Let us consider the space of polynomials $P^N$ of degree $\le N$. If $f\in P^N$ vanishes in $>N$ points, then $f\equiv 0$, but for any $N$ points, or fewer, there exists $f\neq 0$ vanishing at those points. Additionally, if $N < M$ then $P^N \subset P^M$.
My question
Is there a family of function spaces $M^\alpha$, with $\alpha \in [0, 1]$, such that:
- If $f\in M^\alpha$ vanishes in a set of dimension $>\alpha$, then $f \equiv 0$;
- If a set $E\subsetneq\mathbb{R}$ has dimension $\le \alpha$, then there exists $f\neq 0$ vanishing in that set;
- If $\alpha < \beta$ then $M^\alpha \subset M^\beta$; and
- $M^0 \supset \bigcup_{N\ge 0} P^N$ and $M^1 \subset C^\infty$?
Notes
- The word dimension is open to interpretation. Can it be Hausdorff, Minkowski, or your favorite dimension.
- The conditions are not a straitjacket. You can modify them to your convenience, keeping in reasonable agreement with the sense of the question. The conditions may be too open, or very restrictive; maybe, in the second condition $<\alpha$ is a better choice.
- I thought Gevrey functions could somehow fit the bill ... no. There are non-trivial functions with compact support, so 1) does not hold.
- Denjoy–Carleman Theorem makes me think that such a family does not exist. The family cannot be quasi-analytic for $\alpha > 0$. Otherwise, condition 2) would not hold because for any set $E$ with dimension $\alpha > 0$ and any $f$ vanishing in $E$ it holds that $f^{(n)}(x) = 0$ for some $x\in E$ and every $n\ge 0$, so quasi-analyticity implies $f \equiv 0$. On the other hand, anything beyond quasi-analytic seems to admit compact support.
Edits
- Following Christian Remling's comment, one should impose additional restrictions on the set $E$. Maybe $E$ closed works (conditions 1 and 2).
