Is the set of positive definite bilinear forms open in the set of continuous bilinear maps? Let $(E,||\cdot||)$ be a real Banach space, denote by $B = \mathcal{L}(E\times E, \mathbb{R})$ the set of continuous bilinear forms. B is a Banach space with the norm $||\varphi|| = \sup_{||x||\leq 1, \ ||y||\leq 1} |\varphi(x,y)|$.
The set  of (continuous) positive definite bilinear forms is 
$P= \{\varphi\in B\  | \ \forall x\in E\setminus\{0\} \ \varphi (x,x) > 0 \}\subset B$.
If E is finite dimensional then P is an open set in B, I wonder if this is the case also in infinite dimension, if not there are particularly important hypothesis that could make it open?
Thank you.
 A: Consider the case where $E=\ell_1$, and denote $x=(x_i)_{i=1}^\infty$ and $y=(y_i)_{i=1}^\infty$ whenever $x,y\in\ell_1$. Define
$$\phi_n(x,y)=\sum_{i=1}^ni^{-1}2^{-i}x_iy_i\;\;\;\text{ and }\;\;\;\phi(x,y)=\sum_{i=1}^\infty i^{-1}2^{-i}x_iy_i.$$
Then each $\phi_n\in P^c$ (the complement of $P$) and $\phi\in P$. On the other hand,
$$\|\phi-\phi_n\|=\sup|\sum_{i=n+1}^\infty i^{-1}2^{-i}x_iy_i|\leq(n+1)^{-1}\to 0,$$
where the "sup" is taken over all $x,y\in\ell_1$ with $\|x\|,\|y\|\leq 1$. Hence $\phi_n\to\phi$. It follows that the set $P^c$ is not closed, and hence $P$ is not open.
EDIT.
More generally, if $E$ is any separable, infinite-dimensional Banach space, we can find a normalized sequence $(e_i^*)_{i=1}^\infty\subset E^*$ which separates the points of $E$, i.e. such that for every $x\in E$ there is $i\in\mathbb{N}$ with $e_i^*(x)\neq 0$.  (This is because every separable, infinite-dimensional Banach space admits a Markushevich basis.)  Now define
$$\varphi_n(x,y)=\sum_{i=1}^n\frac{e_i^*(x)e_i^*(y)}{i2^i}\;\;\;\text{ and }\;\;\;\varphi(x,y)=\sum_{i=1}^\infty\frac{e_i^*(x)e_i^*(y)}{i2^i}.$$
Then $\varphi_n\in P^c$ and $\varphi\in P$ with $\varphi_n\to\varphi$ as above.
Perhaps one could find a counter-example when $E$ is nonseparable, for example if $E=\ell_\infty$.
