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.



Extension of bounded linear operators
The set $Ext(X\_0, Y)$ will be rarely closed, I believe. (Perhaps you want to use a different symbol for this set, as $Ext$ has its own meaning.)


Suppose that $X$ is a Banach space which is complemented in $X^{\*\*}$, $X$ is isomorphic to $X\oplus X$, $X$ has a Schauder basis and it contains an uncomplemented copy of itself, $X\_0$ say. Then there is a compact operator $K\colon X\_0\to X$ which does not extend to $X$. (This is implicit in in the proof of Lemma 5.7 [here](http://arxiv.org/pdf/1208.4762v3.pdf); apologies for self-advertisement). On the other hand, $X\_0$ has the approximation property, so $K$ can be approximated by finite-ranks which are extendable. 


The classical spaces $\ell\_p$ and $L\_p$ for $p\in [1,2)\cup (2,\infty)$ satisfy the above assumptions. The assumption that $X\cong X\oplus X$ can be dropped but things get messier then.


Note that the above situation with compact operators cannot happen if $Y$ is a $\mathscr{L}\_\infty$-space as in this case we have the [Grothendieck–Lindenstrauss theorem](http://projecteuclid.org/euclid.bams/1183524860) about extensions of compact operators.


