Spectral theory of compact operators

Remarks

The key tool is the Riesz lemma: in an infinite-dimensional normed space, the unit ball contains a sequence of unit vectors at mutual distance at least $1/2$. If $\ker(T - \lambda I)$ were infinite-dimensional for some $\lambda \ne 0$, such a sequence in it would be mapped to $\lambda$ times itself, which has no convergent subsequence, against compactness. A similar argument shows that, for each $\varepsilon > 0$, only finitely many eigenvalues satisfy $\abs{\lambda} \ge \varepsilon$, which gives part 3. Part 2 is the Fredholm alternative: for $\lambda \ne 0$, the equation $Tx - \lambda x = y$ is solvable for every $y$ as soon as the homogeneous equation has only the solution $x = 0$. Part 4 holds because a compact operator on an infinite-dimensional space cannot have a bounded inverse. Example: on $\ell^2$, the diagonal operator $T(x_n) = (x_n / n)$ is compact, and $\sigma(T) = \set{1/n : n \ge 1} \cup \set{0}$, where $0$ is not an eigenvalue. The compactness hypothesis matters: the identity on an infinite-dimensional space has the eigenvalue $1$ with infinite-dimensional eigenspace.

Going deeper

For a student

This is the Riesz–Schauder theory, the abstract form of Fredholm's results on integral equations. It applies to integral operators with continuous kernels, which are compact on spaces of continuous or square-integrable functions. For compact self-adjoint operators on a Hilbert space, it combines with orthogonality to give the spectral theorem for compact self-adjoint operators. In the language of Fredholm theory, part 2 says that $T - \lambda I$ is a Fredholm operator of index $0$ for each $\lambda \ne 0$.