Existence of a bounded right inverse to a linear closed surjective operator 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.
 A: It seems to me that one can prove the existence of a right inverse without using the polar composition: If we endow $D(A)$ with the graph norm, $A$ becomes a continuous linear surjection between Hilbert spaces and thus has a continuous linear right inverse into $D(A)$ which is also continuous as an operator with values in $H$.
This applies also to real Hilbert spaces and in some situations even to Banach spaces, if the range $K$ is projective, e.g., $\ell^1$.
A: 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.
