 Solution

Write $\lambda =a+ib$ with $b\ne 0$. Since $aI-T$ is self-adjoint, $⟨(aI-T)v,v⟩$ is real. The cross term in the squared norm therefore vanishes:This proves the equality and lower bound, and makes both $S=\lambda I-T$ and $S^{\ast }=\overline{\lambda }I-T$ injective.

If $S{v}_n$ converges, the lower bound shows $v_n$ is Cauchy. Completeness gives $v_n\to v$ and boundedness gives $S{v}_n\to Sv$, so the range is closed. Also $(\mathrm{ran}S{)}^{\perp }=\ker S^{\ast }=0$, so the range is dense. Being both closed and dense, it equals $V$. Hence $S$ is bijective, and its lower bound gives