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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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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
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