Fredholm Operators

If $A \in \mathcal{B}(\mathcal{H})$ and $\mathrm{ran}\ A$ is closed, prove that $\mathrm{ran}\ A^*$ is closed without using Theorem VI.1.10. [Hint: Show that there is a bounded operator $B$ on $\mathcal{H}$ such that $BA =$ the projection of $\mathcal{H}$ onto $(\mathrm{ker}\ A)^\perp$.]

Theorem 2.3 states that if $A: \mathcal{H} \to \mathcal{H}'$ is a bounded operator, then the following two statements (among others) are equivalent: (a) $A$ is left semi-Fredholm, that is, there is a bounded operator $B: \mathcal{H}' \to \mathcal{H}$ and a compact operator $K$ on $\mathcal{H}$ such that $BA = 1 + K$; (b) $\mathrm{ran}\ A$ is closed and $\dim \mathrm{ker}\ A < \infty$. Give a direct proof that (b) implies (a).

Let $A, B, C \in \mathcal{B}(\mathcal{H})$ and define $X: \mathcal{H}^{(2)} \to \mathcal{H}^{(2)}$ by the matrix

If $A \in \mathcal{B}(\mathcal{H})$, show that $A\mathcal{M}$ is closed for every closed subspace $\mathcal{M}$ of $\mathcal{H}$ if and only if $A$ has finite rank or $A$ is left Fredholm.

If $\mathscr {X}$ and $\mathscr {Y}$ are Banach spaces and $T \in \mathcal{B}(\mathscr {X}, \mathscr {Y})$ such that the Hamel basis dimension (that is, the algebraic dimension) of $\mathscr {Y}/(\mathrm{ran}\ T)$ is finite, then $\mathrm{ran}\ T$ is closed.

Show that a normal operator is Fredholm if and only if $0$ is not a limit point of $\sigma (N)$ and $\dim \mathrm{ker}\ N < \infty$. (See Proposition 4.5 below.)