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$(x,Ay):=\sum_{k=1}^{\infty}x_k y_k$ and $A:\ell^1 \to c_0^*$ and $x_k\in c_0, y_k\in\ell^1$

I would like to show A is surjective and isometric.

I am not sure about the deifnition of an isometriy, does it mean I have to show that $(x,y)=(Ax,Ay)$ ?

Since $y\in\ell^1$ I know that $\sum_{k=0}^{\infty}|y_k|<\infty$ and because of $x\in x_0$ I know $\lim x_n=0$ This could help me for an infinite sum.

How can I write $(x,y)$ and $(Ax,Ay)$ now?

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Well, the definition says that a linear map $T: (X, \|\cdot\|) \to (Y,\|\cdot\|')$ is an isometry if and only if $\|x\|=\|Tx\|'$ for all $x \in X$.

So you need to show that for $x \in c_0$ and $y \in \ell^1$ it holds that $\|x\|_{c_0} = \|\sum x_k y_k \|_{\ell^1} = \sum |x_k y_k|$. What is your norm on $c_0$?

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The norm should be similiar to that one in en.wikipedia.org/wiki/C_space Now to your equivalences: $||\sum x_k x_k||_{\ell^1}$ should be $\sum\sum|x_k y_k|$ but why did you write only one sum? –  Montaigne Dec 12 '12 at 20:12
    
@Montaigne Why do you write two sums? In your definition in the question there is only one sum. –  Matt N. Dec 12 '12 at 20:55
    
Ok, the right hand side is now clear $\|\sum x_k y_k \|_{\ell^1} = \sum |x_k y_k|$ but what about $\|x\|_{c_0} = \|\sum x_k y_k \|_{\ell^1}$ ? I do not see the equivalence. –  Montaigne Dec 12 '12 at 21:04
    
@Montaigne This equality is what you say you have to show. Assuming that your map is indeed an isometry this equality holds. –  Matt N. Dec 14 '12 at 15:57
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