The Fredholm index is locally constant

Question

Let $X$ and $Y$ be Banach spaces, and let $\mathsf{Fr}(X,Y)\subseteq\mathcal{L}(X,Y)$ denote the set Fredholm operators from $X$ to $Y$. For $F\in\mathsf{Fr}(X,Y)$ we can define the Fredholm index of $F$ as $$\operatorname{ind}(F):=\dim(\ker(F))-\dim(\ker(F^*)).$$ How can I show that the index is locally constant, i.e. that for every $F\in\mathsf{Fr}(X,Y)$ there exists some open set $U\subseteq\mathsf{Fr}(X,Y)$ containing $F$ such that for all $L\in U$ we have $$\operatorname{ind}(L)=\operatorname{ind}(F).$$

I'm not sure how to approach this problem, so any hints would also be appreciated.

Answer 1

We will use two basic results on Fredholm operators: (1) an operator $L\in\mathcal{L}(X,Y)$ is Fredholm if and only if there exists some $R\in\mathcal{L}(Y,X)$ satisfying $$LR=\mathrm{id}_{Y} + K_Y,\qquad RL=\mathrm{id}_X + K_X$$ where $K_X,K_Y$ are some compact operators. (2) if $K$ is a compact operator then $\operatorname{ind}(\mathrm{id}+K)=0$. I will write $\mathcal{L}=\mathcal{L}(X,Y)$. Similarly, I will write $\mathcal{K}=\mathcal{K}(X,Y)$ to denote the set of all compact operatos $X\to Y$.

Note that $\mathcal{K}$ forms a two-sided ideal in $\mathcal{L}$. Now consider $\pi:\mathcal{L}\to\mathcal{L}/\mathcal{K}$ and note that by result (1), $$\mathsf{Fr}(X,Y)=\pi^{-1}((\mathcal{L}/\mathcal{K})^\times),$$ where $(\mathcal{L}/\mathcal{K})^\times$ is the set of units of $\mathcal{L}/\mathcal{K}$, i.e. the set of elements which are invertible.

Now show that $(\mathcal{L}/\mathcal{K})^\times$ is an open set.

Using this and the fact that $\pi$ is continuous gives us that $\mathsf{Fr}(X,Y)$ is open in $\mathcal{L}$. So for every $F\in\mathsf{Fr}(X,Y)$ there is an open set $U\subset\mathcal{L}$ with $F\in U\subset\mathsf{Fr}(X,Y)$.

Now fix an arbitrary $F\in\mathsf{Fr}(X,Y)$. Let $G$ be the "quasi-inverse" of $F$ given in result (1). Use this and (1) once more to

Show that for every $L\in\mathcal{L}$ satisfying $\|L\|\leq\frac{1}{\|G\|}$ we have $F+L\in\mathsf{Fr}(X,Y)$.

Let $R$ denote the quasi-inverse of $F+L$ you find above. Then you should use the composition rule of the Fredholm index and (2) to find that $$\operatorname{ind}(R) + \operatorname{ind}(F+L)=0.$$ Continuing in this way and using the specific form of $R$ you should be able derive that $$\operatorname{ind}(F+L)=\operatorname{ind}(F).$$

Answer 2

From what I remember you need to prove that

$$\textrm{ind}\colon \textrm{Fr}(X,Y) \to \mathbb{R}$$

is continuous and that its image is contained in $\mathbb{Z}$. Given $F \in \textrm{Fr}(X,Y)$, let $k=\textrm{ind}(F)$ and choose an open set on $\mathbb{R}$ containing $k$ but not $k-1$ nor $k+1$, e.g., $(k-1,k+1)$.

Then the open set $U$ you want is the connected component of $\textrm{inf}^{-1}(k)$ that contains $F$.