Fredholm’s alternative

Question

By Fredholm’s alternative I know that if $ K \in \mathcal{K}(X)$ then $ I-K \in \mathcal{B}(X) $ is injective iff it’s surjective and, when X has finite dimension, $K$ is injective iff it is surjective. Is it correct? Why in infinite dimensional spaces the latter does not hold even when the former holds?

Answer 1

Just as you said, the Fredholm alternative states that if $X$ is a Banach space and $K\in\mathcal{K}(X)$ is a compact operator on $X$, then $I-K$ is injective if and only if $I-K$ is surjective.

As a consequence, you can obtain the classical result from Linear Algebra, which states: if $V$ is a finite-dimensional vector space and $L\colon V\to V$ is linear, then $L$ is injective if and only if $L$ is surjective. Indeed, take any Banach space structure on $L$. From general Functional Analysis we know that $V$ becomes a Banach space, and every linear operator on $V$ is compact. Just apply the Fredholm alternative to $K=I-L$.

The idea really is that compact operators are close to finite-rank operators. For example, if $H$ is a Hilbert space then $\mathcal{K}(H)$ is precisely the operator-norm closure of the space of finite-rank operators on $H$.

So a compact modification $I-K$ of the identity map $I$ will behave more-or-less as it would on finite dimension. As an even more particular case, suppose that $K$ is supported on a finite-dimensional subspace $Y$ of $X$. Enlarging $Y$ a bit if necessary, we can assume that $(I-K)(Y)\subseteq Y$. Take a closed complement $Z$ of $Y$ (another subspace of $X$ such that $X=Y\oplus Z$). Then $I-K\colon Y\oplus Z\to Y\oplus Z$ is the identity on $Z$, and takes $Y$ to $Y$. By the finite-dimensional case, $I-K$ is surjective iff it is surjective on $Y$, iff it is injective on $Y$, iff it is injective.

However, there are lots of examples of surjective operators on Banach spaces which are not injective and vice-versa, when dealing with non-compact operators. For example, let $X=\ell^\infty$, the space of bounded real or complex numbers. The left shift $L(x_0,x_1,\ldots)=(x_1,x_2,\ldots)$ is surjective but not injective. The right shift $R(x_0,x_1,\ldots,)=(0,x_0,x_1,\ldots)$ is injective but not surjective.