I'm reading Dirichlet Forms and Symmetric Markov Processes by M. Fukushima, Y. Oshima and M. Takeda. They say a bilinear form $(\mathcal{E},D(\mathcal{E}))$ on a real Hilbert space $H$ is symmetric if $D(\mathcal{E})$ is a dense linear subspace of $H$ and etc...
My question
I don't understand the condition "$D(\mathcal{E})$ is a dense linear subspace of $H$". Why is this condition necessary? I have no idea what they want to do.
Just in case the OP is curious the way I used to be. The density assumption needs to be posed in the first place because a vector space we're dealing with now is no longer a finite-dimensional one.
It is well-known that on a finite-dimensional vector space, a symmetric form can define a linear operator and vice versa (via Riesz representation theorem). We want to carry this correspondence to an infinite-dimensional case. However, in such a situation arises the notion of unbounded operators. And it is very common that Dirichlet form theory deals with unbounded operators.
One important fact about them is any closed unbounded operator cannot be defined everywhere, i.e. the best we can hope is its domain being just dense in an underlying Hilbert space. In order to maintain the correspondence between linear operators and symmetric forms, it makes sense we should allow a symmetric form to be defined just on a dense subspace, not the entire space. You may see how a symmetric form is defined out of a linear operator and vice versa from [Ma, Rockner] Introduction to the Theory of (Non-Symmetric) Dirichlet Forms, page 22-23.
Hope this helps.