In this post, all vector spaces are assumed to be real or complex.

Let $(X, ||\cdot||)$ be a Banach space, $Y \subset X$ a closed subspace. $Y$ is called $\underline{\mathrm{complemented}}$, if there is a closed subspace $Z \subset X$ such that $X =Y \oplus Z$ as topological vector spaces.

If $H$ is a Hilbert space every closed subspace $Y$ is complemented; the orthogonal complement $Y^{\bot}$ is a closed subspace of $H$ and we have $H=Y \oplus Y^{\bot}$. A famous theorem of Lindenstrauß and Tzafriri (which can be found in their article "On the complemented subspaces problem", Isreal Journal of Mathematics, Vol. 9, No.2, pp. 263-269) asserts that the converse is true as well. More precisely, if $(X, ||\cdot||)$ is a Banach space such that every closed subspace is complemented then $||\cdot||$ is induced by a scalarproduct, i.e. $(X,||\cdot||)$ is a Hilbert space.

Now to my question. Can you give me an example of a Banach space $(X,||\cdot||)$, which is not a Hilbert space, and of a closed subspace $Y \subset X$ which is not complemented? It is easily seen that $Y$ must be both infinite-dimensional and infinite-codimensional, for every finite-dimensional and every (closed) finite-codimensional subspace is complemented.

I thought about something like $c_{0} \subset (\ell^{\infty}, ||\cdot||_{\infty})$ the closed subspace of null sequences in the Banach space of bounded sequences but couldn't produce a proof that no closed complement exists in that case. Can you help me either proving that $c_{0}$ is not complemented (if that's true at all) or by giving me a different example?

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    $\begingroup$ Show that a complement to $c_0$ contains a sequence of bounded linear functionals which separates points, while $\ell^\infty / c_0$ doesn't. $\endgroup$ – Mark Feb 11 '12 at 22:23
  • $\begingroup$ So you mean when I consider $\ell^{\infty}$ as the dual of $\ell^{1}$? I will try this one out. Thanks! I might come back to you, when I'm stuck. $\endgroup$ – Nils Matthes Feb 12 '12 at 13:46

Try the following article "A Survey of the Complemented Subspace Problem": https://arxiv.org/abs/math/0501048v1

Your suspicion about $c_0$ is correct. A couple of other examples: The disc algebra (those functions in $C(\mathbb{T})$ which are restrictions of functions analytic in the open unit disc) is closed in $C(\mathbb{T})$ but not complemented. Similarly, in $L^1(\mathbb{T})$, the subspace $H^1(\mathbb{T})$ consisting of functions whose negative Fourier coefficients vanish is closed but not complemented. See Rudin's Functional Analysis (the proof isn't very easy).

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    $\begingroup$ Looks like a good overview of this circle of problems. Thanks for sharing it with me (us)! $\endgroup$ – Nils Matthes Feb 12 '12 at 21:37

The article

Robert Whitley, Projecting $m$ onto $c_0$, The American Mathematical Monthly, Vol. 73, No. 3 (Mar., 1966), pp. 285-286

provides a short proof that $c_0$ is not complemented in $\ell^{\infty}$ by showing that $\ell^{\infty}/c_0$ does not have a countable set $f_n$ of continuous linear functions isolating zero (i.e. $\cap_n\ker f_n=\{0\}$).


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