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There are a lot of questions on this topic see for example Show that $(c_{0})'$ and $(c)'$ are isometrically isomorphic. . So I´m not interested in a proof rather I wonder how to think about the result. If I understand it correctly both $c_0^*$ and $c^*$ are isomorphic to $l^1$. I find this very counter intuitive as $c_0$ and $c$ are not isomorphic but rather $c$ is isomorphic to $c_0 + \mathbb{R}$ . This would mean that $(c_0 + \mathbb{R})^*\simeq c_0^*$

In light of this my question is how to think about dual spaces. How is it possible that $c_0^*$ and $c^*$ are isomorphic or am I misunderstanding the answer in the link above?

Erik
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  • In fact $c_0$ is isomorphic to $c_0\oplus \mathbb R$ essentially because of the Hilbert Hotel Paradox https://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel – Ruy Dec 26 '20 at 15:05
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    I assume that Erik means that the duals are isometrically isomorphic, while the base spaces aren't. – Martin Argerami Dec 26 '20 at 17:19

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Because the way you phrase the question is a bit confusing, let me clarify that there is no isometric isomorphism between $c_0$ and $c$; but there is a Banach space isomorphism.

Regardless, it is very intuitive that the dual of $c_0$ and $c_0+\mathbb R$ is the same. After all $c_0+\mathbb R$ is simply the unitization of $c_0$ when seen as a Banach algebra.

Given any $\psi\in (c_0+\mathbb R)^*$, we have $\psi(a+\lambda)=\psi(a)+\lambda\,\psi(1)$. It is not hard to check that $$ \|\psi\|=\|\psi|_{c_0}\|+|\psi(1)|. $$ This shows that we can see $(c_0+\mathbb R)'$ as the norm-1 direct sum $c_0^*\oplus_1\mathbb R$. So what we are saying is that $\ell^1\oplus_1\mathbb R\simeq\ell^1$ isometrically. Which is trivial to check with the isometric isomorphism $(a,\lambda)\longmapsto (\lambda,a_1,a_2,\ldots)$.

Martin Argerami
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  • Martin, do you mean that the dual of any non-unital Banach algebra is isometrically isomorphic to the dual of the unitization? – Ruy Dec 26 '20 at 22:34
  • I would tend to say no, but I really don't know. It is for a C$^*$-algebra, because you have approximate units. The canonical way to do the unitization for a Banach algebra is to use the norm $|(a,\lambda)|=|a|+|\lambda|$, which in general I would assume does not play nice with taking the dual, and I don't immediately see how to do something like the argument with $\ell^1$. – Martin Argerami Dec 26 '20 at 23:17