Let $E$ be a Banach space. Recall that the collection of all closed linear subspaces of $E$ can be turned into a metric space in a number of ways. In particular, consider the notion of a gap: if $G$ and $H$ are subspace of $E$, then $$g(G,H)=\max\{\sup\limits_{g\in \partial B_{G}} d(g, H),~\sup\limits_{h\in \partial B_{H}} d(h, G)\},$$ where by $\partial B_{G}$ and $\partial B_{H}$ we mean the intersection of the unit sphere with $G$ and $H$, respectively. Note, that $g(G,H)\le h(G,H)\le 2g(G,H)$, where $h(G,H)$ is the Hausdorff distance between $\partial B_{G}$ and $\partial B_{H}$.

Let $F\subset G$ be subspaces of $E$ and let $\varepsilon>0$. Does there exist $\delta>0$ such that every subspace $H$ with $g(G,H)<\delta$ contains a sub-subspace $J\subset H$ with $g(F,J)<\varepsilon$?

  • $\begingroup$ It looks to me that $F\subset G\implies d(F,H)\leq d(G,H)$ so you could take $\delta=\varepsilon$ and $J=H$. $\endgroup$ Commented Dec 16, 2020 at 10:11
  • $\begingroup$ @LiviuNicolaescu seems unlikely: what if $F$ is too small? Note that the distance is symmetrical. $\endgroup$
    – erz
    Commented Dec 16, 2020 at 10:31
  • $\begingroup$ I meant the gap $g(F,H)\leq g(G,H)$. If $ F\subset G$ I believe $g(G,F)=1$. $\endgroup$ Commented Dec 16, 2020 at 11:58
  • $\begingroup$ @LiviuNicolaescu but not only $H$ has to approximate $F$ (which it does at least as well as it approximates $G$), but another way around as well. And if $F$ is too small it does not approximate $H$. In particular, take your example: $g(G,G)=0$, and yet $g(F,G)=1$. $\endgroup$
    – erz
    Commented Dec 16, 2020 at 12:01
  • $\begingroup$ In Hilbert spaces if $F\subsetneq G$, then $g(G,F)=1$. Also if $\dim F<\dim G<\infty$ then $g(G,F)=1$. $\endgroup$ Commented Dec 16, 2020 at 13:38

1 Answer 1


The answer is "No". You can derive this from Lemma 5.9 and Proposition 5.3 in my paper Ostrovskiĭ, M. I. Topologies on the set of all subspaces of a Banach space and related questions of Banach space geometry. Quaestiones Math. 17 (1994), no. 3, 259–319. In that Lemma a collection of subspaces $G_\varepsilon$, $\varepsilon\in(0,1)$ is constructed in the space $X\oplus X/Y$ which converges to $Y\oplus X/Y$ with respect to gap, and is such that all $G_\epsilon$ are isomorphic to $X$.

We get a counterexample in cases where $X/Y$ does not admit an isomorphic embedding into $X$ (such examples are well-known, e.g. $X=\ell_1$ and $X/Y=c_0$). In fact, if there would be a subspace $U$ in $G_\varepsilon$, which is close to $X/Y$, since $X/Y$ is complemented in the whole space, by Berkson's proposition (see Proposition 5.3 in the paper mentioned above), this would imply that $U$ is isomorphic to $X/Y$, a contradiction.

  • $\begingroup$ So do I understand correctly, that Berkson tells us that in a gap-neighborhood of a complemented subspace not only everybody is complemented, but is in fact isomorphic to this subspace? $\endgroup$
    – erz
    Commented Dec 17, 2020 at 3:35
  • $\begingroup$ Yes, Berkson proved that in a gap-neighborhood of a complemented subspace $Z$ all subspaces are isomorphic to $Z$, the size of the neighborhood depends on the infimum of norms of projections onto $Z$. $\endgroup$ Commented Dec 17, 2020 at 4:10

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