Let $X$ be a Banach space and $P$ a projection. Show that the range of any projection is a closed subspace.
Can I use the fact that a Banach space is complete and thus closed and that $P = P^2$ to show that the range of $P$ is in fact closed?
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1Why isn't the answer accepted? – ViktorStein Mar 19 '20 at 22:51
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See https://math.stackexchange.com/questions/1148842/projection-in-hilbert-space-onto-non-closed-subspace?rq=1 – NessunDorma Aug 02 '20 at 12:51
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If the projection is continuous, then the range of $P$ is closed, as it is the kernel of the continuous projection $I-P$.
However not all projections on a Banach space are continuous (as opposed to what happens in the case of a finite dimensional space). So, in general, this is not true. Take for instance the subspace generated by $\{(1,0,\dots), (0,1,0, \dots), \dots\}$ in $\ell^1(\mathbb N)$. You can then extend the set $\{(1,0,\dots), (0,1,0, \dots), \dots\}$ to a basis of $\ell^1(\mathbb N)$ using the Axiom od Choice and then project onto the subspace generated by those elements. As the subspace (which is the range of such projection) has a countable basis, it can't be Banach, hence it can't be closed.
Silvia Ghinassi
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Which projection on that subspace? There is really no difference between Banach and Hilbert spaces here. A projection on a non-closed subspace would not be continuous. But examples of non-continuous linear operators on Banach spaces require the Axiom of Choice. – Robert Israel Jan 12 '16 at 22:19
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@RobertIsrael Yeah, indeed you use the axiom of choice to extend that set to a basis of $X$ and then project. And sorry, I meant finite dimensional, my bad. – Silvia Ghinassi Jan 12 '16 at 22:21