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.