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Let $X$ be a Banach space. Let $Y, F \subseteq X$ be subspaces with $Y$ closed and $F$ finite-dimensional. It follows that $F + Y$ is also a closed subspace of $X$.

Question: Is there an isometric isomorphism of finite-dimensional Banach spaces between $\frac{F+Y}{Y}$ and $\frac{F}{F \cap Y}$?

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  • $\begingroup$ What norm are you proposing to use on the quotient spaces? $\endgroup$ – Rob Arthan Feb 16 '17 at 4:08
  • $\begingroup$ @RobArthan: If $S$ is closed subspace of a Banach space $X$, one usually defines $\|x+S\|$ to be the distance from the coset $x+S$ to $0 \in X$ or, equivalently, the distance from $x$ to $S$. $\endgroup$ – Mike F Feb 16 '17 at 4:10
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    $\begingroup$ @RobArthan: So, in other words, the question is whether $\mathrm{dist}(x,Y) = \mathrm{dist}(x,F \cap Y)$ for all $x \in F$, which looks rather unlikely, now that I write it... $\endgroup$ – Mike F Feb 16 '17 at 4:16
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No, or at least the canonical bijection is typically not an isometry. For instance, let $X=\mathbb{R}^2$ with the Euclidean norm, let $F$ be the span of $(1,1)$, and let $Y$ be the span of $(1,0)$. Then $F\cap Y=0$, so $\frac{F}{F\cap Y}=F$. But the canonical bijection sends the element $(1,1)\in F$ to the coset $(1,1)+Y=(0,1)+Y$ in $\frac{F+Y}{Y}$, and $\|(1,1)\|=\sqrt{2}$ but $\|(0,1)+Y\|=1$.

(In this example, $\frac{F}{F\cap Y}$ and $\frac{F+Y}{Y}$ happen to be isometrically isomorphic, but not by the natural map. Probably you can get a similar example where they are not isometrically isomorphic by any map (say, take a similar example where the quotient spaces end up being 2-dimensional and you have a less symmetrical norm than the Euclidean norm), but proving that no map can be an isometry is probably kind of a mess.)

However, the canonical bijection is always an isomorphism of topological vector spaces. Indeed, the canonical bijection is norm-decreasing as a map $\frac{F}{F\cap Y}\to\frac{F+Y}{Y}$, since given $f\in F$, the distance from $f$ to $F\cap Y$ cannot be smaller than the distance from $f$ to $Y$. So the canonical bijection is a bounded linear bijection $\frac{F}{F\cap Y}\to\frac{F+Y}{Y}$, which is hence a topological isomorphism by the open mapping theorem.

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  • $\begingroup$ Thanks! In hindsight this was clearly naive... but I wanted the lemma so badly that I was psychologically incapable of disproving it! :p $\endgroup$ – Mike F Feb 16 '17 at 4:21

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