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Let $c$ denote the space of convergent sequences in $\mathbb C$, $c_0\subset c$ be the space of all sequences that converge to $0$. Given the uniform metric, both of them can be made into Banach spaces. It can be shown that the dual spaces of them are isometrically isomorphic, i.e. $c^*\cong c_0^*$. Are $c$ and $c_0$ isometrically isomorphic? If not, how can one show the absence of such a isometric isomorphism? Thanks!

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That's a nice question. I must admit that I never asked this myself. After thinking for a bit and getting nowhere, I've looked in some of the obvious places, but I couldn't find a reference addressing it. Pełczyński and Bessaga mention that $c_0$ is not isometrically isomorphic to any space $C(K)$ with $K$ compact see here (last paragraph before 10.9 on page 261) but they give no hint at the proof. It might be proved in H. Elton Lacey's book, but I don't have a copy handy. – t.b. Nov 10 '11 at 4:43
Not an isometry, but $c$ and $c_0$ are isomorphic in an obvious way: Let $(a_n)_{n\in\mathbb N}$ be a convergent sequence of complex numbers. We denote its limit by $a_\ast = \lim_{n \to \infty} a_n$. We obtain an operator $F: c \to c_0$ with $F(a_n) = (a_\ast, a_1 - a_\ast, a_2 - a_\ast, \dots)$. A routine calculation shows $F$ to be linear and continuous with $\Vert F \Vert = 2$. On the other hand we have $G: c_0 \to c$ with $G(a_n) = (a_1 + a_2, a_1 + a_3, a_1 + a_4, \dots)$. Again a routine calculation shows that $FG = id$, $GF = id$ and $\Vert G \Vert = 2$. – Alexander Thumm Nov 10 '11 at 6:53
Alexander's comment above says that the Banach-Mazur distance $d_{BM}(c,c_0)$ of $c$ and $c_0$ is no larger than $4$, where $d_{BM}(X,Y) = \inf \Vert T\Vert\cdot \Vert T^{-1}\Vert$, where the infimum is taken over the set of all isomorphisms of $X$ onto $Y$ (if this set is empty - that is, $X$ is not isomorphic to $Y$ - we write $d_{BM}(X,Y) =\infty$). In general, getting an upper bounds on $d_{BM}(X,Y)$ for isomorphic $X$ and $Y$ is fairly easy. Getting a nontrivial lower bound when $X$ and $Y$ are not known to be isometrically isomorphic seems to be quite difficult in general. It just so... – Philip Brooker Nov 10 '11 at 12:45
happens that the precise value of $d_{BM}(c,c_0)$ is known and equal to $3$; this is due to Michael Cambern; see On mappings of sequence spaces, Studia Math. 30 (1968), 73--77. On a further tangential point Cambern showed in another paper that if $K$ and $L$ are compact Hausdorff spaces with $d_{BM}(C(K),C(L))<2$, then $K$ and $L$ are homeomorphic - hence $d_{BM}(C(K),C(L))=1$. For this, see On isomorphisms of small bound, Proc. Amer. Math. Soc. 18 (1967), 1062--1066. – Philip Brooker Nov 10 '11 at 13:04
up vote 33 down vote accepted

The closed unit ball of $c_0$ has no extreme points. The closed unit ball of $c$ has many extreme points, such as $(1,1,\ldots)$. Since the property of being an extreme point is preserved by isometries, $c$ and $c_0$ are not isometrically isomorphic.

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Of course! I knew that I must have missed something very easy... – t.b. Nov 10 '11 at 10:11
Thanks very much for your nice and simple answer! – Yang Zhou Nov 10 '11 at 15:57
Nice and shocking proof! +1 – Riccardo Nov 28 '13 at 14:35

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