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I am looking for an example of two closed subspaces of a Banach space, such that their sum is not closed.

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2 Answers 2

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Simple examples can be obtained as follows: Let $E$ and $F$ be Banach spaces and suppose $T: E \to F$ is a bounded linear operator with non-closed range. Then $X = E \oplus 0$ and $Y = \operatorname{Graph}(T) = \{(e,Te)\,:\,e \in E\}$ are closed subspaces of $Z = E \oplus F$ with $X + Y$ not closed: If $Te_n \to f \in F \smallsetminus T(E)$ then $(0,Te_n) \in X + Y$ but $(0,f)$ isn't.

For an explicit example, take $E = \ell^1$, $F = \ell^2$ and $T: E \to F$ the obvious inclusion.


Added: It is not too hard to check that for $X$ finite-dimensional or of finite co-dimension the space $X + Y$ is always closed.

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  • $\begingroup$ @Jan: You're welcome. I added a little bit of information to my answer. $\endgroup$
    – t.b.
    Jun 3, 2011 at 18:06
  • $\begingroup$ Very nice!! :-) $\endgroup$ Apr 28, 2012 at 2:26
  • $\begingroup$ Dear Sir Nice Solution.I donot understand last example I know that $l^1\subset l^2$ so inclusion map is welldefined but both spaces are closed why range of map is not closed I donot understand .Please can you help me $\endgroup$ Aug 30, 2019 at 16:18
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Exercise 1.20, p. 40 in Rudin FA shows this for a Hilbert space.

BTW, whenever N,F are subspaces of a TVS, N closed and F finite-dimensional, N+F is closed. This is Theorem 1.42 (p. 30) in Rudin FA.

Rudin: Functional Analysis, 2nd edition, 1991. (Its TVSs are Hausdorff.) https://archive.org/details/RudinW.FunctionalAnalysis2e1991

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