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What is an example of a closed subspace of a Banach space whose linear complement (direct sum decomposition) is not closed?

Martin Sleziak
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JoeDirt
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1 Answers1

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Note that "linear complement" is not unique. Anyway, if $L$ is a closed subspace of $X$ having a closed linear complement $M$ then $L$ is (topologically) complemented, i.e. there is a continuous linear projection onto $L$. This follows from the open mapping theorem: $L\times M\to X$, $(\ell,m)\mapsto \ell+m$ is a bijective continuous linear operator between Banach spaces hence its inverse is continuous.

An example of a closed subspace without closed complement is thus $c_0 \subseteq \ell^\infty$.

Jochen
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  • How do I see that the "linear complement" is not unique? – Siddharth Bhat Mar 31 '23 at 00:32
  • A linear complement of $L$ is another subspace $M$ with $L\cap M={0}$ and $L+M$ is the whole space. For the subspace $L=\mathbb R\times{0}$ of $\mathbb R^2$ every line through the origin and different from $L$ is a linear complement. – Jochen Mar 31 '23 at 06:51
  • Right. All of the subspaces of lines different from L are isomorphic, so it's unique up to iso? – Siddharth Bhat Mar 31 '23 at 10:12
  • Yes for closed complements in the case of Banach spaces. Every closed complement of $L$ is isomorphic to the quotient $X/L$. – Jochen Mar 31 '23 at 16:35
  • Can you give an example of an "open complement" that is not unique up to iso? – Siddharth Bhat Apr 01 '23 at 00:01
  • Subspaces are not doors: Non-closed subspaces need not be open. – Jochen Apr 01 '23 at 06:20