- #1
cogito²
- 98
- 0
I'm wondering if someone can give me an example of an everywhere defined unbounded operator on a (separable for simplicity) Hilbert space in a "constructive" manner. Since it's unbounded, simply a dense definition (i.e. on an orthonormal basis) wouldn't work since you can't extend it by continuity. So a Hamel basis would have to be necessary, but in a Hilbert space a Hamel basis would have to be uncountable so that's not even that easy to "get a hold of". (For that matter, can anyone give me an explicit example of a Hamel basis for the continuous functions on the real numbers?)
I mean if I assumed that the Hamel basis had cardinality at least the continuum (assuming there's nothing between the naturals and the continuum), I could just identify it (or a subset of it) with the reals and then define a multiplication function that multiplied each basis element by its corresponding real number. That would clearly be unbounded and everywhere defined. Although in that case, unless I knew the Hamel basis explicitly, it wouldn't really be that satisfying.
So basically I'm just wondering how necessary the axiom of choice is for the construction of such an operator. Anyone with any wisdom to share?
I mean if I assumed that the Hamel basis had cardinality at least the continuum (assuming there's nothing between the naturals and the continuum), I could just identify it (or a subset of it) with the reals and then define a multiplication function that multiplied each basis element by its corresponding real number. That would clearly be unbounded and everywhere defined. Although in that case, unless I knew the Hamel basis explicitly, it wouldn't really be that satisfying.
So basically I'm just wondering how necessary the axiom of choice is for the construction of such an operator. Anyone with any wisdom to share?