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.

1$\begingroup$ See mathoverflow.net/questions/28415 in particular Nate's ticked answer. $\endgroup$– Robin ChapmanCommented Jul 7, 2010 at 10:34

15$\begingroup$ 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. $\endgroup$– Laurent BergerCommented Jul 7, 2010 at 10:56

3$\begingroup$ Nice exercise, Laurent. It embarrasses me to learn something about Banach spaces from a number theorist. :) $\endgroup$– Bill JohnsonCommented Jul 7, 2010 at 13:28

$\begingroup$ This is in Rudin's FA, Chapter 2 or 3, isn't it? $\endgroup$– Yemon ChoiCommented Jul 7, 2010 at 19:18

1$\begingroup$ @Ayoub Certainly not; just take the kernel of any discontinuous linear functional for a counterexample. (In future, please do not use answer boxes to ask questions. I'm converting this to a comment under the question.) $\endgroup$– Todd Trimble ♦Commented Jan 13, 2019 at 19:21
6 Answers
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 $IT$ will be a projection onto $F$ (unless I've messed something up, which is possible).

$\begingroup$ 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". $\endgroup$ Commented Jul 7, 2010 at 20:08

1$\begingroup$ A couple of comments to make the second paragraph more transparent (The author did express some doubt). Let $F'$ denote the span of $\{y_k\}$. Since $\{x_k\}$ is a basis, one finds that $F' \cap F = \{0\}$. For each $x \in E$ we have $x +F = \sum x_i^*(x+F)\cdot x_i$ and hence there exists $f \in F$ so that $x = f + \sum x_i^*(x+F)\cdot y_i$. Thus $E=F+F'$. $\endgroup$ Commented Jul 4, 2017 at 21:07
No: If you consider a non continuous functional from a Banach Space, its Kernel is onecodimensional 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 banachspace) 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.
As already recalled, a kernel of any noncontinuous linear form is a dense hyperplane, and noncontinuous 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 $RL$.
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.

$\begingroup$ +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 nonseparable case? $\endgroup$ Commented Jul 7, 2010 at 12:25

$\begingroup$ Pete: Just use my comment (or rpotrie's) about noncontinuous 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, discontinuous functional on $\ell^2$, for example). $\endgroup$ Commented Jul 7, 2010 at 12:32

$\begingroup$ @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. $\endgroup$ Commented Jul 7, 2010 at 12:57
It is equi to Choice axiom. If you do not like the Axiom, then every linear functional is continuous. If Not, then HahnBanach theorem is true.
It is up to you.
Oleg Reinov

7$\begingroup$ Not "equivalent" just an implication one way. In ZF, HahnBanach is strictly weaker than Axiom of Choice. $\endgroup$ Commented Jul 12, 2010 at 11:26
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) onecodimensional 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.

2$\begingroup$ 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 innerproduct space (dense in $L^2[1,1]$ for example) and then trying to pick an orthogonal projection: indeed, this does not exist. $\endgroup$ Commented Jul 7, 2010 at 11:18

$\begingroup$ you are right. This is a functional which is not of inner product with any element in the preHilbert space (C[1,1] considered with the $l^2$ norm). $\endgroup$ Commented Jul 7, 2010 at 12:35