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 $\frac{F}{F \cap Y}$?

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.