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If $F: V \to W$ is a continous linear map of Banach spaces, then $\ker(F)$ is closed and $V/\ker(F)$ is again a Banach space (c.f. this answer). As far as I understand it, the image of $F$ is not necessarily a Banach space, but only if the image of $F$ is closed.

I was wondering -- how can we ''algebraify'' taking the closure of $\mathrm{Im}(F)$, i.e.

  1. if there's a way to get a ''natural'' Banach space structure on $\overline{\mathrm{Im}(F)}$; and
  2. how would this induced structure relate to the Banach space $V/\ker(f)$ (so in some sense if there's an ''infinitesimal Banach-thickening of $V/\ker(F)$" that would be isomorphic to $\overline{\mathrm{Im}(F)}$ as a Banach space.)

(Sidenote: I don't know any functional analysis, so this might very well be a very naive/bad-behaved question)

Aaron Wild
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  • What do you mean by "$F$ is closed"? I think that what you really want to say is "the image of $F$ is closed". – Giuseppe Negro Sep 17 '21 at 15:04
  • yeah that's what I meant, sorry. I'll fix it rn. – Aaron Wild Sep 17 '21 at 15:05
  • To clarify, are you asking about the image of a Banach space via a discontinuous linear map? – tomasz Sep 17 '21 at 15:08
  • I'm sorry but why does it need to be discontinous? I was under the impression that the image of a continous map of Banach spaces is not necessarily again a Banach space, but only if it's closed in the domain. – Aaron Wild Sep 17 '21 at 15:13

2 Answers2

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If you take the spaces $\ell_1 $ and $ c_0$ and the map $$F:\ell_1 \to c_0 $$ $$F(x) = x $$ then of course $$||F(x)||_{c_0}\leq ||x||_{\ell_1}$$ hence $F$ is continous linear mapping. The closure of image $\overline{\mbox{im}F}=c_0$ hence it is dense subset of $c_0 $ and $\mbox{ker} F =0.$ So $\ell_1/\mbox{ker} F \equiv \ell_1$ and can not be made a "thicker" to be isomorphic to $c_0.$

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Warning: Heavy use of terminology from category theory

From a category theoretic point of view, it would be better to distinguish the range $F(V)$ and the image which is defined as the kernel of the cokernel. In the category of Banach spaces as objects and continuous linear maps as morphisms, the kernel is indeed the null space $N=\{x\in V: F(x)=0\}$ with the restriction of the norm of $V$, the cokernel is the quotient $W/\overline{F(V)}$ with the quotient norm from $W$ and the image is $\overline{F(V)}$ with the norm from $W$. Finally, $V/N$ is the coimage of $F$. The canonical map between the coimage and the image is as well a monomorphism as an epimorphism of the category but it is not an isomorphisms. In Grothendieck's terminology, the category of Banach spaces is thus not an abelian category.

Jochen
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