Isometric to Dual implies Hilbertable?

Solution 1:

Here's a construction:

Take any reflexive Banach space $X$. Take the direct sum $E = X \oplus X^{\ast}$ and equip it with the norm $\|(x,x^{\ast})\|_E = \left(\|x\|_{X}^2 + \|x^{\ast}\|_{X^{\ast}}^2\right)^{1/2}$. This space is usually denoted by $E=X \oplus_2 X^{\ast}$ for short.

Then $E$ is isometric to its dual space $E^{\ast} = X^{\ast} \oplus_2 X^{\ast\ast}$: An isometric isomorphism is given by $(x,x^{\ast}) \mapsto (x^{\ast},i_{x})$ where $i:X \to X^{\ast\ast}$ is the canonical inclusion (by reflexivity of $X$ the map $i$ is an isometric isomorphism).

The space $E$ won't be isomorphic to a Hilbert space unless $X$ is itself isomorphic to a Hilbert space.

In fact, a (real) Banach space is “Hilbertable” if and only if every closed subspace has a complement by a (very deep) Theorem of Lindenstrauss and Tzafriri.

So: If $X$ is not isomorphic to a Hilbert space, it has a subspace that isn't complemented, and thus so has $E$.