Let $E$ be a separable real Hilbert space. Let $B : E \times E \to \mathbb{R}$ be a non-negative definite bilinear form, that is, a symmetric bilinear function such that $B(e,e) \geq 0 $ for every $e \in E$. Suppose in addition, that $B$ satisfies a finite-trace type condition in the sense: $$ \sum_{n \in \mathbb{N}} B(e_{n} , e_{n}) < \infty, $$ for every complete orthonormal basis of $E$, $(e_{n})_{n \in \mathbb{N}}$.
The question: does this implies that $B$ is bounded? (that is, that there exists a constant $C>0$ such that $B(e,f) \leq C \|e\|_{E} \| f \|_{E}$ for all $e,f \in E $). Hence the point is if supposing some trace and positive-definiteness condition we can imply continuity. Usually the continuity is required before defining the notions of positive definiteness or traces, so I wonder if we could do it in the other sense.
With some work, I have managed to prove that I can find a dense subspace $V \subset E$ (say, the span of a fixed orthonormal basis) where the boundedness is proven. I wonder if the boundedness in a dense subset of $E\times E$ implies a unique bilinear and bounded extension to which we could identify $B$, so we can conclude the result.
Any ideas?
Thanks for your help.