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 $u \in E$.
My proof for the part (C) $\implies$ (A) is much simpler than the one by the author. There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
- (A) $\implies$ (B)
It follows from $\dim N (T) < \infty$ that $N(T)$ has a complement $L$. This means $L$ is a closed subspace of $E$ such that $N(T) \cap L = \emptyset$ and $E = N(T) + L$. Let $P:E\to E$ be the projection onto $N(T)$. Then $N(P)=L$ and $R(P) = N(T)$. Then $P$ is a finite-rank operator and $P=P^2$. We consider a bounded linear operator $$ \varphi:E \to N(T) \times R(T), u \mapsto (Pu, Tu). $$
First, we verify that $\varphi$ is surjective. Let $(u, v) \in N(T) \times R(T)$. There is $x \in E$ such that $v=Tx$. There is a unique pair $(y,z) \in N(T) \times L$ such that $x=y+z$. Then $Ty=0$ and $Pz=0$. Then $P(u+z) =Pu$ and $T(u+z)=v$. Second, we verify that $\varphi$ is injective. Let $u\in E$ such that $(Pu, Tu)=(0, 0)$. Then $u \in L$ and $u\in N(T)$. Then $u=0$.
Because $F$ is a Banach space and $R(T)$ closed in $F$, we get $R(T)$ is also a Banach space. By open mapping theorem, $\varphi^{-1}$ is continuous. The claim then follows.
- (B) $\implies$ (C)
We just pick $G:=E$ and $Q:=P$.
- (C) $\implies$ (A)
We consider a bounded linear operator $$ \varphi:E \to F \times G, u \mapsto (Tu, Qu). $$
By Theorem 2.21 in the same book, $R(\varphi)$ is closed and $N(\varphi) = \{0\}$. Then $R(\varphi)$ is itself a Banach space. By open mapping theorem, $\varphi:E \to R(\phi)$ is a homeomorphism. Then a subset is open (resp. closed) in $R(\varphi)$ IFF its inverse image is open (resp. closed) in $E$.
We have $\varphi^{-1}(R(T) \times \{0\})= N(Q)$ is closed in $E$, so $R(T) \times \{0\}$ is closed in $R(\varphi)$. On the other hand, $R(\varphi)$ is closed in $F \times G$. Then $R(T) \times \{0\}$ is closed in $F \times G$. Then $R(T)$ is closed in $F$.
We have $\varphi^{-1}(\{0\} \times R(Q))= N(T)$ is closed in $E$, so $\{0\} \times R(Q)$ is closed in $R(\varphi)$. On the other hand, $R(\varphi)$ is closed in $F \times G$. Then $\{0\} \times R(Q)$ is closed in $F \times G$. Then $R(Q)$ is closed in $G$. Then $Q$ is a finite-rank operator. Then $\dim R(Q) < \infty$. We have $\varphi^{-1}(\{0\} \times R(Q))= N(T)$, so $\dim N(T) = \dim R(Q)$. This completes the proof.
