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I want to prove the following fact:

$C[0,1]$ doesn't contain a complemented subspace which is isomorphic to $l^1$

Here is the definition of complemented subspaces.

All I can do with this problem is to look at the dual spaces. Assume $C[0,1]$ contains a complemented subspace, say $X\cong l^1$, such that $C[0,1] \cong X \oplus Y$. By taking dual on both sides, we have $C[0,1]^* \cong X^* \oplus Y^*$ and we can apply Riesz–Markov–Kakutani representation theorem on $C[0,1]^*$. It seems that such an approach does't simplify this problem at all.

Every hint, solution or reference will be appreciated!

Community
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No One
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    Your approach does simplify the problem: the space $M[0,1]=C[0,1]^*$ has cotype 2, whereas $\ell^\infty$ does not; so $\ell^\infty$ cannot be isomorphic to a subspace of $M[0,1]$. – Etienne Oct 28 '15 at 22:30
  • @Etienne Thank you! But this is really the first time I see the concept of cotype. Could you please show me some references in which those results are proved?($\ell^\infty$ does not have cotype 2 and $M[0,1]$ has the cotype 2, if they are hard to prove) – No One Oct 28 '15 at 22:48
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    For $\ell^\infty$, this is easy. For $M[0,1]$, observe that any finite-dimensional subspace of $M[0,1]$ is isometric a subspace of some $L^1(\mu)$ space, and use the fact that $L^1(\mu)$ has cotype 2 (which is proved in any book on Banach space theory, for example Kalton-Albiac). – Etienne Oct 29 '15 at 08:54
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    It's shown in the introductory article of the Handbook of the Geometry of Banach Spaces that an infinite dimensional complemented subspace with an unconditional basis of a $C(K)$ space is isomorphic to $c_0$. – David Mitra Oct 29 '15 at 09:32
  • This is even better! – Etienne Oct 29 '15 at 09:49
  • @TomekKania Sorry about my late acceptance~~~~Thank you so much! – No One Nov 28 '15 at 18:21

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Note that $C[0,1]$ is an $\mathscr{L}_\infty$ space, then so does all its complemented subspaces. On the other hand $\ell_1$ is an $\mathscr{L}_1$-space. It is remains to recall that no infinite dimensional Banach space can be $\mathscr{L}_1$ and $\mathscr{L}_\infty$-space at the same time.

To learn more on $\mathscr{L}_p$-spaces see New classes of $\mathscr{L}_p$-spaces. J. Bourgain.

Norbert
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If $\ell_1$ were complemented in $C[0,1]$, $\ell_\infty$ would be isomorphic to a (complemented) subspace of $C[0,1]^*$. The dual space $C[0,1]^*$ is however weakly sequentially complete being isomorphic to an uncountable $\ell_1$-sum of $L_1$'s. Certainly $\ell_\infty$ is not weakly sequentially complete.

Tomasz Kania
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  • Thank you so much! But could you provide me some references for the fact you mentioned about $C[0,1]^*$? – No One Nov 03 '15 at 22:51
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    This is explained in Albiac and Kalton's Topics in Banach space theory. See also http://www.ams.org/journals/tran/1967-126-02/S0002-9947-1967-0206732-8/S0002-9947-1967-0206732-8.pdf – Tomasz Kania Nov 04 '15 at 18:38
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The purpose of this answer is to provide an alternative point of view for new visitors rather than providing an answer for the question owner for their question that was asked 6 years ago.

$C(K)$ has the property (V) of Pelczynski, and so does every quotient of it. If there existed a projection $P:C(K)\to\ell^1$, then $\ell^1$ would have property (V). This is a contradiction, since $\ell^1$ does not have property (V).

Definitions: A bounded linear operator $T:X\to Y$ between two Banach spaces is called to fix a copy of $c_0$ if there exist two subspaces $V\subseteq X$, $W\subseteq Y$, both of which are isomorphic to $c_0$ as Banach spaces, such that $T:V\to W$ is a Banach space isomorphism.

A Banach space $X$ is said to have the property (V) of Pelczynski if every bounded linear operator $T:X\to Y$ into another Banach space $Y$ is weakly compact whenever $T$ does not fix a copy of $c_0$.

$\ell^1$ doesn't have property (V): $\ell^1$ contains no subspace isomorphic to $c_0$. Thus, the identity map $I:\ell^1\to\ell^1$ doesn't fix a copy of $c_0$. If $\ell^1$ had property (V), then $I$ would be weakly compact, so the unit ball of $\ell^1$ would be weakly compact, and thus $\ell^1$ would be reflexive.

Onur Oktay
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