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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 $X$?

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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.

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  • $\begingroup$ A minor issue: you probably need $Z$ closed here, as $Y$ should be Banach. $\endgroup$ – student Feb 28 '12 at 14:32
  • $\begingroup$ @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? $\endgroup$ – Norbert Feb 28 '12 at 14:55
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    $\begingroup$ 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. $\endgroup$ – t.b. Feb 28 '12 at 16:20
  • $\begingroup$ 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 $\endgroup$ – Norbert Feb 28 '12 at 16:25
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    $\begingroup$ I would like to point you to another question which is related to this one: math.stackexchange.com/questions/2119896/… 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 $\endgroup$ – Sebastian Bechtel Feb 1 '17 at 14:32

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