I am learning functional analysis by myself and am confused about the definition of complementary subspaces of Banach spaces. In the book I use (Exercise 2.2.1 in "Analysis Now" by Gert Pederson), it seems complementary subspaces are defined to be closed subspaces. I somehow can show that even without closed-ness in the definition, complementary subspaces are closed.

Algebraically, complementary subspaces are the same as they are in the linear algebra \begin{equation} X = X_1 + X_2 \end{equation} and \begin{equation} X_1 \cap X_2 = \phi \end{equation} So any vector $x\in X$ has a unique decomposition $x=P_1 x + P_2 x$ with $P_1$ and $P_2=1-P_1$ being the projection operators. Define a new norm $\lVert x \rVert^{'}= \lVert P_1x \rVert + \lVert P_2x \rVert$, one can verify it is a norm and dominates the original norm $\lVert\rVert$. By the open mapping theorem two norms are equivalent and give the same topology. It's clear from the new norm $\lVert\rVert^{'}$ that the projection operators are bounded (continuous). Then the subspaces $X_1 = P_2^{-1}\{0\}, X_2=P_1^{-1}\{0\}$ are closed, being the preimage of a one-point (closed) set.

  • 2
    $\begingroup$ "By the open mapping theorem..." it doesn't work like that. The theorem is for Banach spaces. $\endgroup$
    – Jakobian
    Nov 27, 2023 at 9:58
  • $\begingroup$ Every subspace of a vector space has an algebraic complement. But not every subspace of a normed space must have a complement. Thus algebraic complementary subspaces of normed space aren't necessarily closed. So the assumption is important. $\endgroup$
    – Jakobian
    Nov 27, 2023 at 10:00
  • $\begingroup$ For example, $c_0$ is not complemented in $\ell^\infty$. $\endgroup$
    – Jakobian
    Nov 27, 2023 at 10:06
  • $\begingroup$ @Jakobian Thanks for the comment. The space $X$ with the new norm $\lVert\rVert^{'}$ must be Banach for me to use the open mapping theorem. If the subspaces are not closed, this is not guaranteed. $\endgroup$ Nov 27, 2023 at 10:08
  • $\begingroup$ Yes. Your new normed space is Banach iff the subspaces are closed. So you essentially used what you wanted to prove. $\endgroup$
    – Jakobian
    Nov 27, 2023 at 10:15


You must log in to answer this question.

Browse other questions tagged .