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

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 $$A=0$$?

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 $$A=0.$$