When is a Banach space a Hilbert space?

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?

• 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
• 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
• 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

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

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

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).

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]

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.

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 (Inner Product Structures: Theory and Applications By V.I. Istratescu), the other you can hopefully get at your library (Characterizations of Inner Product Spaces by Dan Amir).

From the point of view of manifolds and curvature the following result is valid:

A Banach space is a Hilbert space if and only if it is a NPC (non-positive curvature) space. http://www.iam.uni-bonn.de/fileadmin/WT/Inhalt/people/Karl-Theodor_Sturm/papers/paper41.pdf

S. Kwapien proved in "Isomorphic characterizations of inner product spaces by orthogonal series with vector valued coefficients" (Studia Mathematica 44, 1972) that the Banach spaces in which the vector valued Khintchine inequality is valid, are Hilbert spaces. To be more precise, a Banach space $X$ is isomorphic to a Hilbert space if and only if $X$ has type $2$ and cotype $2$, i.e., if there are constants $c,C>0$ such that for all $n\in\mathbb N$ and all $x_1,\dots,x_n\in X$ $$c \Big(\sum_{i=1}^n\|x_i\|^2\Big)^{1/2} \leq \bigg(\int_0^1 \Big\| \sum_{i=1}^nr_i(t)x_i \Big\|^2 \bigg)^{1/2} \leq C \Big(\sum_{i=1}^n\|x_i\|^2\Big)^{1/2}$$