# If a subspace $F$ is contained in a subspace $G$, and $H$ is close to $G$, can we choose a subspace of $H$ close to $F$?

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 $$g(F,J)<\varepsilon$$?

• It looks to me that $F\subset G\implies d(F,H)\leq d(G,H)$ so you could take $\delta=\varepsilon$ and $J=H$. Dec 16, 2020 at 10:11
• @LiviuNicolaescu seems unlikely: what if $F$ is too small? Note that the distance is symmetrical.
– erz
Dec 16, 2020 at 10:31
• I meant the gap $g(F,H)\leq g(G,H)$. If $F\subset G$ I believe $g(G,F)=1$. Dec 16, 2020 at 11:58
• @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 $g(F,G)=1$.
– erz
Dec 16, 2020 at 12:01
• In Hilbert spaces if $F\subsetneq G$, then $g(G,F)=1$. Also if $\dim F<\dim G<\infty$ then $g(G,F)=1$. Dec 16, 2020 at 13:38

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.
• 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 $Z$. Dec 17, 2020 at 4:10