Past exam of the analysis course of the University of Oxford / 2011 / 3 / b / Solution

Linearity and $\Vert Tf{\Vert }_{\infty }\le 2\pi \Vert f{\Vert }_{\infty }$ show that the integral defines a bounded linear operator. Inductively,For $k=0$ this is immediate. If it holds for $k$, then $|\sin s|\le 1$ gives $|T^{k+1}f(t)|\le {\int }_0^t{s}^k\Vert f{\Vert }_{\infty }/k!ds$, which proves the next case. Thus $\Vert T^k\Vert \le (2\pi {)}^k/k!$.

For every $\lambda \ne 0$, the series $A_{\lambda }={\sum }_{k\ge 0}{\lambda }^{-k-1}{T}^k$ converges in operator norm, dominated by $|\lambda {|}^{-1}{e}^{2\pi /|\lambda |}$. Multiplying its partial sums by $\lambda I-T$ on either side gives the identity minus a remainder ${\lambda }^{-N-1}{T}^{N+1}$, which tends to zero by the factorial estimate. Therefore $A_{\lambda }$ is the bounded inverse of $\lambda I-T$.

Finally every function in the image of $T$ vanishes at zero, so $T$ is not onto the continuous functions. Hence $0$ belongs to its spectrum and