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.


2 Answers 2


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; 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 about extensions of compact operators.


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 $C$.

  • $\begingroup$ 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 $\|A\|\leqslant C\|A\|$. $\endgroup$ Mar 15, 2014 at 20:56
  • $\begingroup$ 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 $\leqslant C$? $\endgroup$ Mar 15, 2014 at 21:01
  • $\begingroup$ @Ivan. Sorry; my mistake. Probably you also need that $Y$ has the approximation property or that $X$ has the bounded approximation property. $\endgroup$ Mar 16, 2014 at 17:53

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