The Fredholm Index

Prove Lemma 3.6: let $\mathcal{M}$ and $\mathcal{N}$ be two closed subspaces of the Hilbert space $\mathcal{H}$.

If $A \in \mathcal{B}(\mathcal{H})$ and $\mathrm{ran}\ A$ is closed, show that $\mathrm{ran}\ A^{(\infty )}$ is closed. If $A \in \mathcal{SF}$ and $\mathrm{ker}\ A = (0)$, show that $A^{(\infty )} \in \mathcal{SF}$ and $\operatorname {ind} A^{(\infty )} = -\infty$ or $0$.

Does the unilateral shift of multiplicity $1$ have a square root?

Show that for every $n$ in $\mathbb {Z} \cup \left\{ \pm \infty \right\}$ there is an operator $A$ in $\mathcal{SF}$ such that $\operatorname {ind} A = n$.

If $A \in \mathcal{SF}$, then for every $n \ge 1$, $A^n \in \mathcal{SF}$ and $\operatorname {ind} A^n = n(\operatorname {ind} A)$.

If $A: \mathcal{H} \to \mathcal{H}'$ is a left semi-Fredholm operator, then there is a finite rank operator $F: \mathcal{H} \to \mathcal{H}'$ such that $\mathrm{ker}(A + F) = (0)$ and $\operatorname {ind}(A + F) = \operatorname {ind} A$.

If $A$ is a Fredholm operator in $\mathcal{B}(\mathcal{H})$, prove that the following statements are equivalent.

1. $\operatorname {ind} A = 0$.

2. There is a compact operator $K$ such that $A + K$ is invertible.

3. There is a finite rank operator $F$ such that $A + F$ is invertible.

If $U$ is the unilateral shift of multiplicity $1$ and $\pi : \mathcal{B}(\mathcal{H}) \to \mathcal{B}(\mathcal{H})/\mathcal{B}_0(\mathcal{H})$ is the natural map, show that $\pi (U)$ is normal in $\mathcal{B}(\mathcal{H})/\mathcal{B}_0(\mathcal{H})$ but there is no normal operator $N$ such that $U - N$ is compact. (See Exercise IX.8.14.)