Let $\mathcal H$ be a separable Hilbert space. Assume that $Q$ is a non-negative compact self-adjoint operator $\mathcal H\to\mathcal H$.
Assume there is a dense subspace $U$ of $\mathcal H$ such that, for all non-zero $x\in U$ $$\langle Qx,x\rangle >0.$$
Can I claim that $U$ is positive on the whole space $\mathcal H$, i.e., for any non-zero $x\in\mathcal H$, $\langle Qx, x\rangle>0$?
I think that all I need to prove is that the range of $Q$ is dense in $\mathcal H$, since in this case the compactness and self-adjointness of $Q$ imply that $Q$ is injective and so, since it is even non-negative, it must be positive.
EDIT: I think the property is false, since any dense subset $U$ that does not contain the nullspace $N$ of $Q$, would satisfy the property that $\langle Qx,x\rangle >0$ if $x\in U$. Now is there always a subspace $U$ such that $U\cap N = \{0\}$? I guess so...
Let $U$ be a proper dense subspace of $H$ and let $e_0$ be a unit vector not in $U$. Consider an orthonormal basis $\{e_n\}_{n\in {\bf N}}$ which includes the chosen vector $e_0$ and let $T$ be the compact operator specified by $$ T(e_n) = \left\{\matrix{ 0, & \text{ if } n=0, \cr e_n/n, & \text{ if } n>0. }\right. $$ Since the kernel of $T$ coincides with $\mathbb Ce_0$, we see that the restriction of $T$ to $U$ is injective. So, for every nonzero $x$ in $U$, one has that $$ \langle T^2(x), x\rangle = \|T(x)\|^2 > 0. $$ However, since $$ \langle T^2(e_0), e_0\rangle = 0, $$ we see that $Q:=T^2$ is not positive.