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We know that any projection on a closed subspace of a Hilbert space is bounded. Is it true for any Banach space? Any help would be appreciable.

Shailesh
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  • Remark: Even in a Hilbert space, not every projection onto a closed subspace is continuous (if the dimension is infinite), but orthogonal projections are. In a Hilbert space, there exist continuous projections onto every closed subspace, which doesn't hold in general Banach spaces. – Daniel Fischer Aug 11 '13 at 14:18

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If the projection $P \colon E \to F$, where $E$ is Banach and $F$ a closed subspace of $E$, is continuous (bounded), then we have the decomposition

$$E \cong \ker P \oplus F.$$

Thus a necessary condition for the existence of a continuous projection onto a closed subspace $F$ is that $F$ is complemented. That condition is of course also sufficient, if $E \cong F \oplus G$ with a closed subspace $G$, the projection along $G$ is continuous.

Not all closed subspaces in a Banach space are complemented, in general, hence in general, a continuous projection onto $F$ need not exist.

Daniel Fischer
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  • When F is complemented, is the continuity of the projection along G trivial? – dezdichado Dec 07 '15 at 09:34
  • That depends on what you can build on, @Eza. It is clear that the projection $\pi_F \colon F \oplus G \to F,; (f,g) \mapsto f$ is continuous (where the norm on $F\oplus G$ can be any of the standard product norms, $\lVert (f,g)\rVert = \max {\lVert f\rVert, \lVert g\rVert}$, $\lVert (f,g)\rVert = \lVert f\rVert + \lVert g\rVert$, $\lVert (f,g)\rVert = \bigl(\lVert f\rVert^p + \lVert g\rVert^p\bigr)^{1/p}$). Then the continuity of the projection along $G$ in $E$ follows from the fact that $E \cong F \oplus G$, where $\cong$ means topological isomorphism. – Daniel Fischer Dec 07 '15 at 09:51
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    If we spell it out, with the topological isomorphism $\alpha \colon E \to F \oplus G$ and the injection $\iota_F \colon f \mapsto (f,0)$, the projection to $F$ along $G$ in $E$ is $\alpha^{-1} \circ \iota_F \circ \pi_F \circ \alpha$, which, as a composition of continuous maps, is continuous. – Daniel Fischer Dec 07 '15 at 09:59
  • But $F \oplus G=E$ already has a norm, the original norm, and it may not coincide with any of the standard norms on direct sums. Why would it be equivalent to one of those for an arbitrary complemented subspace $F$? That is what is needed to show that $\alpha$ is, indeed, a topological isomorphism. – Conifold Aug 30 '23 at 00:18