Problem 9
Let \(A\) be a bounded linear operator mapping a Banach space E into itself. Show that if \(A^{*}\) is completely continuous, then so is \(A\).
Problem 11
Give an example of a completely continuous operator \(A\) mapping a Hilbert space \(\boldsymbol{H}\) into itself, such that \(\boldsymbol{A}\) has no eigenvectors. Reconcile this with Theorem \(7 .\) Hint. Let \(A\) be the operator in \(l_{2}\) such that $$ A x=A\left(x_{1}, x_{2}, x_{3}, \ldots, x_{n}, \ldots\right)=\left(0, x_{1}, \frac{x_{2}}{2}, \ldots, \frac{x_{n-1}}{n-1}, \ldots\right) $$ Then \(A x=A x\) implies $$ \mathrm{Ax},=0, \mathrm{Ax},=x_{1}, \lambda x_{3}=\frac{x_{2}}{2}, \ldots, \mathrm{Ax},=\frac{x_{n-1}}{n-\mathrm{L}}, \ldots $$ and hence \(\mathrm{x}=0\). Comment. This situation differs from the finite-dimensional case, where every linear operator (self-adjoint or not) has at least one eigenvector.
