 Solution

For a bounded operator on a complex Banach space, the resolvent set is $\rho (T)=\{\lambda :\lambda I-T\text{ is bijective with bounded inverse}\}$, and the resolvent operator is $R_T(\lambda )=(\lambda I-T{)}^{-1}$. The spectrum is its complement $\sigma (T)=\mathbb{C}\setminus \rho (T)$; the point spectrum consists of $\lambda$ with $\ker (\lambda I-T)\ne 0$. For a real Banach space these definitions use its complexification.

If $|\lambda |>\Vert T\Vert$, the norm-convergent Neumann series gives $R_T(\lambda )={\lambda }^{-1}{\sum }_{n\ge 0}(T/\lambda {)}^n$, so the spectrum is bounded. If ${\lambda }_0\in \rho (T)$, factorThe second factor is invertible by a Neumann series whenever $|\lambda -{\lambda }_0|\Vert R_T({\lambda }_0)\Vert <1$. Thus the resolvent set is open and the spectrum is closed and bounded.

The point spectrum need not be closed. On ${\ell }^2$, define $T{e}_n=e_n/n$. Every $1/n$ is an eigenvalue, but zero is not: $Tx=0$ forces every coordinate of $x$ to vanish. No other eigenvalues occur, so ${\sigma }_p(T)=\{1/n:n\ge 1\}$ is not closed.

Finally, if $T$ is self-adjoint and $Tv=\lambda v$, $v\ne 0$, then $⟨Tv,v⟩$ is real and equals $\lambda \Vert v{\Vert }^2$ under the convention linear in the first argument. Therefore every eigenvalue of a self-adjoint operator is real.