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Let $\mathcal{A}:X\to Y$ be continuous linear operator, $X$ and $Y$ are Banach spaces. Let $\text{Im} \mathcal{A}=Y$.

Is $\ker\mathcal{A}$ a complemented subspace of $X$?

sas
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1 Answers1

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No, not in general. For example, if $X$ is not isomorphic to a Hilbert space then $X$ contains a noncomplemented subspace $Z$. The quotient map $X \longrightarrow X/Z$ is surjective, continuous and linear, but the kernel (which is $Z$ of course) is not complemented.

Philip Brooker
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  • A minor issue: you probably need $Z$ closed here, as $Y$ should be Banach. – student Feb 28 '12 at 14:32
  • @Philip Brooker As far as I know this result is due to Lindenstrauss and Tzafriri. Given a normed space X which is not isomorphic to Hilbert space, how to construct non-complementable subspace? – Norbert Feb 28 '12 at 14:55
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    Probably the easiest example of a non-complemented subspace of a Banach space is $c_0 \subset \ell^\infty$. This short note by Whitley in the Monthly gives a very short proof. So you can take $Z = c_0$ and $X = \ell^{\infty}$ to give a specific example. – t.b. Feb 28 '12 at 16:20
  • No, I just wanted to get more or less constructive algorithm for arbitrary space $X$ which is not isomorphis to Hilbert space. Anyway thanks for example – Norbert Feb 28 '12 at 16:25
  • @Leandro: you are right that $Z$ is closed. However, in practice (i.e. in journal articles) Banach space theorists usually omit the word 'closed' and simply write 'subspace' whenever they mean 'closed subspace'; well, that is my experience anyway. – Philip Brooker Feb 28 '12 at 20:30
  • @Norbert, I do not know an algorithm as such off the top of my head, but then I can't remember the details of the Lindenstrauss-Tzafriri result, so I don't know if it is likely that such a thing exists. I'll have a look at it when I get a chance (quite busy today). – Philip Brooker Feb 28 '12 at 20:33
  • Thanks, for your attention. You will have some progress there I will make a distinct question, which you can answer. – Norbert Feb 28 '12 at 21:18
  • @Norbert: My previous comment wasn't directed at you :) Anyway, this crucial lemma in the proof of the result (from the old LNM 338 edition of Lindenstrauss-Tzafriri) amounts to such a recipe or "algorithm", I think: page 1, page 2. – t.b. Mar 01 '12 at 11:58
  • @t.b. Thanks for the reference! – Norbert Mar 01 '12 at 12:14
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    I would like to point you to another question which is related to this one: http://math.stackexchange.com/questions/2119896/can-every-closed-subspace-be-realized-as-kernel-of-a-bounded-linear-operator-fro If the restriction of this question to the case that $X=Y$ would result in a positive answer, this would result in a negative answer to the question I have linked above. @Norbert – Sebastian Bechtel Feb 01 '17 at 14:32