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 $S \in \mathcal K(E, F)$.
- Proce that the set of all operators $T \in \mathcal L(E, F)$ satisfying (A) is open in $\mathcal L(E, F)$.
- Let $F_0$ be a closed linear subspace of $F$ and let $S \in \mathcal K(F_0, F)$. Prove that $R(I+S)$ is closed. Here $I$ is the identity map.
There are possibly subtle mistakes that I could not recognize in below attempt of (4). Could you please have a check on it?
My proof is quite long. Is there a shorter alternative approach? Surprisingly, question (4) seems "easy" that that author does not include the solution.
Because $F_0$ is closed subspace of a Banach space, it is itself a Banach space. Let $E_1 := N(I+S)$. Then $E_1$ is closed subspace of $F_0$.
First, we prove that $\dim E_1 < \infty$. Let $B_{E_1}$ and $B_{F_0}$ be the closed unit balls of $E_1$ and $F_0$ respectively. Let $x\in B_{E_1}$. Then $|x|_{F}=1$ and $Sx=-x$. Then $x \in S(B_{F_0})$. So $B_{E_1} \subset S(B_{F_0})$ and thus $B_{E_1} \subset \overline{S(B_{F_0})}$. Because $S$ is compact, $\overline{S(B_{F_0})}$ and thus $B_{E_1}$ are compact. Then $E_1$ is finite dimensional.
Let $(u_n) \subset F_0$ and $f\in F$ such that $f_n := (I+S)u_n \to f$. We will prove that $f \in R(I+S)$. Let $$ d(u_n, E_1) := \inf_{x \in E_1} |u_n-x|_F \quad \forall n \ge 1. $$
Because $\dim E_1 < \infty$, there is $v_n \in E_1$ such that $d(u_n, E_1) = |u_n-v_n|_F$. Then $f_n = (I+S)(u_n-v_n) \to f$.
Second, we prove that $(u_n-v_n)_n$ is bounded. Assume the contrary that there is a subsequence $(u_{n_k}-v_{n_k})_k$ such that $|u_{n_k}-v_{n_k}|_F \xrightarrow{k \to \infty} \infty$. Let $$ w_n := \frac{u_n-v_n}{|u_n-v_n|_F} \quad \forall n \ge 1. $$
Then $$ (I+S)w_{n_k} = \frac{(I+S)(u_{n_k}-v_{n_k})}{|u_{n_k} - v_{n_k}|_F} = \frac{f_{n_k}}{|u_{n_k} - v_{n_k}|_F} \xrightarrow{k \to \infty} 0. $$
We have $w_n \in B_{F_0}$ for all $n\ge 1$. Because $S$ is compact, there exist $y\in F$ and a subsequence (still denoted by $(w_{n_k})_k$ for simplicity) such that $Sw_{n_k} \xrightarrow{k \to \infty} y$. On the other hand, $(I+S)w_{n_k} \xrightarrow{k \to \infty} 0$. Then $w_{n_k} \xrightarrow{k \to \infty} -y$. Then $(I+S)(-y)=0$ and thus $-y \in E_1$. Then $$ d(w_{n_k}, E_1) \xrightarrow{k \to \infty} d(-y, E_1)=0, $$ which contradicts $$ \begin{align*} d(w_n, E_1) &= d \left ( \frac{u_n-v_n}{|u_n-v_n|_F}, E_1 \right ) \\ &= \frac{d \left (u_n, E_1 \right )}{|u_n-v_n|_F} \quad \text{because} \quad v_n \in E_1 \\ &=1. \end{align*} $$
Then $(u_n-v_n)_n$ is indeed bounded. Because $S$ is compact, there exist a subsequence $(u_{n_k}-v_{n_k})_k$ and $z \in F$ such that $S (u_{n_k}-v_{n_k}) \xrightarrow{k \to \infty} z$. We have $(I+S)(u_n-v_n) \to f$. Then $u_{n_k}-v_{n_k} \xrightarrow{k \to \infty} f-z$. Notice that $F_0$ is closed, so $g:=f-z \in F_0$. Clearly, $f =(I+S)g \in R(I+S)$. This completes the proof.
