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I know that for a Banach space $X$ and a linear functional $T:X\rightarrow\mathbb{R}$ in its dual $X'$ the following holds: \begin{align}T \text{ is continuous } \iff \text{Ker }T \text{ is closed}\end{align} which probably holds for general operators $T:X\rightarrow Y$ with finite-dimensional Banach space $Y$. I think the argument doesn't work for infinite-dimensional Banach spaces $Y$. Is the statement still correct? I.e. continuity of course still implies the closedness of the kernel for general Banach spaces $X,Y$ but is the converse still true?

Gerry Myerson
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Julian
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    You're right: when $Y$ is finite-dimensional, the above result holds (see http://math.stackexchange.com/a/135805/19379). Also, if you assume $X$ is finite-dimensional, the above result also holds (you can try to prove it; basically, use the fact that, given a basis, the coordinate functions are linear functionals). – M Turgeon Jul 12 '12 at 21:56

2 Answers2

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The result is false if $Y$ is infinite dimensional. Consider $X=\ell^2$ and $Y=\ell^1$ they are not isomorphic as Banach spaces (the dual of $\ell^1$ is not separable). However they both have a Hamel basis of size continuum therefore they are isomorphic as vector spaces. The kernel of the vector space isomorphism is closed (since is the zero vector) but it can not be continuous.

azarel
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    One should note that the topological dual of $\ell^1$ is not separable. However if we wish to use top. duals to separate the spaces, $\ell^2$ is self-dual, while $\ell^1$ is not. This property separates them even in strange models of automatic continuity. – Asaf Karagila Jul 12 '12 at 22:00
  • Is it possbile to construct a concrete unbounded operator which has a closed kernel? – Julian Jul 12 '12 at 22:04
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    @Julian: If your space is a Banach space then it is not possible to construct an unbounded operator explicitly. Indeed it is consistent without the axiom of choice that every linear operator from a Banach space into a normed space is continuous, i.e. bounded. – Asaf Karagila Jul 12 '12 at 22:40
  • @Julian: No, you cannot "concretely" construct an unbounded, everywhere defined operator on a Banach space. If I recall correctly, it is consistent with ZF+DC that no such operator exists, so you have to use the axiom of choice. – Nate Eldredge Jul 12 '12 at 22:40
  • @AsafKaragila That's a really nice result do you have a reference? – azarel Jul 12 '12 at 22:49
  • @azarel: You can start with this answer of mine and follow the links. It follows from either Lebesgue measurability or Baire property (Garnir proved it from the former, and it was later proved the latter holds - at least for separable spaces). There is a paper by Wright about continuous operators on Hilbert spaces. I once wrote a minor observation that $\ell^1$ is reflexive in such models, but I did not finish it (it's near done, though) when I discovered it was already in Schecter's book. – Asaf Karagila Jul 13 '12 at 07:28
  • @AsafKaragila Thanks – azarel Jul 13 '12 at 17:42
  • I recall a particular paper by Brunner called something like "Garnir's Dream Space" or something like that. It was where I started reading on the topic. – Asaf Karagila Jul 13 '12 at 17:53
  • The spaces are not only not isomorphic in the sense of Banach spaces (linearly isometric) but even not isomorphic in the sense of topological vector spaces (linearly homeomorphic). – C-star-W-star Jul 11 '16 at 23:04
  • Besides, nice answer! – C-star-W-star Jul 11 '16 at 23:07
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No, this is not true. Here is a counterexample for every infinite-dimensional Banach space $Y$.

Take a discontinuous functional $\phi\in Y^*$ (this is always possible if $Y$ is infinite-dimensional, by the axiom of choice). Let $A:=\ker \phi$, and $B$ any algebraic supplement. Note that $A$ is non-closed, and $B$ is closed in view of $\dim B=\text{codim} A=1$.

Consider the projection $p:Y\to Y$ onto $A$, w.r.t. the algebraic direct sum decomposition $Y=A\oplus B$. Now $\ker p=B$ is closed, but $\ker(1-p)=A$ is not, so $p$ is discontinuous.

wildildildlife
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