27
$\begingroup$

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?

$\endgroup$
  • 8
    $\begingroup$ 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
19
$\begingroup$

Try the following article "A Survey of the Complemented Subspace Problem": http://m.mathnet.or.kr/mathnet/kms_tex/986009.pdf

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).

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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. $\cap_n\ker f_n=\{0\}$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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