Timeline for Extension of bounded linear operators

5 events

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Mar 16, 2014 at 17:53	comment	added	Bill Johnson		@Ivan. Sorry; my mistake. Probably you also need that $Y$ has the approximation property or that $X$ has the bounded approximation property.
Mar 15, 2014 at 21:01	comment	added	Ivan Feshchenko		Question: the restriction of $T$ to $E$ is a finite dimensional operator with norm $\leqslant 1$, but this operator is defined on $E$, not on $X_0$. So, why the restriction of $T$ to $E$ has an extension to $X$ with the norm $\leqslant C$?
Mar 15, 2014 at 20:56	comment	added	Ivan Feshchenko		Remark: I don't understand why "Otherwise, as Tomek pointed out, there will be a non extendable compact operator", but the existence of $C$ follows from your idea of considering the restriction map and the open mapping theorem. Indeed, let $\sigma:B(X,Y)\to B(X_0,Y)$ be the restriction map. Then, as you pointed out, $Ran(\sigma)=Ext(X_0,Y)$. Suppose $Ext(X_0,Y)$ is closed. We apply the open mapping theorem to $\sigma$ regarded as a mapping from $B(X,Y)$ to $Ext(X_0,Y)$. Conclusion: there is $C$ such that for any $A_0\in Ext(X_0,Y)$ there is an extension $A:X\to Y$ with $\\|A\\|\leqslant C\\|A\\|$.
Mar 12, 2014 at 22:26	history	edited	Bill Johnson	CC BY-SA 3.0	added 5 characters in body
Mar 12, 2014 at 21:49	history	answered	Bill Johnson	CC BY-SA 3.0