Let $(X,\|\cdot \|)$ be a Banach space over the field $\mathbb{K}$. Assume there exists a bounded, coercive bilinear form $b:X\times X \to \mathbb{K}$. Is $(X,\|\cdot \|)$ necessarily a Hilbert space?
I, so far, can get only that there exists a Hilbert space $(X,\|\cdot \|^\prime, b^\prime )$ where the norm $\|\cdot \|^\prime$ is equivalent to the original one $\|\cdot \|$ ($b^\prime$ is just obtained by symmetrizing $b$).
It seems clear to me how to proceed in case $(X,\|\cdot \|^\prime, b^\prime )$ is separable via orthonormal bases, but the case that it is not separable should be covered as well.
The question Banach Space with coercive bilinear form is a Hilbert Space deals with the same question but it stops at the equivalent norm and not at the same norm.
Best Answer
Certainly not. As a trivial counterexample, take any norm on $\mathbb{R}^2$ which fails the parallelogram law, e.g. the $\ell^1$ norm. This is not a Hilbert space, but the bilinear form $b$ defined by the Euclidean inner product is bounded and coercive; you can prove this directly, or from the fact that all norms on $\mathbb{R}^2$ are equivalent.