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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?

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re: Leonid's comment; Another isomorphic characterisation of Hilbert spaces is that a Banach space $X$ is isomorphic to a Hilbert space if and only if every closed linear subspace of $X$ is complemented (that is, is the range of a continuous linear projection on $X$). I believe this result is due to Lindenstrauss and Tzafriri. Another result along these lines is that a separable infinite dimensional Banach space $X$ is isomorphic to $\ell_2$ if and only if every infinite dimensional closed subspace of $X$ is isomorphic to $X$. I believe that this result is due to Tim Gowers. –  Philip Brooker Jan 9 '10 at 2:02
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Characterizing Hilbert spaces isomorphically is a very interesting topic in Banach space theory. Another one is that every nuclear operator on the space has absolutely summable eigenvalues. Open is whether a Banach space all of whose subspaces have an unconditional basis must be isomorphic to a Hilbert space. A non characterization is that there are Banach spaces non isomorphic to a Hilbert space all of whose subspaces have a Schauder basis. –  Bill Johnson Jan 9 '10 at 8:14
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I wonder if the algebra $\mathcal{B}(X)$ of all bounded linear operators on the Banach space $X$ is a $C*$-algebra with the operator norm if and only if $X$ is isometrically isomorphic to a Hilbert space. There are many isomorphic variants one could ask in this direction too. On a related note, the Eidelheit theorem ($\mathcal{B}(X)$ and $\mathcal{B}(Y)$ are isomorphic as Banach algebras if and only if $X$ and $Y$ are isomorphic as Banach spaces) gives an isomorphic characterisation of Hilbert spaces, though admittedly it is probably not easy to check. –  Philip Brooker Mar 31 '10 at 3:46
    
Have only just noted this question from Philip Brooker, since the original question was bumped by a new answer. I think I've seen a proof (in work of Daws) that if E and F are Banach spaces and we have a closed-range unital homomorphism from A(E) into B(F), then E is isomorphic to a weakly complemented subspace of F. If I have remembered this correctly, then it would answer the natural isomorphic variant of Philip's question. –  Yemon Choi Mar 2 '11 at 10:14
    
Yemon, thanks for bringing that to my attention. –  Philip Brooker Mar 22 '11 at 23:58

7 Answers 7

up vote 15 down vote accepted

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

A real Banach space $(X, \|\cdot\|)$ 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

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For the really lazy among us: google.com/search?q=%22is+a+hilbert+space+if%22+banach –  Tom LaGatta Mar 31 '10 at 5:24

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

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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:

  1. 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);
  2. 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).

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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]

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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.

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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.

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There is a book by Amir from 1986 on this very topic. It lists hundreds of results.

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