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I know from Show that $(l_1)^* \cong l_{\infty}$ that $l^∗_1$ is isometrically isomorphic to $l_{\infty}$. Indeed we can show that a map $L: l_{\infty} \rightarrow (l_1)^*$ given by $$L(x)(y) = \sum_{n \in \mathbb{N}}{x_ny_n}$$

where $y = (y_n)_{n \in \mathbb{N}} \in l_1$, is indeed an isometric isomorphism.

But my question is how to construct a natural map $l_1 → l_{\infty}^∗$ and show whether the map constructed is an isomorphism~ hmm

GNUSupporter 8964民主女神 地下教會
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Homaniac
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    The map $\tilde L:y\mapsto (x\mapsto L(x)(y) )$ is the "obvious choice" that embeds $\ell^1 \hookrightarrow (\ell^\infty)^* $ but the dual of $\ell^\infty$ is not isomorphic to $\ell^1$. A Banach limit (https://en.wikipedia.org/wiki/Banach_limit) is an element of $(\ell^\infty)^*$ that is not in the image of $\ell^1$ under this map. See also https://math.stackexchange.com/questions/47395/the-duals-of-l-infty-and-l-infty – Calvin Khor Mar 02 '19 at 22:32

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$\ell^{1}$ is separable and $({\ell^{\infty}})^{*}$ is not. So there cannot be an isomorphism between these. The natural map is $(a_n) \to f$ where $f(b_n)=\sum a_nb_n$.

Kavi Rama Murthy
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