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Let $l^2$ be a Hilbert space of infinite sequences $(z_0, z_1, \cdots)$ with finite $\sum_{i=0}^{\infty} |z_i|^2$.

Are there any simple example of unbounded linear opearator $T: l^2 \to l^2$ with $D(T)=l^2$?

falagar
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    By $D(T)$, do you mean the domain of $T$ in the usual sense for unbounded operators; or are you just looking for an everywhere-defined, unbounded linear map from $\ell^2$ to itself? – Yemon Choi Jul 09 '10 at 09:36
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    Dear Yemon Choi, I'm confused: What's the difference between the two things you mention? – Rasmus Jul 09 '10 at 10:00
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    Good point, Rasmus. For some reason when I write that I thought there was a distinction, but a quick check in Rudin tells me I was mistaken. (I think I was thinking of closed operators, in which case every closed operator with full domain is necessarily bounded by the Closed Graph Theorem.) – Yemon Choi Jul 09 '10 at 10:15
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    My understanding is that you can't do this. But, it would be interesting to know how tight this is logically: are there models of ZF where every linear map $\ell^2\rightarrow\ell^2$ is bounded? – Matthew Daws Jul 09 '10 at 10:27
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    You probably know this already, but $T$ of course cannot be symmetric by the Hellinger-Toeplitz theorem. http://en.wikipedia.org/wiki/Hellinger%E2%80%93Toeplitz_theorem – Willie Wong Jul 09 '10 at 10:49
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    Actually, you immediately have unbounded linear operators on a normed spaces as soon as you have a Hamel basis, and as you know, in general the existence of a Hamel basis on a linear space is ensured by the Zorn lemma. Then, if $(x_i)$ is any Hamel basis and $(y_i)$ is any family of vectors indicized on the same set, there is a unique linear map sending $x_i$ to $y_i$, and it is certainly unbounded if e.g. the $y_i$ are chosen in such a way that $|y_i|/|x_i|$ is unbounded. – Pietro Majer Jul 09 '10 at 13:12
  • Almost identical question has been posted also at MSE: Discontinuous linear functional – Martin Sleziak Oct 21 '12 at 10:37

1 Answers1

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No there aren't any simple, or even any constructive, examples of everywhere defined unbounded operators. The only way to obtain such a thing is to use Zorn's Lemma to extend a densely defined unbounded operator. Densely defined unbounded operators are easy to find.

Zorn's lemma is applied as follows. Let $A$ be an operator on a domain $\mathcal D$. Consider the set $E$ of all extensions of $A$, that is the collection of operators $A'$ on domains $\mathcal D' \supset \mathcal D$ that agree with $A$ when restricted to $\mathcal D$. Then $E$ is partially ordered by inclusion on domains. Furthermore, any linear chain has an upper bound, by taking unions of domains. So there is a maximal element by Zorn. Finally, suppose the maximal element $A$ is defined on a domain $\mathcal D'$ that is not all of $\ell^2$. Let $v$ be any vector in the complement of $\mathcal D'$. Define an extension of $A'$ on $\mathcal D'+\{a v\}$ by, say, mapping $v$ to zero. This contradicts maximality, so any maximal element is globally defined.

Jeff Schenker
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    I'm not sure I understand: How exactly do I use the Hahn-Banach Theorem to extend a densely defined unbounded operator? Hahn-Banach is usually used to extend bounded functionals! – Matthew Daws Jul 09 '10 at 10:26
  • You are right. It is Zorn's lemma that you need. I changed the post. – Jeff Schenker Jul 09 '10 at 12:28
  • It is worth noting that applying Zorn to get the extension used absolutely no topology. This works on any infinite dimensional vector space. – Jeff Schenker Jul 09 '10 at 12:35
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    And in Solovay's model of ZF, where every set in a Polish space has the property of Baire, there is no unbounded linear map from one Banach space to another. – Gerald Edgar Jul 09 '10 at 13:20
  • @Gerald: Thanks, that answers my question above! – Matthew Daws Jul 09 '10 at 14:32
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    @JeffSchenker: Does that mean that if I have a simple (constructive) example of an everywhere defined operator on a Banach space X, then it must be necessarily bounded? Can you give me a reference for this? (I am assuming, of course, that AC holds). – Ruben A. Martinez-Avendano Aug 04 '16 at 19:16
  • What is the motivation for considering a dense domain? In your argument, this assumption is not needed. – John Mars Dec 04 '21 at 20:55
  • This argument for the existence of an extension seems rather abstract. We can also simply extend an algebraic basis of $D(A)$ to a basis of the whole space and assign arbitrary images to the new basis vectors. – Christian Remling Sep 26 '23 at 18:24