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Let $X$ be a Banach space and consider the canonical embedding in it's bidual $X^{**}$, namely $j:X\to X^{**}, \; x\mapsto j(x)$, where $j(x)(x^*)=x^*(x)$ for $x^*\in X^*$.

My question: Why is $j(X)$ weak*-dense in $X^{**}$ for $X=l^1(G)$, where $G$ is a locally compact group (or is true for arbitrary Banach spaces $X$ as well)? (1)

To prove this, for all $u\in X^{**}$ one has to find a sequence $(v_n)_n\subseteq j(X)$ such that $v_n(x^*)\to u(x^*)$ for all $x^*\in X^*$, and $v_n=j(x_n)$ for a $x_n\in X$ (since $(v_n)_n\subseteq j(X)$ ).

I only know Goldstine ( https://en.wikipedia.org/wiki/Goldstine_theorem ):If $B\subseteq X$ is the closed unit ball, then $j(B)$ is weak*-dense in $B^{**}$, where $B^{**}$ is the closed unit ball in $X^{**}$.

So, in the case $\|u\|_{op}>1$, is it possible to consider $\hat{u_n}:=\frac{1}{\|u\|_{op}}u_n$ and reduce to Goldstine, so that (1) is true?

Or how to justify (1)?

Daniel Buck
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    Be careful: In general, density does not imply sequential density! The weak$^$-density of $j(X)$ in $X^{}$ follows from the Hahn-Banach theorem because the dual of $(X^{},\sigma(X^{},X^{}))$ is $X^$ (more precisely, every element of that dual is the evaluation in some element of $X^$). – Jochen Sep 20 '16 at 07:14
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    @ Jochen Could you please elaborate it? – Anupam May 11 '17 at 10:04

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