I'm trying to show that the dual of Banach space $l^1$ is isometrically isomorphic $l^{\infty}$. I've defined a linear map $F: (l^1)^* \to l^{\infty}$ by $F(y)(x) = \sum x_n y_n$. So far I've shown that this map is well-defined and linear in x and y. Now I'm trying to show that it is norm-preserving.
I can easily show that $|F(y)| \le \|y\|_{\infty} \|x\|_1$ as this is immediate from the definition. I'm now trying to show the other direction of the inequality.
So far I have: Suppose $M$ is such that $|F(y) (x)| \le M\|x\|_1$. I want to try and choose a particular sequence $x$ so that $M$ must necessarily be greater than $\|y\|_{\infty}$ and then the inf over all $M$ must then be greater than $\|y\|_{\infty}$ and I will have the result. The only thing I have tried so far is the sequence $X = (1,1,...1,0,...)$ with 1's up to the nth place but unfortunately I cannot get this to work.
Thanks
Firstly, for any $y=\{y_1, y_2, \cdots\} \in l^{\infty}$, $f_y(x) = \sum_{n=1}^{+\infty}x_ny_n$ for $x= \{x_1, x_2,\cdots\} \in l^1$ is an element in the dual space, and $\|f_y\|_{(l^1)^*} = \|y\|_{l^\infty}$(take $e_n = \{0,0,\cdots, 0,1,0,\cdots\}\in l^1$ where the $1$ is in the $n$th position to check it )
Secondly, for any $g \in (l^1)^*$, define $y = \{g(e_1), g(e_2), \cdots\}$, it's obvious that $y \in l^\infty$(since $\|y\|_{l^\infty} \leq \|g\|_{(l^1)^*}$) and $g= f_y$, so we conclude $(l^1)^* = l^\infty$.
Finally it's easy to check $\|f_y - f_{y'}\|_{(l^1)^*} = \|y-y'\|_{l^\infty}$ since $f_y - f_{y'} = f_{y-y'}$ and we repeat the first step, so it's isometric