I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,
Let $E,F$ be Banach spaces. Let $\mathcal L(E, F)$ be the space of bounded linear operators from $E$ to $F$, and $\mathcal K(E, F)$ its subspace consisting of compact operators. Let $T \in \mathcal L(E, F)$. We denote by $R(T)$ and $N(T)$ the range and kernel of $T$.
- Prove that the following properties are equivalent:
- (A) $\dim N(T) < \infty$ and $R(T)$ is closed.
- (B) There exist a constant $C>0$ and a finite-rank operator $P \in \mathcal L(E, E)$ such that $P=P^2$ and $|u|_E \le C (|Tu|_F + |Pu|_E)$ for all $u \in E$.
- (C) There exist a Banach space $G$, an operator $Q \in \mathcal K(E, G)$, and a constant $C>0$ such that $|u|_E \le C (|Tu|_F + |Qu|_G)$ for all $u \in E$.
- Assume that $T$ satisfies (A). Prove that $(T+S)$ also satisfies (A) for every $S \in \mathcal K(E, F)$.
- Prove that the set of all operators $T \in \mathcal L(E, F)$ satisfying (A) is open in $\mathcal L(E, F)$.
There are possibly subtle mistakes that I could not recognize in below attempt of (3). Could you please have a check on it?
Let $\| \cdot \|$ be the operator norm of $\mathcal L(E, F)$. Let $\mathcal A$ be the set of all operators $T \in \mathcal L(E, F)$ satisfying (A). Let $T \in \mathcal A$. We will find $\varepsilon>0$ small enough such that $B(T, \varepsilon) \subset \mathcal A$. Here $B(T, \varepsilon) := \{S \in \mathcal L(E, F) : \|S-T\| < \varepsilon\}$.
By (1), there exist a Banach space $G$, an operator $Q \in \mathcal K(E, G)$, and a constant $C>0$ such that $|u|_E \le C (|Tu|_F + |Qu|_G)$ for all $u \in E$. Then for $S \in \mathcal L(E, F)$ and $u \in E$, $$ \begin{align*} |u|_E &\le C (|Su|_F + |(T-S)u|_F + |Qu|_G) \\ &\le C (|Su|_F + |Qu|_G) + C\|T-S\| \cdot |u|_F. \end{align*} $$
The claim then follows from picking $\varepsilon < \frac{1}{C}$ and from (1).