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A little while ago, I asked this question and it all got sorted out. But, now I am asked to show that $J$ is not onto, if $\Omega = \mathcal{l}^1$. We haven't learned anything about reflexive spaces, nor that $(\mathcal{l}^1)^* = \mathcal{l}^\infty$. The only hint that I was given was that I should show that the the Riesz Representation Theorem doesn't hold when $p = 1$, by using the following:

Let $(\Omega,||\cdot|| )$ be a normed space, and $x_0 \in \Omega$, and $x_0^* \in \Omega^*$. Then, $$||x_0|| = \sup\{ |\langle x_0, x^* \rangle| : ||x^*||_* \leq 1\}$$ and, $$||x_0^*||_* = \sup\{ |\langle x, x_0^* \rangle| : ||x|| \leq 1\}$$ and that the supremum is attained.

This seems like what I will have to end up doing is showing that $(\mathcal{l}^1)^* = \mathcal{l}^\infty$ and proceeding from there. However, that is hard for me to imagine, as most problems I am given are not so difficult.

I get that we are suppose to show that for some $y \in (\mathcal{l}^1)^{**}$ there does not exists an $x \in \mathcal{l}^1$ for which $y = \langle x, x^* \rangle$. But, what is the strategy to do this? How does the hint I was given play a role?

Martin Sleziak
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I don't think there is a really elementary proof that $J$ is not onto.

It is well-known that one cannot write explicitly functionals in the dual of $\ell^\infty$ that do not come from $\ell^1$.

The usual argument to show that the dual of $\ell^\infty$ is not $\ell^1$ goes by looking at separability.

Finally, I have no idea how to use the suggestion you were given. The norm in a space can be calculated both by using the dual and the predual, so I don't see how it could notice that in the double dual you are using $J(\ell^1)$ and not all of it.

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Martin Argerami
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