 Solution

For a positive operator, $⟨Tx,x⟩\ge 0$. We show directly that every nonreal or negative real $\lambda$ lies in the resolvent set.

If $\mathrm{Im}\lambda \ne 0$, the Cauchy-Schwarz inequality and reality of $⟨Tx,x⟩$ giveThus $T-\lambda I$ is bounded below, hence injective with closed range: a convergent sequence of outputs has Cauchy preimages, whose limit maps to that output. Its adjoint is $T-\bar{\lambda }I$, which has trivial kernel by the same estimate. The identity $\mathrm{Ran}(A{)}^{\perp }=\ker A^{\ast }$, proved by the same inner-product calculation as in (i), makes its range dense. Being closed and dense, it is all of $X$, and the lower bound makes the inverse bounded.

If $\lambda <0$ is real, positivity gives $⟨(T-\lambda I)x,x⟩\ge (-\lambda )\Vert x{\Vert }^2$, so the same argument applies with lower bound $-\lambda$. Here the operator is self-adjoint, so its range is dense because its own kernel is zero. Hence