Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 22G (Linear Analysis)

Suppose the spectrum were empty. The resolventwould then be an entire operator-valued function. For $|\lambda |>\Vert T\Vert$,because the series converges in operator norm and multiplication by $\lambda I-T$ telescopes to $I$. Hence $\Vert R(\lambda )\Vert \le (|\lambda |-\Vert T\Vert {)}^{-1}$.

For fixed $x,y\in {\ell }^2$, the scalar function $⟨R(\lambda )x,y⟩$ is entire, bounded outside a disc by the estimate and bounded inside by compactness. Liouville's theorem makes it constant, and its limit at infinity makes that constant zero. This for all $x,y$ would imply $R(\lambda )=0$, impossible for an inverse. Thus the spectrum is nonempty.

This is the unilateral right shift. It is not compact because the orthonormal sequence $(T{e}_n)=(e_{n+1})$ has no convergent subsequence. Its norm is one. For $|\lambda |>1$ the Neumann series gives a resolvent; for $|\lambda |<1$, $\bar{\lambda }$ is an eigenvalue of the adjoint left shift, so $\lambda$ lies in the spectrum of $T$. Closedness supplies the boundary. Hence

This is the unilateral left shift. It is not compact because $S{e}_{n+1}=e_n$. For every $|\lambda |<1$,is an eigenvector with eigenvalue $\lambda$. The norm bound and closedness then give

The weighted shift is compact because truncating after the first $N$ weights gives finite-rank operators converging in norm, as $1/n\to 0$. Moreoverso its spectral radius is zero. Since the spectrum is nonempty,

Again the weights tend to zero, so finite-rank truncations prove compactness. Products of $m$ consecutive weights are bounded by $1/(m+1)!$, making the spectral radius zero. Therefore