Let $X=(X,\|\cdot\|)$ be a Banach space and suppose that $F\subset X$ is a finite-dimensional subspace. There is then an equivalent norm $|\cdot|$ on $F$ such that $|\cdot|$ is induced by an inner product on $F$ (i.e. $|\cdot|$ will satisfy the parallelogram law) and it follows that \begin{equation} c_{(F,|\cdot|)}|x|\leq\|x\|\leq C_{(F,|\cdot|)}|x| \end{equation} for some constants $c,C>0$ and for all $x\in F$. Here is my main question: is there a name for the following property?

There exists $M\geq 1$ such that for every finite-dimensional subspace $F\subset X$, there is an equivalent norm $|\cdot|$ on $F$ that is induced by an inner product on $F$ and is such that $1\leq \frac{C}{c}\leq M$.

Clearly, any Hilbert space has this property by taking $M=1$ and $|\cdot|=\|\cdot\|$. Are there examples of non-Hilbert spaces that have this property? Is this property related somehow to the type/cotype of $X$?

  • $\begingroup$ The property has no quantifier on $c,C$. Probably you define $c,C$ as functions of $(F,|\cdot|)$ in the first paragraph but you could be more explicit. $\endgroup$
    – YCor
    Dec 4, 2021 at 17:08
  • $\begingroup$ Yes, this is exactly what I meant. $\endgroup$
    – JWP_HTX
    Dec 4, 2021 at 17:12
  • $\begingroup$ This is finite representability. Please see the definition in this question: mathoverflow.net/questions/151758/… $\endgroup$
    – Onur Oktay
    Dec 4, 2021 at 17:48
  • $\begingroup$ See also books.google.ca/… $\endgroup$
    – Onur Oktay
    Dec 4, 2021 at 17:55

1 Answer 1


Your condition implies that $X$ is isomorphic to a Hilbert space with isomorphism constant at most $M^2$. The distance condition implies that both type 2 constant and cotype 2 constant of $X$ is bounded by $M$. By Kwapien theorem the Banach-Mazur distance of $X$ to a Hilbert space is bounded by type 2 constant times cotype 2 constant.

As a reference see e.g., Albiac-Kalton.

  • $\begingroup$ Got it. I had a feeling something like this was the case. Thanks, Bunyamin $\endgroup$
    – JWP_HTX
    Dec 4, 2021 at 18:05
  • 2
    $\begingroup$ That is a perfectly fine answer, but using Kwapien's Theorem is a bit of an overkill. It is an elementary compactness argument that the condition implies that X is isomorphic to a Hilbert space with isomorphism constant at most M. $\endgroup$ Dec 4, 2021 at 22:19

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