A self-adjoint operator without eigenvalues and with spectrum equal to {0}

Question

Let $A$ be a self-adjoint operator on a Hilbert space $H$. We know that the spectrum of $A$ ( $\sigma(A)$) can be decomposed into an essential spectrum ($\sigma_{ess}(A)$) and a set of eigenvalues ($\sigma_e(A)$) : $\sigma(A)=\sigma_{ess}(A)\cup\sigma_e(A)$.
The question is: if we know that $\sigma(A)=\sigma_{ess}=\{0\}$ can we prove that $A=0$?

Answer 1

If $A: H \to H$ is self-adjoint, then $A$ is bounded (Hellinger Toeplitz !).

If $r(A)= \max\{|\lambda|: \lambda \in \sigma(A)\}$ denotes the spectral radius of $A$, then it is well-known that $r(A)=||A||$, if $A$ is self-adjoint.

Hence, if $\sigma(A)=\{0\}$, then $r(A)=0$, hence $A=0.$