MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $D(T)=l^2$?

share|cite|improve this question
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 9 '10 at 9:36
Dear Yemon Choi, I'm confused: What's the difference between the two things you mention? – Rasmus Bentmann Jul 9 '10 at 10:00
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 9 '10 at 10:15
You probably know this already, but $T$ of course cannot be symmetric by the Hellinger-Toeplitz theorem. – Willie Wong Jul 9 '10 at 10:49
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 9 '10 at 13:12
up vote 10 down vote accepted

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.

share|cite|improve this answer
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 9 '10 at 10:26
You are right. It is Zorn's lemma that you need. I changed the post. – Jeff Schenker Jul 9 '10 at 12:28
I got it, thank you! – falagar Jul 9 '10 at 12:33
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 9 '10 at 12:35
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 9 '10 at 13:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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