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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.

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    $\begingroup$ Does something go wrong if you simply quotient out the kernel? Then the induced operator is bijective. $\endgroup$
    – Mateusz Wasilewski
    Commented Jun 19, 2020 at 12:37
  • $\begingroup$ But then how do you know that the quotient operator is still closed? $\endgroup$
    – an_ordinary_mathematician
    Commented Jun 19, 2020 at 13:09
  • $\begingroup$ Ok now I see even if the operator is not bounded the kernel is closed, right? $\endgroup$
    – an_ordinary_mathematician
    Commented Jun 19, 2020 at 13:12
  • $\begingroup$ The kernel is closed but I don't think it has to be complemented in $D_A$. I smell a counterexample (at least if you want $R$ to be bounded) but I don't have time to think about it now ... $\endgroup$
    – Nik Weaver
    Commented Jun 19, 2020 at 13:22
  • $\begingroup$ Maybe differentiation on $L^2(0,\infty)$ would work as a counterexample? The right inverse would essentially have to be the antiderivative and unboundedness of the domain should show that it is unbounded. I don't have time to check the details now, sorry. $\endgroup$
    – Mateusz Wasilewski
    Commented Jun 19, 2020 at 14:09

2 Answers 2

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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.

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  • $\begingroup$ You are assuming H is a Hilbert space. The notation may suggest it, but this was never specified explicitly. $\endgroup$
    – Michael Renardy
    Commented Jun 20, 2020 at 0:10
  • $\begingroup$ Yes, I thought it was obvious from the notation that we're talking about Hilbert spaces. $\endgroup$
    – Nik Weaver
    Commented Jun 20, 2020 at 0:48
  • $\begingroup$ Yes I forgot to mentioned that the spaces are assumed to be Hilbert. $\endgroup$
    – an_ordinary_mathematician
    Commented Jun 20, 2020 at 7:25
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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., $\ell^1$.

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