Problem 5

Exercise 2.5 (Perturbed compact operators have closed range)   Let $K : X \to X$ be a compact operator and suppose $A=I+K$ has a trivial kernel. Then $A(X)$ is closed.

Proof. We can show $A(X)$ is weakly closed. Suppose $\{Ax_n\}_{n=1}^\infty \subseteq A(X)$ converges weakly. We will find a subsequence $\{x_{n_k}\}$ such that $Kx_{n_k}\to Kx$.

Let $\phi \in X^*$. By assumption the limit $\vert\phi(Ax_n)\vert$ is finite, so that we can deteremine

$$\sup_{n\geq 1}\vert\phi(x_n)\vert=\sup_{n\geq 1}\vert\phi(A^{-1}Ax_n)\vert \leq\Vert A^{-1}\Vert\sup_{n\geq 1}\vert\phi(Ax_n)\vert < \infty$$ (2.19)

where $\Vert A^{-1}\Vert:=\Vert(A\vert _{A(X)})^{-1}\Vert<\infty$ because the open mapping theorem applies once we realize the trivial kernel makes $A\vert _{A(X)}$ a surjection. This means the sequence is uniformly bounded, so by the compactness of $K$, select a subsequence $Kx_{n_k}\to Kx$.

Now it remains to prove that

$$\phi(Ax_{n_k}) \to \phi(Ax) \quad \forall \phi\in X^*$$ (2.20)

We will need

$$\lim_{k\to\infty} \vert\phi(x_{n_k} - x)\vert = \lim_{k\to\infty} \vert\phi(A^{-1}Ax_{n_k} - A^{-1}Ax)\vert \leq \Vert A^{-1}\Vert \lim_{k\to\infty} \vert\phi(Ax_{n_k} - Ax)\vert$$ (2.21)

$$\leq \Vert A^{-1}\Vert \lim_{k\to\infty} \vert\phi((K+I)x_{n_k} - (K+I)x)\vert$$ (2.22)

$$\leq \Vert A^{-1}\Vert \lim_{k\to\infty} \bigg[\phi(Kx_{n_k} - Kx)\vert + \vert\phi(x_{n_k}-x)\vert\bigg]$$ (2.23)

The first limit equals zero since $Kx_{n_k}\to Kx$, leaving us with

$$\lim_{k\to\infty} \vert\phi(x_{n_k} - x)\vert \leq \Vert A^{-1}\Vert \lim_{k\to\infty} \vert\phi(x_{n_k} - x)\vert$$ (2.24)

This inequality proves that this limit equals zero since $\Vert A^{-1}\Vert<1$. Therefore,

$$\lim_{k\to\infty} \phi(Ax_{n_k}) = \lim_{k\to\infty} \phi(x_{n_k} + Kx_{n_k}) = \phi(x + Kx) = \phi(Ax)$$ (2.25)

See Lemma 7.3.1 of [4] for a proof that does not suppose $\ker A=0$. $\qedsymbol$