When a Banach space is a Hilbert space? - MathOverflow

When a Banach space is a Hilbert space?

2010-01-08T21:46:10Z

Let $\mathcal{X}$ be a Banach space. It is a well known fact that $\mathcal{X}$ is a Hilbert space (i.e. the norm comes from an inner product) if the parallelogram identity holds.

Question: Are there other (simple) characterizations for a Banach space to be a Hilbert space?

Answer by Steve Huntsman for When a Banach space is a Hilbert space?

2010-01-08T22:05:02Z

From http://www.springerlink.com/content/731n45405715p2m3/

a real Banach space (X, ‖ · ‖) is a Hilbert space if and only if for any three points A,B,C of this space not belonging to a line there are three altitudes in the triangle ABC intersecting at one point.

Many other references show when Googling

"is a hilbert space if" banach

Answer by Jonas Meyer for When a Banach space is a Hilbert space?

2010-01-08T22:16:56Z

Yes, there are many (simple) characterizations of when a normed space is an inner product space. Here are two book references, one with Google preview, the other you can hopefully get at your library.

Answer by Ady for When a Banach space is a Hilbert space?

2010-03-31T03:07:08Z

Just two isometric/isomorphic characterizations:

A Banach space $X$ is [isometric to] a Hilbert space if and only if there exists a Banach space $Y$ and a symmetric bilinear mapping $f:X\times X\rightarrow Y$ satisfying

$||f(x,z)||$ $=$ $||x||\cdot||z|$| for all $x,z$ $\in$ $X$.

[J. Becerra Guerrero & A. Rodriguez-Palacios]

A Banach space is [isomorphic to] a Hilbert space iff it is uniformly homeomorphic to a Hilbert space. [Per Enflo]

Answer by BigBill for When a Banach space is a Hilbert space?

2010-04-08T16:27:16Z

Bessaga and Pelczynski wrote a survey on Banach spaces. The chapter 4 is devoted to this question.

http://matwbn.icm.edu.pl/ksiazki/or/or2/or214.pdf

Answer by Garrisi Daniele for When a Banach space is a Hilbert space?

2011-03-02T08:30:14Z

More characterisations are in the book of Haim Brezis (Analyse fonctionnelle), at the appendix of Chapter 5. I will copy two of these below, toghether with the references:

If $ \dim(E)\geq 2 $ and every subspace $ X\subset E $ of dimension $ 2 $ is the image of a bounded projector $ P $ such that $ \|P\| = 1 $, then $ E $ is isometric to a Hilbert space (Kakutani, Japanese Journal of Mathematics, 1939);
if $ \dim(E)\geq 3 $ and the map $ T $, defined as the identity on the unit ball and as $ u/\|u\| $ when $ \|u\|\geq 1 $, is lipschitzian with constant $ 1 $, then $ E $ is isometric to a Hilbert space (de Figueiredo; Karlovitz, Bulletin of the American Mathematical Society, 1967).

Also, if $ E $ is isomorphic to all its infinite-dimensional subspaces, then it is isomorphic to a separable Hilbert space (Gowers, Annals of Mathematics, 2002).

Answer by Valerio Capraro for When a Banach space is a Hilbert space?

2011-10-07T09:50:28Z

In this simple note http://arxiv.org/abs/0907.1813 (to appear in Colloq. Math.), Rossi and I proved a characterization in terms of "inversion of Riesz representation theorem".

Here is the result: let $X$ be a normed space and recall Birkhoff-James ortogonality: $x\in X$ is orthogonal to $y\in X$ iff for all scalars $\lambda$, one has $||x||\leq||x+\lambda y||$.

Let $H$ be a Hilbert space and $x\rightarrow f_x$ be the Riesz representation. Observe that $x\in Ker(f_x)^\perp$, which can be required using Birkhoff-James orthogonality:

Theorem: Let $X$ be a normed (resp. Banach) space and $x\rightarrow f_x$ be an isometric mapping from $X$ to $X^*$ such that

1) $f_x(y)=\overline{f_y(x)}$

2) $x\in Ker(f_x)^\perp$ (in the sense of Birkhoff and James)

Then $X$ is a pre-Hilbert (resp. Hilbert) space and the mapping $x\rightarrow f_x$ is the Riesz representation.