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let $(X,|| || )$ be a norm linear space. And $M$ be a closed subspace of norm linear space .does there exist a closed subspace $N$ such that $X=M \oplus N $ . I know such an subspace $N$ exist .but i am not conform about such an $N$ is closed or not .

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up vote 8 down vote accepted

In short, the answer is no in general. A well-known counterexample is given by $c_0$ in $\ell^\infty$. This was first proved by Phillips in 1940. There is a long history behind that, known as the complementary subspace problem for a Banach space $X$.

In 1971, Lindenstrauss and Tzafriri proved that given a Banach space $X$, every closed subspace is topologically complemented in $X$ if and only if $X$ is isomorphic to a Hilbert space.

Here are some details, to begin with a clarification of the notion of complement in a normed vector space, and in particular in a Banach space.

1) If two subspaces $M,N$, not necessarily closed, satisfy $X=M\oplus N$, we say that $M$ is algebraically complemented by $N$ in $X$. A stronger notion is that of topologically complemented. But these coincide in Banach spaces for pairs of closed subspaces.

If $X=M\oplus N$ algebraically, with $M,N$ subspaces of $X$, we have an isomorphism $T:M\times N\longrightarrow X$ by $T(x,y)=x+y$. Putting, say, the norm $\|(x,y)\|:=\|x\|+\|y\|$ on $M\times N$, and denoting by $P:x+y\longmapsto x$ the projection onto $M$ parallel to $N$, the following are equivalent:

  1. $T$ is a homeomorphism
  2. both $M$ and $N$ are closed and $P$ is bounded.

In this case, we say that $M$ is topologically complemented by $N$ in $X$.

If $X$ is a Banach space, and if $M,N$ are two closed subspaces of $X$, the closed graph theorem yields: $X=M\oplus N$ algebraically if and only if topologically.

2) So when $X$ is a Banach space, you are asking whether every closed subspace is topologically complemented. This is of course true when $X$ is a Hilbert space as it suffices to take $N=M^\perp$. Therefore it is true if $X$ is isomorphic to a Hilbert space. It is also true in a normed vector space with the additional assumption that $M$ has finite dimension of finite codimension.

A much more difficult result which appeared in 1971 after a long history of partial results is due to Lindenstrauss and Tzafriri: every infinite-dimensional Banach space which is not isomorphic to a Hilbert space contains a closed subspace which is not topologically complemented.

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Sorry for my error. Posting via smartphone is a humbling experience. –  Kyle Schlitt Jul 7 '13 at 23:53
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@Kyle You'll find a detailed treatment of the fact that c0 is not complemented in linfty in Albiac-Kalton, Topics in Banach Space Theory, GTM. Together with many other things. –  1015 Jul 8 '13 at 0:02

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