4

Background and context

I've been reading several posts related to the existence of unbounded everywhere-defined linear operators on Banach spaces. To name a few:

and then

To briefly summarise what I understood from these:

  • All unbounded operators encountered in practice are going to not be defined everywhere (and generally be defined on a dense subset of a Banach space), and the "interesting theory" concerns closed/closable operators.
  • For closed operators, we know by the closed graph theorem that an unbounded operator can't have closed domain, and thus can't be defined on a full Banach space. For closable operators, I assume one just considers the closure, and then the above applies.
  • If we lift completeness (so work more generally in normed vector spaces), it's not hard to find examples of unbounded operators with full domain.
  • For non-closable operators, examples of unbounded everywhere-defined operators exist but require some form of the axiom of choice.

Question

Assuming all the above is correct, my question mostly revolves around the last point. Focusing on non-closable linear operators in Banach spaces, is there any way to make statements of the form "everywhere-defined unbounded operators defined like such and such can't exist"? I ask this because my (poor) understanding of matters related to the axiom of choice is that it is generally only relevant when making constructions in infinite-dimensional, generally uncountable, spaces.

I would therefore be tempted to say that any linear operator constructed in a "standard way", where we don't employ gimmicks disturbing the axiom of choice, can't be everywhere-defined and unbounded. But of course, I'm not sure how to make precise sense of this. The example here seems to work for arbitrary normed spaces, and therefore it can't be a matter of just considering some simple class of Banach space.

For example, a common way to define linear operators in $\ell^p$ spaces is to take a countable (Schauder) basis, and define the action of the operator on those elements. For example, defining $Te_n=n e_n$ on the subset of $x\in\ell^2$ (or perhaps more generally $x\in\ell^p$) that gives finite $\|Tx\|_p$. Can this "type" of construction (assuming there's a meaningful way to talk about such "type") ever produce unbounded everywhere-defined operators? Can similar procedures work in other standard spaces like $L^p$? Or are there better ways to make statements about the inexistence of "sufficiently nice" unbounded everywhere-defined linear operators between Banach spaces?

glS
  • 7,963
  • It is known that, without the axiom of choice, it is consistent (with ZF) that all linear operators between Banach spaces are bounded. See Asaf’s comment under this question. – David Gao Aug 27 '24 at 17:25
  • 1
    (The example in this answer you cited requires the existence of a Hamel basis, which in turn requires the axiom of choice. As for what you’re proposing in the final paragraph, that always produces an unbounded closed operator, whence never everywhere defined.) – David Gao Aug 27 '24 at 17:32
  • @DavidGao interesting, thanks! How do you see that it always produces a closed operator? Thinking a bit more about this I figure what I had in mind was something about defining operators on spaces that have a Schauder basis (which apparently is referred to as "having the $\Pi$ property"), via the action on the basis elements. Are you saying this always leads to closed operators? – glS Aug 27 '24 at 17:46
  • also thanks for the link to the MO question. That led me to https://mathoverflow.net/a/31166/84108, which is highly relevant to this question and I didn't see before – glS Aug 27 '24 at 17:50
  • 1
    Defining operators “via the action on the basis elements” is overly broad. I just meant diagonal operators are always closed. Namely, if ${e_n}$ is a Schauder basis and ${\alpha_n}$ is a sequence of numbers, then consider the operator $T$ defined by $T(\sum_nc_ne_n)=\sum_n\alpha_nc_ne_n$, whenever both $\sum_nc_ne_n$ and $\sum_n\alpha_nc_ne_n$ converge. Such $T$ is always closed, which follows by noting coordinate functionals are bounded, so if $x_m\to x$ and $Tx_m\to y$, then $(x_m)_n\to(x)_n$ and $(Tx_m)_n=\alpha_n(x_m)_n\to(y)_n$ for each $n$. Thus, $(y)_n=\alpha_n(x)_n$, i.e., $Tx=y$. – David Gao Aug 27 '24 at 18:00
  • 1
    But defining operators by arbitrary actions on a Schauder basis don’t always lead to closed operators, not even closable operators. For example, if ${e_n}$ is the standard basis of $\ell^2$, defining $Te_n = e_1$ does not result in a closable operator - $\frac{1}{n}(e_1 + \cdots + e_n) \to 0$, but $T(\frac{1}{n}(e_1 + \cdots + e_n)) = e_1 \not\to 0$. – David Gao Aug 27 '24 at 18:04

0 Answers0