Let $A:D\_A \subseteq H \to K$ a linear closed surjctive operator between two Hilbert spaces $H$ and $K$.


One would expect that in such a situation there must exist a bounded right inverse of $A$, namely an operator $R:K \to H$ such that $AR=Id\_K$. In fact this is certainly true if $A$ is bijective but the proof doesn't seem to go through with the hypothesis of surjectivity.


Any ideas what is going on in this situation ?


EDIT: Although the answers cover my original question, i think it is quite natural at this point to ask wether this is true if $H$ and $K$ are more generally Banach space instead of Hilbert spaces.



Existence of a bounded right inverse to a linear closed surjective operator
It's still true in the unbounded case, and you can see this using polar decomposition. Write $A = BU$ where $B$ is some positive unbounded operator on $K$ and $U$ is the orthogonal projection from $H$ onto a closed subspace $H\_0$ followed by some isometry from $H\_0$ onto $K$. We can take $B$ to be a multiplication operator, $B = M\_f$, on $K = L^2(X)$, and then the fact that $BU$ is surjective implies that $f$ must be bounded away from zero. So $1/f$ is a bounded function and thus $B^{-1} = M\_{1/f}$ is a bounded operator. Finally, $U^\*B^{-1}$ is the desired bounded right inverse.


