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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 $l_p\subset l_2$.

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?

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 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.

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  • $\begingroup$ In the question "positive definite" has to be expunged, right? $\endgroup$ Jul 28 '14 at 21:06
  • $\begingroup$ Well, $E$ has to be positive definite anyway. $\endgroup$
    – Conifold
    Jul 28 '14 at 21:24
  • $\begingroup$ A doubt: what's your definition of "positive definite"? Thank you. $\endgroup$ Jul 28 '14 at 21:56
  • $\begingroup$ $E(x,x)\geq0$ for any $x\in X$ as in the linked paper. $\endgroup$
    – Conifold
    Jul 28 '14 at 23:26
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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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  • $\begingroup$ Thanks. Can you recommend references on symmetric bilinear/quadratic forms on Banach spaces? Those that study the order especially. $\endgroup$
    – Conifold
    Jul 28 '14 at 23:41
  • $\begingroup$ There are for instance these papers: ARONSZAJN, Quadratic forms on vector spaces, and LANCE Quadratic forms on Banach spaces; though I do not know if they go in the direction you want. $\endgroup$ Jul 29 '14 at 9:38

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