TITLE: Extension of bounded linear operators
QUESTION [5 upvotes]: Let $X,Y$ be Banach spaces, and let $X_0$ be a subspace of $X$ (by subspace I mean a closed linear set). Consider the set $Ext(X_0,Y)$ of all bounded linear operators $A_0:X_0\to Y$ which have an extension to a bounded linear operator $A:X\to Y$ 
(not necessarily with the same norm).
Question: is $Ext(X_0,Y)$ closed in $B(X_0,Y)$ (with the usual operator norm)?
Equivalently, is the set of all bounded linear operators $A_0:X_0\to Y$ which have no extension to a bounded linear operator $A:X\to Y$ open in $B(X_0,Y)$?
If $Ext(X_0,Y)$ is not closed (in general), for which $X,X_0;Y$ this set is closed in $B(X_0,Y)$?
I will be very grateful for any remarks and comments.

REPLY [6 votes]: Tomek answered your question, but here are some further comments. $Ext(X_0,Y)$ is the image of $L(X,Y)$ under the restriction map, the kernel of which is a closed subspace, so $Ext(X_0,Y)$ is complete under a stronger norm. Thus the only way $Ext(X_0,Y)$ can be complete in the operator norm is for the two norms to be equivalent. (Otherwise, as Tomek pointed out, there will be a non extendable compact operator.)  Now every finite rank operator from $X_0$ to $Y$ is extendable, so for $Ext(X_0,Y)$ to be complete in the operator norm, there must be a constant $C$ s.t. every finite rank norm one operator from $X_0$ to $Y$ has an extension to an operator of norm at most $C$.  This condition is also sufficient if $Y$ is reflexive (or, more generally, is complemented in $Y^{**}$).  Indeed, direct the finite dimensional subspaces of $X_0$ by inclusion.  Given $T:X_0 \to Y$ of norm one, consider for $E\subset X_0$ finite dimensional an extension $T_E$ of the restriction of $T$ to $E$ with $\|T_E\|\le C$.  This net has a subnet that converges in the weak$^*$ operator topology to an extension of $T$ to an operator from $X$ to $Y^{**}$ that has norm at most $C$.