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 $Ry=\tilde{R}y$.
PREVIOUSThe Riemann–Lebesgue Lemma
NEXTSheet3 Q3