Let $A$ be a self-adjoint operator in a complex separable Hilbert space that is unbounded, but bounded from below, i.e.
$$\exists m>-\infty, ~\forall f\in D(A) \subsetneq \mathcal H , \langle Af, f \rangle \geq m \vert\vert f\vert\vert^2$$
Question: does the spectrum of $A$ have an infimum (lower bound)? My "gut feeling" is yes, it does, because it's completely counterintuitive if it didn't. So how can one prove it for an arbitrary $A$? I need this proof for an article I am writing, but I couldn't find it in the books that I've searched in.
I can even assert that $\inf \sigma (A) = \min \sigma (A)$, in other words the infimum (whose existence I am questioning) is actually a spectral value.
We prove that $-(m+\varepsilon) \notin \sigma(A)$ for $\varepsilon > 0$. We consider operator $B = A + (m+\varepsilon)I$ and prove that $B$ is invertible.
Obviously, $B$ is self-adjoint and $\langle Bf,f\rangle \ge \varepsilon\|f\|^2$. It follows that $\|Bf\| \|f\| \ge \langle Bf,f\rangle \ge \varepsilon \|f\|^2$ and, therefore, $\|Bf\| \ge \varepsilon \|f\|$.
Also $\ker B = \{0\}$ and $\overline{\operatorname{Ran}(B)} = (\ker B)^\perp = H$. Thus, operator $B^{-1}$ is bounded ($\|B^{-1}\| \le \frac{1}{\varepsilon}$) and defined on a dense subspace $\operatorname{Ran}(B)$. It follows that $B^{-1}$ is defined on $H$, since it is a closed operator (the well known statement is that a densely defined closed bounded operator is defined on a whole space). Thus, $B$ is invertible.