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Is every finite codimensional subspace of a Banach space closed? Is it also complemented? I know how to answer the same questions for finite dimensional subspaces, but couldn't figure out the finite codimension case.

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See in particular Nate's ticked answer. – Robin Chapman Jul 7 '10 at 10:34
Here's a fun exercise: take $E$ to be a Banach space and let $H$ be a codimension one subspace. Prove that $H$ is dense in $E$ if and only if $E \setminus H$ is arcwise connected. – Laurent Berger Jul 7 '10 at 10:56
Nice exercise, Laurent. It embarrasses me to learn something about Banach spaces from a number theorist. :) – Bill Johnson Jul 7 '10 at 13:28
This is in Rudin's FA, Chapter 2 or 3, isn't it? – Yemon Choi Jul 7 '10 at 19:18
up vote 9 down vote accepted

It's a standard result that a linear functional from a Banach space to the underlying field (real or complex numbers) is continuous if and only if the its kernel is closed. Notice that its kernel is of codimension one. So, use the axiom of choice to find a discontinuous linear functional, and you have found a codimension one subspace which isn't closed. (As I was typing this, rpotrie got the same answer...)

As for complementation: well, this only makes sense for closed finite codimension subspaces. But then it's a perfectly reasonable question, and the answer is "yes". If F is of finite codimension in E, then by definition we can find a basis $\{x_1,\cdots,x_n\}$ for E/F. For each $k$ let $x_k^*$ be the linear functional on $E/F$ dual to $x_k$, so $x_k^*(x_j) = \delta_{jk}$. Then let $\mu_k$ be the composition of $E \rightarrow E/F$ with $x_k^*$. Finally, pick $y_k\in E$ with $y_k+F=x_k$. Then the map $$T:E\rightarrow E; x\mapsto \sum_k \mu_k(x) y_k$$ is a projection of $E$ onto the span of the $y_k$, and $I-T$ will be a projection onto $F$ (unless I've messed something up, which is possible).

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to be precise, the question about complementation makes sense for any linear subspace, why. I would rather say: "a complemented linear subspace is necessarily closed". – Pietro Majer Jul 7 '10 at 20:08

No: If you consider a non continuous functional from a Banach Space, its Kernel is one-codimensional and dense.

For example take $l^2(\mathbb{Z})$ and consider the sequence $e_i$ ($(0,..., 1, 0....)$ where the $1$ is in the $i$-th position). Complete this to a base (which exists by Zorn's lemma, and it is uncountable since $l^2$ is a banach-space) and consider the subspace generated by the $e_i$ toghether with the elements of the base except one of them. This gives a dense codimension one subspace.

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This is very related to rpotries construction: take a dense, proper subspace and pick a basis $(v_i)$ of that subspace. Now we can adjoin $(w_j)$ such that $(v_i) \cup (w_j)$ is a basis of the whole space. Now the span of all $v_i,w_j$ but finitly many $w_j$ is a dense subspace of finite codimension. So it can not be closed.

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+1: This seems to be a distillation of the core content of the previous answers. (In fact, after reading them I came to the same conclusion.) Moreover: by the Baire Category Theorem, an algebraic (or "Hamel") basis of an infinite dimensional Banach space is uncountable, whereas -- by definition! -- a separable Banach space admits a dense subspace of countable dimension. So nonclosed codimension one subspaces exist in every separable Banach space. Can someone address the non-separable case? – Pete L. Clark Jul 7 '10 at 12:25
Pete: Just use my comment (or rpotrie's) about non-continuous linear functionals: that works in any Banach (or indeed, otherwise) space. Sure, you need to use the Axiom of Choice a bit, but that's true in the separable case as well (try to write down an everywhere defined, dis-continuous functional on $\ell^2$, for example). – Matthew Daws Jul 7 '10 at 12:32
@MD: You're right. I didn't stop to think (and am not expert on Banach spaces) so didn't see that it is obvious that any infinite dimensional Banach space admits an unbounded linear functional: just choose an infinite linearly independent set $\{e_n\}$ of norm one elements, define $L(e_n) = n$ and extend to the whole space by AC. – Pete L. Clark Jul 7 '10 at 12:57

As already recalled, a kernel of any non-continuous linear form is a dense hyperplane, and non-continuous forms exist in infinite dimension as a consequence of the existence of Hamel basis. That said, it's worth recalling a relevant fact in the affirmative direction, which is a corollary of the open mapping theorem:

A linear subspace in a Banach space, of finite codimension, and which is the image of a Banach space via a linear bounded operator, is closed.

Btw, the property of being complemented has also a particular characterization for those subspaces that are images of operators: the image of $R:X\to Y$ is complemented if and only if $R$ is a right inverse, meaning that there is a bounded operator $L:Y\to X$ such that $LR=I$. A linear projector onto the subspace it is then $RL$.

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Another counterexample: Take $C[-1,1]$ as the Banach Space. Define $ T(f)=\int_{0}^{1} f - \int_{-1}^{0} f $. Clearly $T$ is a (continuous) linear functional whose kernel is a (closes) one-codimensional subspace. But there is no complementary subspace: in fact, the complementary subspace should morally be generated by the function which is identically $1$ on $[0,1]$ and $-1$ on $[-1,0]$, but this function is not continuous.

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I don't agree: pick some function g with T(g)=1. (g(x)=x+1 works I think). Then define a map $P:C[-1,1]\rightarrow C[-1,1]$ by $P(f) = f - T(f)g$. If $T(f)=0$ then $P(f)=f$, while also $TP(f) = T(f) - T(f)T(g) = 0$ for all $f$. So $P$ is a projection onto the kernel. It seems like you are trying to treat $C[-1,1]$ as an inner-product space (dense in $L^2[-1,1]$ for example) and then trying to pick an orthogonal projection: indeed, this does not exist. – Matthew Daws Jul 7 '10 at 11:18
you are right. This is a functional which is not of inner product with any element in the pre-Hilbert space (C[-1,1] considered with the $l^2$ norm). – Keivan Karai Jul 7 '10 at 12:35

It is equi to Choice axiom. If you do not like the Axiom, then every linear functional is continuous. If Not, then Hahn-Banach theorem is true.

It is up to you.

Oleg Reinov

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Not "equivalent" just an implication one way. In ZF, Hahn-Banach is strictly weaker than Axiom of Choice. – Gerald Edgar Jul 12 '10 at 11:26

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