Let $X$ be a separable Banach space. Is this property equivalent to the approximation property?

There exists a chain $X_n$ of finite-dimensional subspaces of $X$, each being a range of some projection $P_n$, such that $\bigcup X_n$ is dense in $X$ and for every $x\in X$ we have $P_nx \to x$ as $n\to \infty$?

Does this depend on the choice of a chain of finite dimensional subspaces with dense union?

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    $\begingroup$ (For readers): Wiki: approximation property $\endgroup$
    – YCor
    Dec 9, 2019 at 16:53
  • $\begingroup$ This question is related to the $\pi$-property but seems (as far as I can see) weaker. $\endgroup$ Dec 9, 2019 at 17:56
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    $\begingroup$ @MatthewDaws: this seems equivalent to the $\pi$-property to me (by the uniform boundedness principle). $\endgroup$ Dec 9, 2019 at 18:30
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    $\begingroup$ To answer the second question: yes, it depends on the chain, as a space satisfying this for every chain is easily seen to have the property that there is a constant $C$ such that every finite dimensional subspace is $C$-complemented. This characterizes spaces isomorphic to Hilbert spaces. $\endgroup$ Dec 9, 2019 at 18:34

1 Answer 1


It seems to me that the property which you describe is equivalent to the $\pi$-property. An interesting example, showing that it does not follow from the approximation property was discovered by Charles Read (1958-2015). Unfortunately Charles never published this nice paper, but you can find it on my web page: http://facpub.stjohns.edu/ostrovsm/ReadDifferentFormsOfAP1989UnpublishedManuscript.pdf

The second question was already answered by Mikael de la Salle (see above).

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    $\begingroup$ P.S. If you mean the approximation property of Grothendieck, a counterexample was found much earlier in: Figiel, T.; Johnson, W. B. The approximation property does not imply the bounded approximation property. Proc. Amer. Math. Soc. 41 (1973), 197–200. $\endgroup$ Dec 9, 2019 at 21:39

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