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.
[Math] Subspaces of finite codimension in Banach spaces
banach-spaces
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I believe the answer to your first question is no. The counterexample I have in mind is related to the peculiar fact (first proved by Enflo, Lindenstrauss and Pisier) that being isomorphic to a Hilbert space isn't a "three-space property". More specifically, there is an example of a Banach space $E$ with an uncomplemented subspace $F$ such that both $F$ and $E/F$ are isomorphic to $\ell^2$. You can also find examples where $F$ and $E/F$ are isomorphic to any $\ell^p$. This is due to Kalton and Peck.
The answer to the first question is "no". You can see this with specific examples, but here is a more conceptual approach: Take $Y$ an uncomplemented subspace of $X$ and in $Z:= X\oplus_1 X$ identify $Y\oplus 0$ with $0 \oplus Y$ in the obvious way; that is, mod out from $Z$ the subspace $\{(y,-y) | y \in Y\}$. $X\oplus 0$ and $0 \oplus X$ are norm one complemented in the resulting quotient space of $Z$ but their intersection $Y \oplus 0 = 0 \oplus Y$ is not complemented. (This is just a categorical push out construction specialized to the appropriate category of Banach spaces.)
The answer is yes if the subspaces are norm one complemented and the space $X$ is uniformly convex. This is intuitive, because if $P$ is a norm one projection on a uniformly convex space and $x$ is not in the range of $P$, then $\|Px\| < \|x\|$, since otherwise all vectors on the line segment from $x$ to $Px$ would have norm $\|x\|$. Hence one guesses that playing ping pong with two norm one projections $P$ and $Q$ will produce a norm one projection onto $PX\cap QX$. To see that this works without doing any computations or calculating rates of convergence (at the risk of making experts cringe), set $P_1=P$, $P_{2n} = QP_{2n-1}$, $P_{2n+1}=PP_{2n}$. Let $x\in X$ and let $a=a(x)$ be the limit of the nonincreasing sequence $\|P_n x\|$. I claim that $\|P_{n+1}x - P_{n}x\| \to 0$. Indeed, (1/2)$\|P_{n+1}x + P_{n}x\|$ also converges to $a$, so the claim follows from the uniform convexity of $X$. Let $V$ be a limit in the weak operator topology of some subnet of $P_{2n}$. By the claim, the corresponding subnet of $P_{2n+1}$ also converges to $V$ in the weak operator topology. From this it is evident that $V$ is a norm one projection onto $PX\cap QX$.
ADDED 13 Jan. 2012: Notice that in the first construction $X$ can be uniformly convex, in which case $Z$ (and therefore also every quotient of $Z$) is isomorphic to a uniformly convex space.
Best Answer
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).