c_0^*} = \sup_{\lVert y\rVert=1} \|(Tx)(y)\| = \sup_{\lVert y\rVert=1} |\sum x_k y_k| \leq \sup_{\lVert y\rVert=1}\sum |x_k| |y_k| = \sup_{\lVert y\rVert=1}\sup_{i} |y_i| \sum |x_k| = \sum |x_k| = \lVert x\rVert_1$
$\Rightarrow$ T is bounded $\Rightarrow$ T is continuous with $\lVert T\rVert_{L(l^1, c_0^*)}$
(iii) T is an isometry: $\lVert Tx \rVert = \sup |\langle Tx, y\rangle|$.
Let $y_k^N = x_k/|x_x|$ when $x_k = 0$ and $k \leq N$, and 0 otherwise. Then $y^N \in c_0$ with $\lVert y^N\rvert=1$, and $|\langle Tx, y^N\rangle| = |\sum^N x_k y_k^N| = \sum^N |x_k| \rightarrow \lVert x\rVert_{l^1} \Rightarrow \lVert Tx\rVert = \lVert x\rVert \Rightarrow$ T is an isometry.
(iv) T is an isomorphism, i.e., $T^{-1} \in L(c_0^*, l^1)$ (T is bijective and $T^{-1}$ is linear and continuous)
- T is injective: let $x, y \in l^1, \eta \in c_0$.
$\langle Tx, \eta \rangle = \sum x_k\eta_k = \sum y_k\eta_k = \langle Ty, \eta \rangle \Rightarrow x=y \Rightarrow$ T is injective
- T is surjective: the idea is to prove that for any $y^* \in c_0$ there exists $x\in l^1$ s.t. $(Tx)(y) = y^*(y)$.
Let $y=(0, \dots)\in c_0$. Then take $x = (0, \dots) \in l^1$. Since $y^* (y)=0$ (because $y^* $ is linear), then it follows $(Tx)(y) = y^* (y)$
Suppose there exists $i_0$ s.t. $y_{i_0}\neq 0$. Then take $x=(0,\dots, y_{i_0}/|y_{i_0}|, 0, \dots)\in l^1$ and $(Tx)(y) = \sum x_ky_k = y^* (y) \Rightarrow $ T is surjective
$\Rightarrow$ T is bijective
By the bounded inverse theorem $T^{-1}\in L(c_0^*, l^1)$