Every finite-dimensional subspace is one-complemented

Let $X$ be a Banach space. It is known that if every closed subspace of $X$ is one-complemented, then $X$ is isometrically isomorphic to a Hilbert space.

Now if every finite-dimensional subspace of $X$ is one-complemented, is it true that is $X$ isometrically isomorphic to a Hilbert space?

• Perhaps you should explain (or link to an explanation of) what "one-complemented" means. – Omnomnomnom Feb 10 '15 at 12:19
• @Omnomnomnom I suppose one-complemented means "complemented by projection of norm one". – Norbert Feb 10 '15 at 22:06

Yes, this is true and you may restrict yourself to two-dimensional subspaces! That is, $X$ is isometric to a Hilbert space if and only if every two-dimensional subspace is 1-complemented. This is due to Kakutani (1939) in the real case, and Bohnenblust (1941) in the complex case.

References:

P. Bohnenblust, A characterization of complex Hilbert spaces, Portugaliae Math., 3 (1942), 103–109.

S. Kakutani, Some characterizations of Euclidean space, Japan. J. Math. 16 (1939), 93–97.

• Good references! – Norbert Nov 26 '15 at 16:01

It is known that if every closed subspace of X is one-complemented, then X is isometrically isomorphic to a Hilbert space.

This is true for spaces of dimension at least three, right? So, the argument of Norbert does not work. You can not prove that a two dimensional vector space is a Hilbert space by proving that all the one dimensional subspaces are one-complemented.

I never heard of result you stated. I take it for granted.

Thanks to the parallelogram law it is enough to show that any two-dimensional subspace is a Hilbert space. Now take any two-dimensional subspace $E$ of $X$. From assumption any $1$-dimensional subspace of $E$ is one-complemented in $X$ and a fortiori in $E$. Therefore, from result you stated in the beginning of the question we get that $E$ is a Hilbert space.

• No need to take it for granted :-) – Tomek Kania Nov 26 '15 at 13:36