Let $X$ be Banach space and $Y$ a closed subspace of $X$. Assume that there exist a closed "subset" $Z$ of $X$ with the properties:

$Z\cap Y=\{0\}$ and every $x\in X$ can be written in a unique form as $x=y+z$ with $y\in Y$ and $z\in Z$

Can we conclude that $Y$ is complemented in $X$?

Edit: I'm not asking if $Z$ is a complement of $Y$ in $X$. Indeed, $Z$ does not need to be linear. What I am asking is if we can find a set closed linear $W\subset X$ such that $W$ is a complement of $Y$ in $X$.

  • $\begingroup$ You'd want $Z\cap Y = \{0\}$, not $\emptyset$. Otherwise you'd have problems with $x \in Y$. My gut feeling says you can't conclude that $Y$ is complemented, but it has been wrong before. $\endgroup$ – Daniel Fischer Sep 12 '13 at 12:14
  • $\begingroup$ Yes @DanielFischer, let me fix it, thank you. $\endgroup$ – Tomás Sep 12 '13 at 12:16
  • $\begingroup$ I'm not sure about your question: are you asking if the definition of topologically complementary sets was weakened, then it continues to be true that $Z$ is a complement of $Y$? Otherwise the answer is "no" in general, because by definition $Z$ must be a (closed) linear subspace. $\endgroup$ – Federico Sep 12 '13 at 14:13
  • $\begingroup$ @Federico, Im using here the usual definition of a space being complemented. I will write it there $\endgroup$ – Tomás Sep 12 '13 at 14:26


Take $X = \mathbb R^2$ with your favorite norm (say $l^2$ norm). Let $Y$ be the $y$-axis, a closed linear subspace. Let $Z$ be the graph of a continuous function $h$ with $h(0)=0$. Say $h(x) = x^3$. Every point of $X$ is uniquely the sum of a point on the graph plus a point of $Y$. SO ... the set $Z$ need not be the complement.

Can we imitate this using $X$ a Banach space and $Y$ a non-complemented subspace? Make a function $h$ choosing one element from each equivalence class?

phrase this variant in a different way ... If $T : X \to U$ is a surjective continuous linear map of Banach spaces, and if there is a continuous section (that is, a continuous $V : U \to X$ with $V(T(u))=u$ for all $u$), then must there exist also a continuous linear section?

  • $\begingroup$ Sorry, I think there is a misundertanding here. This is my hypothesis, now I want to know if it is possible to find a complement (in the usual definition) of $Y$ in $\mathbb{R}$. In this case this is obvious. $\endgroup$ – Tomás Sep 12 '13 at 14:19
  • $\begingroup$ Yes that's my question, but I don't see what is the relation of your comment with your addendum. $\endgroup$ – Tomás Sep 12 '13 at 14:40
  • 2
    $\begingroup$ You do not gave an answer to my question and seems not interested in answer my comments. I think that if you want to make a comment, you can use the proper place to do it, which is not in the answer place. (-1). $\endgroup$ – Tomás Sep 12 '13 at 15:20

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