Closed subspace of Banach space [duplicate]

Let S closed subspace of E, where E is Banach space.

(1): Is there a decomposition of E in direct sum, E = S $\oplus$ N ?

(2): If (1) holds, we have that N has linear homeomorphism with E/S.

I know to prove (2), but I don't know if (1) is true.

Others questions:

a) In case that (1) is false, if S has finite dimension, it can be true (1)?

b) In case that (1) is false. If E/S has finite dimension, where E/S has linear homeomorphism with Y, is there a decomposition of E in direct sum, E = Y $\oplus$ S ?

Thanks!

(In my question, I don´t ask for complemented, i.e., my N is not necessarily closed.)

For (1), it has a negative answer. Suppose it's true, so for all S closed subspace of X, Banach space, we have that exists N subspace such that E = S $\oplus$ N. But, by this argument A question about complement of a closed subspace of a Banach space, N is closed. So, every closed subspace is complemented, it's a contradiction with this Example of a closed subspace of a Banach space which is not complemented? .

For (a), I found a positive answer, here: Does there exist a Banach space with no complemented closed subspaces? , we have that "Finite-dimensional subspaces are always complemented; this follows from the Hahn–Banach theorem.".

But (b) remains open. I saw this If the quotient of a subspace of a banach space is finite, is it a closed subspace? , however it's a different question.

marked as duplicate by copper.hat, Milo Brandt, user147263, user228113, LeucippusJan 13 '16 at 0:47

• If (1) holds, then $N$ is the kernel of bounded linear operator, hence closed. – user147263 Jan 13 '16 at 2:48
• I'm assuming that the decompositions are supposed to be topological, i.e. that $E = S \oplus N$ means that the map $\alpha\colon (s,n) \mapsto s+n$ is a topological isomorphism $S\times N \to E$. We call $S$ a complemented subspace of $E$ if there is a subspace $N$ such that $E = S \oplus N$, and any such $N$ is called a complement of $S$, and is naturally itself a complemented subspace of $E$. If $E$ is a Hausdorff TVS, then every complemented subspace $S$ is closed, since $S$ is the image of $S\times \{0\}$ under the homeomorphism $\alpha$, and $S\times \{0\}$ is closed. – Daniel Fischer Jan 14 '16 at 12:21
• In a Haudorff locally convex TVS, so in particular in a normed space (completeness plays no role), a) every finite-dimensional subspace is complemented, and b) every closed subspace $S$ of finite codimension is complemented. In the situation of b), every algebraic complement of $S$ is a topological complement. Proof of b): If $N$ is an algebraic complement of $S$, then the map $\alpha \colon S\times N \to E$ is a continuous linear isomorphism. Let $\pi \colon E \to E/S$ be the quotient map. Then $\pi\lvert_N \colon N\to E/S$ is a linear isomorphism (also continuous). – Daniel Fischer Jan 14 '16 at 12:21
• Since $E/S$ is a finite-dimensional Hausdorff TVS, the linear isomorphism $(\pi\lvert_N)^{-1}\colon E/S\to N$ is continuous. Thus the map $p_N \colon E \to N,\; p_N = (\pi\lvert_N)^{-1}\circ \pi$ is continuous. It's also a projection, $p_N\circ p_N = p_N$, since $\pi \circ (\pi\lvert_N)^{-1} = \operatorname{id}_{E/S}$. Then $\beta \colon E \to S\times N,\; \beta(x) = (x - p_N(x), p_N(x))$ is continuous. But $\beta = \alpha^{-1}$, so $\alpha$ is indeed a topological isomorphism between $S\times N$ and $E$. – Daniel Fischer Jan 14 '16 at 12:22