Does the spectrum of a bounded from below self-adjoint operator have a lower bound?

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.
• I can follow your proof up until $\overline{\mathcal{Ran}(B)} = \left(\mathfrak{Ker} B \right)^\perp$ and later $\vert\vert B^{-1}\vert\vert \leq \frac{1}{\epsilon}$. Can you prove these two statements? Commented May 8, 2021 at 0:39