Let $(X, \|\cdot\|)$ be a Banach space, $Y \subset X$ a closed subspace. $Y$ is called complemented if there is a closed subspace $Z \subset X$ such that $X =Y \oplus Z$ as topological vector spaces.

Show that if a closed subspace $Y$ is not complemented, then the identity map on $Y$ can not be extended to a continuous linear function on $X$.

Hi, I think I should do this by contradiction, but I don't know how, I wish you could help me! Thank you so much!


1 Answer 1


Suppose that the identity map $\iota\colon Y\to X$ can be extended to a bounded linear operator $P\colon X\to Y$. Then $P$ is an idempotent. Indeed, for $x\in X$ we have $Px\in Y$ and so $PPx=\iota Px=Px$. Thus, $P$ is a bounded projection with ${\rm im}\, P = Y$. Let $Z = {\rm im}\, (I_X - P)$. Then $Z$ is closed and $X=Y\oplus Z$ as $I_X = P + I_X - P$.

  • $\begingroup$ Thank you for answering, but why is Z closed? $\endgroup$
    – user392559
    Mar 12, 2017 at 2:50
  • 1
    $\begingroup$ I've already noticed that it's because Z is the kernel of P, thank you so much! $\endgroup$
    – user392559
    Mar 12, 2017 at 3:51

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