Decomposition of Hilbert space related to index of an operator

Question

Let $A\in\mathcal{L}(H),$ is a bounded operator on the Hilbert space $H$. Then the ascent and descent of A are given by

$$\operatorname{asc}(A):= \inf_{p\in\mathbb N}\{N(A^p)=N(A^{p+1})\},\quad\operatorname{dsc}(A)=\displaystyle \inf_{p\in\mathbb N}\{R(A^p)=R(A^{p+1})\}\,.$$ If $\operatorname{asc}(A)$ and $\operatorname{dsc}(A)$ are finite, then prove that

1. $\operatorname{asc}(A)=\operatorname{dsc}(A)$ (Common value is called Index of A, denoted by $ind(A)$)
2. $R(A^{\operatorname{ind}(A)})$ is closed.
3. $H=R(A^{\operatorname{ind}(A)})\oplus N(A^{\operatorname{ind}(A)})$

I got this problem from a pdf paper. I don't know how to prove this. Since the index of Fredholm operator is

$$\operatorname{ind}(A)= \dim\ker A-\operatorname{codim}\operatorname{ran} A $$ Is there any relation between these two definitions of index ? I am also confused with the definitions of asc and dsc. are they always positive integer ? Kindly make me clarify.

Thanks.

Answer 1

Point 1. can be proven by induction, using the following induction steps:

If $R(A^p)=R(A^{p+1})$ and $N(A^{p+1})=N(A^{p+2})$ then $N(A^p)=N(A^{p+1})$.

Let $x\in N(A^{p+1})$, then $A^px \in R(A^p)$ and so there is a $y_x$ so that $A^px=A^{p+1}y_x$. But then $0=A^{p+1}x=A^{p+2}y_x$ and so $A^{p+1}y_x=0$, giving $A^px=0$. Then $N(A^p)\subseteq N(A^{p+1})$, the other inclusion is true anyway.

If $N(A^p)=N(A^{p+1})$ and $R(A^{p+1})=R(A^{p+2})$ then $R(A^p)=R(A^{p+1})$.

Let $x\notin R(A^{p+1})$ and $x=A^py_x$ for some $y_x$, this is equivalent to $A^{p+1}y-A^py_x\neq0$ for all $y$, or $Ay-y_x\notin N(A^p)$. Since $N(A^p)=N(A^{p+1})$ you get $A^{p+2}y-A^{p+1}y_x=A^{p+1}(Ay-y_x)\neq0$ for all $y$. But this implies that $A^{p+1}y_x$ is not in $R(A^{p+1})$, a contradiction.

If both descent and ascent are finite then an induction on the first statement shows $asc(A)\leq dsc(A)$, while an induction on the second shows $dsc(A)\leq asc(A)$.

For point 2.:

Note first that $ind(A)=0$ if and only if $A$ is an isomorphism. In this case $R(A)=H$ is closed. If $ind(A)\neq0$ we replace $A$ by $A^{ind(A)}$ in order to assume $ind(A)=1$.

First check that $R(A)\cap N(A)=\{0\}$. If $Ax=0$ and $x=Ay_x$ then $A^2 y_x=0$ and so by $N(A)=N(A^2)$ you also have $x=Ay_x=0$, giving this step.

With this let $\widetilde A: H/N(A)\to H/N(A)$, $[x]\mapsto [Ax]$, which is well defined and (by the previous step) injective. Since its injective you have $asc(\widetilde A)=0$. On the other hand you have $$R(\widetilde A^2)=\{ [A^2x]\mid x\in H\} = \{ [Ax]\mid x\in H\} = R(\widetilde A).$$ So $dsc(\widetilde A)\leq 1$ is finite, then $dsc(\widetilde A)=asc(\widetilde A)=0$ and $R(\widetilde A) = H/N(A)$. But you also have $$R(\widetilde A) = R(A)/N(A) \subseteq H/N(A)$$ and then the projection $R(A)\to H/N(A)$ is bijective and continuous, which is only possible if $R(A)$ is Banach in particular it must be closed.

For point 3:

As before assume $ind(A)=1$. We have already seen $R(A)\cap N(A)=0$ and that the map $R(A)\to H/N(A)$ is a bijection. This implies the topological decomposition $H=N(A)\oplus R(A)$. Note that this is not an orthogonal decomposition, as the example $$A=\begin{pmatrix}1&1\\0&0\end{pmatrix}$$ shows.

About the relation to the Fredholm index:

If ascent and descent are both finite then the Fredholm index (if it is defined!!) is $0$. For by the decomposition: $$H=N(A^{ind(A)})\oplus R(A^{ind(A)})$$ you have that $A$ restricts to an automorphism on $R(A^{ind(A)})$ and a nilpotent endomorphism on $N(A^{ind(A)})$. If $N(A)$ is finite dimensional then $N(A^{ind(A)})$ must also be finite dimensional and so the Fredholm index is described entirely by what $A$ is doing on the finite dimensional $N(A^{ind(A)})$, and so it is $0$.

This is the strategy used by Riesz in his 1916 paper to show that the Fredholm index of an operator $I+K$ is zero (by constructing these ascending and descending sequences).