Past exam of the analysis course of the University of Oxford / 2011 / 3 / a

For a bounded linear operator on a complex Banach space, its spectrum isThe complement is the resolvent set. For a real space, one may use its complexification when speaking of the complex spectrum; the algebraic arguments below apply over the underlying scalar field as well.

The bounded operators on a Banach space form a Banach algebra in the operator norm. If $\Vert T\Vert <1$, the Neumann series $A={\sum }_{k=0}^{\infty }{T}^k$ converges absolutely in that norm. Its partial sums satisfyTaking limits proves $\boxed{(I-T{)}^{-1}=\sum _{k\ge 0}{T}^k}$.

If $|\lambda |>\Vert T\Vert$, factor $\lambda I-T=\lambda (I-T/\lambda )$ and apply the same argument. Thus such $\lambda$ are in the resolvent set and

If $T$ has a bounded inverse, it is onto and $\Vert x\Vert \le \Vert T^{-1}\Vert \Vert Tx\Vert$, giving the claimed lower bound. Conversely a positive lower bound makes $T$ injective. Together with surjectivity it gives an everywhere-defined linear inverse with norm at most $1/\epsilon$. Hence it is bounded, proving the equivalence.

If $R,S$ are individually invertible, their product has inverse $S^{-1}{R}^{-1}$. Conversely assume $RS=SR$ and put $A=RS$, with $A$ invertible. Both factors commute with $A$ and therefore with $A^{-1}$. Direct multiplication givesThese are bounded two-sided inverses, so an invertible commuting product has both factors invertible. The commutation assumption is used when verifying the second side of each inverse.

The factorizationhas commuting factors because both are polynomials in $T$. If ${\lambda }^n\notin \sigma (T^n)$, their product is invertible. The preceding part would then make $T-\lambda I$ invertible, contradicting $\lambda \in \sigma (T)$. ThereforeThis proves the required inclusion without assuming the full spectral mapping theorem.