Past exam of the analysis course of the University of Oxford / 2010 / A / 3

The spectrum of a bounded operator is $\sigma (T)=\{\lambda \in \mathbb{C}:T-\lambda I\text{ has no bounded inverse}\}$. The point spectrum consists of those $\lambda$ for which $\ker (T-\lambda I)\ne 0$, namely the eigenvalues. The spectral radius is $r(T)={\sup }_{\lambda \in \sigma (T)}|\lambda |$.

If $\Vert T\Vert <1$, the Neumann series $S={\sum }_{n=0}^{\infty }(-T{)}^n$ converges in the Banach space of bounded linear operators, since its term norms are bounded by a convergent geometric series. Multiplying the finite sums on either side by $I+T$ gives $I-(-T{)}^{N+1}$, which tends in operator norm to $I$. Therefore

If $|\lambda |>\Vert T\Vert$, write $T-\lambda I=-\lambda (I-T/\lambda )$. The Neumann series applies to $T/\lambda$, so $\lambda$ is outside the spectrum of a bounded operator. Thus $\sigma (T)\subseteq \{\lambda :|\lambda |\le \Vert T\Vert \}$ and $r(T)\le \Vert T\Vert$.

The polynomial spectral mapping theorem gives $\sigma (T^m)=\{{\lambda }^m:\lambda \in \sigma (T)\}$, and henceTaking $m$th roots and then the limit, whose existence is allowed, provesOnly this inequality is required; the reverse inequality is the additional content of the full spectral radius formula.

For the multiplication operator, $\Vert Tf{\Vert }_{\infty }\le \Vert g{\Vert }_{\infty }\Vert f{\Vert }_{\infty }=\Vert f{\Vert }_{\infty }$. Taking $f=1$ gives equality, so $\Vert T\Vert =1$.

An eigenfunction obeys $(g(t)-\lambda )f(t)=0$ at every $t$. If $\lambda \in [0,1)$, its level set is the single point $t=\lambda$, and a continuous function supported at a single point is zero. If $\lambda \notin [0,1]$, that level set is empty. For $\lambda =1$, every continuous function vanishing on $[0,1]$ is an eigenfunction, and nonzero examples exist, such as $f(t)=\max \{t-1,0\}$. Thus ${\sigma }_p(T)=\{1\}$.

If $\lambda \notin [0,1]$, multiplication by $(g-\lambda {)}^{-1}$ is a bounded inverse of $T-\lambda I$. If $\lambda \in [0,1]$, choose $t_0$ with $g(t_0)=\lambda$. Every function in the range of $T-\lambda I$ vanishes at $t_0$, so the range omits the constant function one. Hence it is not onto. Therefore

The unilateral shift operator $R$ is not onto, since its outputs have first coordinate zero. The left shift operator $L$ is not injective, since $L{e}_1=0$. Both have operator norm one, and $LR=I$. ConsequentlyIf $\Vert A\Vert <1$, then $\Vert LA\Vert <1$, so the Neumann series makes $I+LA$ invertible. If $R+A$ were invertible too, the identity above would givea product of invertible operators, contradicting the nontrivial kernel of $L$. ThusThis is norm stability of noninvertibility for the unilateral right shift.