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Suppose a (separable) Banach space $X$ has the following property: If $P:X\to X$ is a bounded projection different from $I$ such that $\|P\|=1$, then $\|I-P\|=1$. Does this imply that $X$ is a Hilbert space?

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    $\begingroup$ Do you know the answer for $\dim(X)=2$? $\endgroup$
    – YCor
    Nov 10, 2022 at 8:41
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    $\begingroup$ This might be a very silly question, but I guess you want to rule out the identity mapping as $P$, right? $\endgroup$
    – Hannes
    Nov 10, 2022 at 11:03
  • $\begingroup$ I don't know the answer for dimension 2. Indeed, I edited the question to exclude identity. $\endgroup$
    – Markus
    Nov 10, 2022 at 11:46
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    $\begingroup$ Somewhat related (perhaps): arxiv.org/abs/math/0110171 $\endgroup$ Nov 10, 2022 at 17:58
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    $\begingroup$ Dimension two is the general case since an affirmative answer for dimension two implies that if $X$ satisfies the condition, then all two dimensional subspaces are Hilbert spaces. $\endgroup$ Nov 17, 2022 at 20:38

1 Answer 1

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The answer is positive if $\dim X \geq 3$. Moreover it suffices to check the property for rank $1$ projections. See characterization (18.14) in the book [1].

However the characterization fails in dimension $2$. A rank $1$ projection is of norm $1$ iff every $x$ in the range and $y$ in the kernel are Birkhoff-orthogonal, which means that $$ \forall t \in \mathbf{R}, \, \, \, \|x\| \leq \|x+ty\| .$$ There are non-Euclidean $2$-dimensional spaces (known as "Radon planes", see [2]) where Birkhoff-orthogonality is a symmetric relation. Such spaces have the desired property.

[1] Amir, Dan, Characterizations of inner product spaces, Operator Theory: Advances and Applications, Vol. 20. Basel - Boston - Stuttgart: Birkhäuser Verlag. VII, 200 p. DM 72.00 (1986). ZBL0617.46030.

[2] Alonso, Javier; Martini, Horst; Wu, Senlin, On Birkhoff orthogonality and isosceles orthogonality in normed linear spaces, Aequationes Math. 83, No. 1-2, 153-189 (2012). ZBL1241.46006.

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