Range of $I + A$ where $A$ is a compact operator on a Hilbert space.

Question

Let $A$ be a linear compact operator on a Hilbert space $H$.

Prove that the range of $A + I$ is a closed subspace of $H$.

I do not know where to start. Could you give me some tips please.

Answer 1

The result is valid both in a Banach space and in a Hilbert space. In a Banach space it would rely on $Z=\ker(I+A)$ being finite dimensional and therefore admitting a closed complement $G$, i.e. $H=Z\oplus G$. In a Hilbert space (as remarked by the OP) you may simply choose $G=Z^\perp$, the orthogonal complement to the closed subspace $Z$.

It then suffices to show that $F=(I+A)G$ is closed. Thus let $y\in \overline{F}$ we need to show that there is $x\in G$ with $(I+A)x=y$. Since $y\in \overline{F}$ there is a sequence $x_n\in G$ so that $y_n=(I+A)x_n\rightarrow y$.

Suppose first that the sequence $(x_n)$ admits a bounded subsequence. Then $Ax_n$ admits a convergent subsequence $Ax_{n_k}\rightarrow z$ so that $x_{n_k}=y_{n_k}-A x_{n_k}\rightarrow x$ is convergent as well. As $G$ is closed $x\in G$ and by continuity $(I+A)x=y$.

If on the other hand $|x_n|\rightarrow \infty$ then $e_n=x_n/|x_n|\in G$ is of norm one and $(I+A)e_n\rightarrow 0$. Using compactness we see as before that there is a convergent subsequence $e_{n_k}\rightarrow z\in Z=\ker(I+A)$. But $e_{n_k} \rightarrow z\in G$ as $G$ was closed. Contradiction.

Answer 2

Note that $N = \ker(I + A)$ is finite dimenisional, and therefore closed. It follows that the induced map $$ (I + A)^{\sim} : H/N \to H $$ is an injective map from a Banach space. In order to prove that its range is closed, it suffices to show that there exists a constant $C > 0$ such that $\|(I + A)^\sim x\| \geq C\|x\|$.

Suppose to the contrary that no such constant exists. It would follow that there exists a sequence of vectors $(x_k)_{k \in \Bbb N} \subset H/N$ such that $\|x_k\| = 1$, but $\|x_k\| \to 0$. Using these vectors, we can build a sequence $(y_k)$ in the unit ball of $H$ such that $Ay_k$ fails to have a convergent subsequence, which means that $A$ fails to be compact.