$J: \ell^1 \to c_0^*$ where $y \mapsto y^T$ and $y^T : x \mapsto y \cdot x$ for $x \in c_0$ and $y \in \ell^1$ show $J$ is 1-1.

Question

I have shown that $J$ is an onto, linear, and an isometry. However, I'm having trouble showing that this map is 1-1 too. What I have so far is that:

For $x,y \in \ell^1$ suppose that $J(x) = J(y)$ then $x^T = y^T$ so that for any $u \in c_0$ we have that $x \cdot u = y \cdot u$ so that $\sum_{n=0}^{\infty} x_n u_n = \sum_{n=0}^{\infty} y_n u_n$

I don't think this directly implies that $x_n = y_n$ for all $n$. Maybe there's another approach. Thank you for any advice/help.

Answer 1

Isometries are injective!! To see it just notice that if $J(x)=J(y)$ then $$ 0=||J(x)-J(y)||_{c_0^*}=||J(x-y)||_{c_0^*}=||x-y||_{\ell^1} $$ and so $x=y$.