Welcome to MathOverflow

Question

6

1

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 $D(T)=l^2$?

flag	asked Jul 9 2010 at 9:08 falagar 1,703●5●23

1

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 2010 at 9:36

1

Dear Yemon Choi, I'm confused: What's the difference between the two things you mention? – Rasmus Jul 9 2010 at 10:00

1

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 2010 at 10:15

2

You probably know this already, but $T$ of course cannot be symmetric by the Hellinger-Toeplitz theorem. en.wikipedia.org/wiki/… – Willie Wong Jul 9 2010 at 10:49

3

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 2010 at 13:12

show 2 more comments

Jeff Schenker · Accepted Answer · 2010-07-09 12:27:08Z UTC

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.

Unbounded linear operator defined on $l^2$

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.