Sheet3 B1 Right Inverse Of Surjective Linear Operator

 
Let $X$ and $Y$ be real Hilbert spaces and let $T ∈ ℬ(X, Y)$ be surjective. Show that there exists a unique bounded linear operator $R ∈ ℬ(Y, X)$ such that \[ T R=I_Y \text{ and }\|R T x\|≤\|x\| \text{ for all } x ∈ X . \] %[Hint: Follow the proof 3.11 that for operators between Hilbert spaces $S X$ is closed iff $S^* Y$ is closed] Proof. Define $Z=\ker(T)^⟂⊆X,A=T|_Z∈ℬ(Z,Y)$. $A$ is injective since $Z∩\ker(T)=\{0\}$. $A$ is surjective since $Tx=T(P^Zx)=A(P^Zx)∀x∈X$ where $P^Z$ is orthogonal projection onto $Z$. $A$ is bijective, hence $∃A^{-1}∈ℬ(Y,Z)$ by Inverse Mapping Theorem. Define $R∈ℬ(Y,X),Ry=A^{-1}y∀y∈Y$, then $TR=AA^{-1}=I_Y$. $RT(x)=RA(P^Zx)=P^Zx$, so $‖RT(x)‖≤‖x‖∀x∈X$. Uniqueness: Say $\tilde{R}$ is another possible candidate. For all $y∈Y$, \begin{align*}‖P^Z\tilde{R}y‖&≥‖\tilde{R}TP^Z\tilde{R}y‖ &\text{since }&\|x\|≥\|\tilde{R}T x\|\text{ for all } x ∈ X \\&=‖\tilde{R}T\tilde{R}y‖&\text{since }&TP^Z=T \\&=‖\tilde{R}y‖&\text{since }&T\tilde{R}=I_Y&\end{align*} but $‖\tilde{R}y‖≥‖P^Z\tilde{R}y‖$ by Pythagoras, so $\tilde{R}y∈Z$, so $$ARy=y=T\tilde{R}y=A\tilde{R}y$$ but $A$ is bijective, so $Ry=\tilde{R}y$.