Past exam of the functional analysis I course of the University of Oxford / 2023 / 3 / a

Writing $\mathcal{B}(X)$ for the Banach space of bounded linear operators on $X$, the spectrum isHere $I$ is the identity operator. Since $X$ is a Banach space, the bounded inverse theorem makes this equivalent to failure of bijectivity of $\lambda I-T$.

The spectral radius isFor a nonzero complex Banach space, the spectrum of a bounded linear operator is nonempty and compact, so the supremum is a maximum. If the zero space is allowed, its empty spectrum is assigned spectral radius zero.

The Neumann series criterion states that $\Vert A\Vert <1$ for a bounded linear operator on a Banach space implieswith convergence in the operator norm. In particular, $\Vert T\Vert <1$ suffices for invertibility of $I-T$.

If $|\lambda |>\Vert T\Vert$, apply this criterion to $A=T/\lambda$. It givesThus no such $\lambda$ belongs to the spectrum, and

For each integer $j\ge 1$, if $|\lambda {|}^j>\Vert T^j\Vert$, then ${\lambda }^{j}I-T^j$ is invertible by the same argument. The commuting factorizationthen gives a bounded two-sided inverse for $\lambda I-T$, namely the displayed sum multiplied by $({\lambda }^{j}I-T^j{)}^{-1}$. Hence every $\lambda \in \sigma (T)$ satisfies $|\lambda {|}^j\le \Vert T^j\Vert$. Taking the supremum over the spectrum and then the infimum over $j$ proves