# “Almost” Hilbert spaces

This question is a bit (very?) vague. Is there some notion of how "close" a Banach space is to being a Hilbert space?

What I have in mind is something like a real or complex valued function on (equivalence classes) of Banach spaces, which gives $0$ for all Hilbert spaces and some non-zero value for other spaces, which tells you how close it is to being a Hilbert space. Of course, one could always put some arbitrary function, but is there something 'natural'?

For instance, is $L^3$ somehow closer to being a Hilbert space than $L^{10}$, or $L^\infty$?

• I've never heard of one, but the first thing that springs to mind is how close the parallelogram law is to holding, perhaps taking some sort of supremum of $$2(\|x\|^2 + \|y\|^2) - (\|x + y\|^2 + \|x - y\|^2)$$ over a suitable set of pairs $x,y$. – Matt Rigby Sep 15 '14 at 18:35
• For example, we could assume $||x||=||y||=1$ (otherwise the supremum is $\in \{0,\infty\}$). – Martin Brandenburg Sep 15 '14 at 18:37
• In the spirit of the above, there is a characterization of Hilbert spaces in terms of their modulus of smoothness (example: books.google.com/… Corollary 2.7.10). This is a bit like asking does the unit ball of your Banach space have the right shape to be the unit ball of a Hilbert space. – Tom Cooney Sep 15 '14 at 20:03
• You mentioned equivalence classes: equivalence up to isometry? or up to isomorphism? – user147263 Sep 15 '14 at 22:32
• I meant isometry, not isomorphism. – user56914 Sep 15 '14 at 23:55

Generally, people tend to think of distances between normed spaces in multiplicative terms, because it fits the way how composition of operators works. That is, the smallest value of the distance is $1$ and the triangle inequality has multiplication instead of addition. If you don't like this, take the logarithm.
The Banach-Mazur distance does not directly answer your question, since it is infinite whenever the Banach space $X$ is not isomorphic to a Hilbert space. But one can restrict consideration to all $n$-dimensional subspaces of $X$, and take supremum of their BM distances to a Euclidean space. For example, in Banach space projections and Petrov-Galerkin estimates by Ari Stern one finds this supremum for $n=2$, called the Banach-Mazur constant of $X$, denoted $C_{BM}(X)$. The author notes:
There are various other such “geometric constants” for normed vector spaces; see Kato and Takahashi [Some recent results on geometric constants of Banach spaces] for a survey of recent results on several of these constants. Generally, these constants lie between $1$ and $2$, equaling $1$ in the case of an inner product space, and equaling $2$ for the most pathological spaces, such as non-reflexive spaces. One of the oldest and best-known is the von Neumann–Jordan constant, dating to the 1935 paper of Jordan and von Neumann [On inner products in linear, metric spaces] (see also Clarkson [The von Neumann–Jordan constant for the Lebesgue spaces]), which measures the degree to which the norm satisfies (or fails to satisfy) the parallelogram law.
The von Neumann-Jordan constant of $X$ is defined as $$C_{NJ}(X) = \sup \left\{\frac{\|x+y\|^2+\|x-y\|^2}{2(\|x\|^2+\|y\|^2)}: \|x\|+\|y\|>0 \right\}$$ The aforementioned paper of Clarkson has the proof that $C_{NJ}(L^p)=2^{2/\min(p,p')-1}$ where $p'=p/(p-1)$. In particular, $$C_{NJ}(L^2)=1, \quad C_{NJ}(L^3)=2^{1/3}, \quad C_{NJ}(L^{10})=2^{4/5}$$
Theorem 3.4 of Stern's paper shows that $C_{NJ}(X)\le C_{BM}(X)$ for every $X$.