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Suppose that $X$ and $Y$ are Banach spaces. Is $\{f\in B(X,Y):f\ \text{has a left inverse}\}$ an open subset of $B(X,Y)$?

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    $\begingroup$ In fact, both the set of left and right inverses in B(X,Y) are open sets. The set of surjective operators as well -- while, of course the injective operators do not make an open set, unless it's empty. $\endgroup$ – Pietro Majer Aug 20 '14 at 15:22
  • $\begingroup$ @Pietro Would you please explain why the set of injective operators is not necessarily open? $\endgroup$ – Aurora Aug 20 '14 at 19:00
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    $\begingroup$ Just think to Id/n... $\endgroup$ – Pietro Majer Aug 20 '14 at 19:41
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    $\begingroup$ If $X=Y$ is of finite dimension, injective operators are surjective. One thus needs infinite dimensional spaces to have the set of injectivions not open. Anyway, the "right" question would be not for injections but for monohomomorphisms, i.e., injections with closed range (surjections between Banach spaces are always epihomomorphisms, i.e. open). $\endgroup$ – Jochen Wengenroth Aug 21 '14 at 12:13
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Yes. Let $U=\{A\in B(X,Y);\text{there is }B\in B(Y,X)\text{ so that }BA=I_X\}$. (You probably want to have the left inverse in $B(Y,X)$. Otherwise you can get counterexamples from compact injections.)

Take any $A\in U$. Then there is $B\in B(Y,X)$ so that $BA=I_X$. Suppose $C\in B(X,Y)$ satisfies $\|C\|\leq\frac12\|B\|^{-1}$. Then $B(A+C)=I_X+BC\in B(X,X)$. Now $\|BC\|\leq\|B\|\cdot\|C\|\leq\frac12$, so $I_X+BC$ has a continuous inverse by Neumann series. Since $(I_X+BC)^{-1}B(A+C)=I_X$, the operator $A+C$ has a continuous left inverse. Therefore $A$ is an interior point of $U$.

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