If a operator $L$ in a Hilbert space is self-adjoint, then $L$ is coercive

Question

Let $H=(H,(\cdot, \cdot))$ be a Hilbert space and $L:D(L) \subset H \longrightarrow H$ a linear operator densely defined. If $L$ is self-adjoint operator, then $L$ is coercive, that is, there exists $C>0$ such that
$$(L(x),x)\geq C ||x||^2,\: \forall \: x \in D(L)?$$

I don't know if that's true. I couldn't prove it or set a counterexample.

Answer 1

You inequality would imply that $$ \langle (L-(C+\epsilon)I)f,f\rangle \ge \epsilon \|f\|^2,\;\; f\in H,\;\; \epsilon > 0. $$ And that would force $(L-CI+\epsilon I)$ to be invertible for every $\epsilon > 0$, assuming that $L=L^*$. So, if your conjecture were true, then every self-adjoint operator would have a spectrum that is bounded below.