This is a question about characterizing Hilbert spaces in terms of quadratic forms. Let $X$ be a real Banach space and $E$ a bounded quadratic form on it, it is called positive definite if $E(x,x)\geq0$ for all $x\in X$. If moreover $E(x,x)\geq a\|x\|^2$ for some $a>0$ then $E(x,x)^\frac12$ is an equivalent norm on $X$ and $X$ is isomorphic to a Hilbert space. On the other hand if $E$ is only strictly positive definite, i.e. $E(x,x)>0$ for $x\neq0$, then this need not be the case. For example, $E(x,x)=\sum\_{i=1}^\infty x\_i^2$ is a bounded strictly positive form on $l\_p$ for $1\leq p<2$ because $l\_p\subset l\_2$. 



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> **Question:** Suppose that on a real Banach space $X$ there exists a bounded positive definite quadratic form $E$ which is an order unit for quadratic forms, i.e. for any bounded quadratic form $Q$ on $X$ $E(x,x)\geq b\,|Q(x,x)|$ for some $b>0$. Does it follow that $X$ is isomorphic to a Hilbert space? Does it follow if $X$ is additionally assumed reflexive?
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Of course, the question can be rephrased equivalently in terms of symmetric positive definite operators $X\to X^\*$ since $Q(x,x)=\langle Kx,x\rangle$ for such an operator. It is proved [here](http://arxiv-web3.library.cornell.edu/abs/0805.4721) that $X$ is isomorphic to a Hilbert space if and only if there exists a positive definite symmetric isomorphism $K:X\to X^\*$. They seem to claim in Remark 3.2 that isomorphism can be weakened to injective when $X$ is reflexive, but that contradicts the example above with $K:l\_p\to l\_q$ being the inclusion.



Positive definite quadratic forms on Banach spaces
If the bounded 2-form $E$ satisfies $E(x,x)\ge b |Q(x,x)|$ for any bounded 2-form $Q$ and for some $b>0$ depending on $Q$, then actually $E(x,x)\ge a\|x\|^2$ for some $a >0$. 


Indeed, suppose the latter does not hold. Then for some sequence $(x\_n)\_n\subset X$ there holds $E(x\_n,x\_n)<4^{-n}\|x\_n\|^2$. Let, for any $n\in\mathbb{N}$, $x^\*\_n\in X^\*$ be a norm $1$ linear form such that $\langle x^\*\_n,x\_n\rangle=\|x\_n\|$, given by the Hahn-Banach theorem. Consider the bounded symmetric 2-form $Q(x,y):=\sum\_n 2^{-n}\langle x^\*\_n,x \rangle\langle x^\*\_n,y\rangle$. Then $Q(x\_n,x\_n)\ge 2^{-n}\|x\_n \|^2$ and 
$E(x\_n,x\_n)<2^{-n}Q(x\_n,x\_n)$, so that $E$ does not satisfy the former property either
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