Of an Hilbert Space $H$ and an (self-adjoint) operator $A$ on it,
(1) When $A$ has only real and positive / non-negative eigenvalues, does it imply positivity of $A$?
(2) When $A$ has only real and positive / non-negative spectrum, does it imply positivity of $A$?
I can only find cases where $H$ is finite, I want to know whether these statements are true on a general Hilbert space.
What is the base field of $H$ ? $\mathbb R$ or $\mathbb C$ ? I'll suppose the latter because it is the usual assumption and makes things easier. Is $A$ continuous (or equivalently, bounded) ? I'll suppose yes, because this case is simpler, far more frequent in math. discussion and I know little about the unbounded case. Then:
(2) has answer yes. Under some definitions (in C$^*$-agebras, which include the algebra of continuous endomorphisms of $H$ as a special case), it's by definition ($A$ self-adj. with positive real spectrum <=> $A$ positive) ... so I imagine you have got the definition specific to operators in Hilbert spaces: ($x|A(x)) \ge 0$ for all $x \in H$, and there is a theorem saying that this is equivalent. Now you probably are looking for a proof of the equivalence. The implication in one direction is relatively simple although not trivial, but what you need goes in the other direction, and that is quite a difficult story, unless one knows (and may use) a theorem from which one concludes easily ... I find no place where the theorem is named (may-be Wikipedia doesn't have it at all); it says that $A$ is positive iff it is a product $B^*B$ where B is some bounded linear operator on $H$ ... this is just a general theorem for C$^*$-algebras applied to our special case; the proof I know is in the general case, may-be there exists a simpler proof in the special case, but I don't know about that.
(1) is obviously true when $H$ is finite-dim. because it is then equivalent to (2). I guess it is wrong in the infinite-dim. case; in principle your hyp. in (1) is satisfied in case $A$ has no eigenvalue at all; it is probably easy to find an example where self-adj. $A$ has a spectral value < 0 but no eigenvalue, this will not be positive due to the converse of (2). If you want a case having an eigenvalue, you will probably have it by taking $H = \mathbb C . x \oplus H_1$ where $x \in H$ is $\neq 0$ and $H_1$ is the hyperplane orthogonal to $x$, then you transfer the example without eigenvalue to $H_1$ and define your operator on $\mathbb C . x$ as identity ...