Real spectrum of self-adjoint operators

Remarks

Proof idea: let $\lambda = a + ib$ with $a, b$ real and $b \ne 0$. Because $T - a I$ is self-adjoint, $\norm{(T - \lambda I)x}^2 = \norm{(T - aI)x}^2 + b^2 \norm{x}^2 \ge b^2 \norm{x}^2$. So $T - \lambda I$ is injective with closed range, and the orthogonal complement of its range is $\ker(T - \bar\lambda I) = \set{0}$ by the same estimate; hence $T - \lambda I$ is invertible.

Example: the multiplication operator $(Tf)(x) = x f(x)$ on $L^2[0,1]$ is self-adjoint, and its spectrum is $[0, 1]$, a real interval, although $T$ has no eigenvalues.

Self-adjointness is needed: the unilateral shift on $\ell^2$ is not self-adjoint, and its spectrum is the closed unit disc. The finite-dimensional case is the fact that eigenvalues of a self-adjoint operator are real. The same conclusion holds for unbounded self-adjoint operators, and for self-adjoint elements of a C*-algebra.

Going deeper

For a student

In infinite dimensions the spectrum can contain points that are not eigenvalues, so the finite-dimensional argument with an eigenvector does not suffice; the proof uses a lower bound instead. The result is the first step towards the spectral theorem for self-adjoint operators, which represents $T$ as an integral of $\lambda$ over $\sigma(T) \subseteq \R$. The numbers $m$ and $M$ are the ends of the closure of the numerical range, and $\norm{T} = \max(\abs{m}, \abs{M})$.