Let $X, Y$ be Banach spaces s.t. $X^{*}\simeq Y^{*}$. Is it true that $X\simeq Y$? This is clearly true for finite dimensional spaces, but I can't certain that this holds for infinite dimensional case since $X^{**}$ is not isomorphic to $X$ in general.
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1This may be of interest: https://mathoverflow.net/questions/77383/a-separable-banach-space-and-a-non-separable-banach-space-having-the-same-dual-s – Math Helper May 10 '18 at 03:39
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The answer is no.
Consider the Banach spaces $c_0$ and $c$.
We have $\ell^1 \cong (c_0)^*$ via the map
$$(a_n)_{n=1}^\infty \in \ell^1\mapsto f \in (c_0)^* \quad\text{ given by }\quad f((x_n)_{n=1}^\infty) = \sum_{n=1}^\infty a_nx_n, \forall(x_n)_{n=1}^\infty \in c_0 $$
Also, we have $\ell^1 \cong c^*$ via the map
$$(a_n)_{n=0}^\infty \in \ell^1\mapsto f \in c^* \quad\text{ given by }\quad f((x_n)_{n=1}^\infty) = a_0 \cdot \left(\lim_{n\to\infty} x_n\right) + \sum_{n=1}^\infty a_nx_n, \forall(x_n)_{n=1}^\infty \in c $$
However $c_0 \not\cong c$, which is shown here.
mechanodroid
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Nice. I assume $c$ is the space of bounded sequences with a limit, right? – Adrián González Pérez May 11 '18 at 11:41
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@AdriánGonzález-Pérez Yes, $c$ is the space of convergent sequences, and $c_0$ is the space of sequences converging to $0$. – mechanodroid May 11 '18 at 15:03