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!

  • 9
    $\begingroup$ 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. $\endgroup$ – t.b. Nov 10 '11 at 4:43
  • 5
    $\begingroup$ 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 $\Vert G \Vert = 2$. $\endgroup$ – Alexander Thumm Nov 10 '11 at 6:53
  • 4
    $\begingroup$ 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... $\endgroup$ – Philip Brooker Nov 10 '11 at 12:45
  • 9
    $\begingroup$ 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. $\endgroup$ – Philip Brooker Nov 10 '11 at 13:04
  • $\begingroup$ A recent question on MathOverflow about this: $c_0$ is not isometrically isomorphic to $c$. $\endgroup$ – Martin Sleziak May 17 '18 at 14:09

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.

  • $\begingroup$ Of course! I knew that I must have missed something very easy... $\endgroup$ – t.b. Nov 10 '11 at 10:11
  • $\begingroup$ Thanks very much for your nice and simple answer! $\endgroup$ – YZhou Nov 10 '11 at 15:57
  • 2
    $\begingroup$ Nice and shocking proof! +1 $\endgroup$ – Riccardo Nov 28 '13 at 14:35

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.