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Fix $ k \in \mathbb{N} $ and let $ H^k $ be the $k$-dimensional Hausdorff measure on $\ell^\infty $. Also, if $ V $ is a subspace of $ \ell^\infty $, we denote the projection onto $ V $ by $ \pi_V $. Let $ A $ be a subset of $ \ell^\infty $ such that $ H^k(\pi_V(A)) = 0 $ for all finite dimensional subspaces $ V $ of $ \ell^\infty $.

Can we conclude that $ H^k(A) = 0 $?

Any help would be appreciated.

Thanks.

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    $\begingroup$ For $\ell^2$ the answer seems to be no, for orthogonal projections. For $\ell^\infty$ it is not clear which metric you want to consider on $A$ and which projections $\endgroup$
    – Anton Petrunin
    Commented Nov 10, 2013 at 8:42
  • $\begingroup$ The standard metric on $ \ell^\infty $ which is the supremum metric. And you're right. I should've been more careful about the projections. I meant to say projection onto the first m components. $ \pi_V(x_1, x_2, \dots) = (x_1, \dots, x_m) $. $\endgroup$
    – Axiom
    Commented Nov 10, 2013 at 17:41
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    $\begingroup$ In this case I think the answer should be no (see my answer below). Maybe assuming the stronger condition $H^k(\pi(A))=0$ for any projector $\pi$ with finite dimensional range allows a positive answer? $\endgroup$
    – Pietro Majer
    Commented Nov 10, 2013 at 21:24

1 Answer 1

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For any $m\in\mathbb{N}$ let $\pi_m$ denote the projector on the space $V_m:=\{x\in \ell_\infty\, : \, x_j=0,\,\, \forall j>m\}$. Let $S$ denote the left shift operator on $ \ell_\infty$, that is $Sx:=( x_2,x_3,\dots)$ for any $x:=(x_1,x_2\dots)\in \ell_\infty$. Consider the bounded linear operator (a norm-$2$ projector onto $V_m$, indeed)

$$P_m:=\pi_m(I-S^m/2)^{-1}=\pi_m \sum_{j=0}^\infty 2^{-j}S^{jm}\, .$$ Finally, consider the set $A:=\{0,1\}^\mathbb{N_+}=\{x\in \ell_\infty\, : \forall j\,\, x_j\in\{0,1\} \} $. For any $m$ the set $\pi_m(A)=\{0,1\}^m$ is finite, so $\mathcal{H}^k(\pi_m(A))=0$ for any $k>0$. On the other hand, $P_m(A)=[0,2]^m$, so $2^m=\mathcal{H}^m(P_m A)\le \|P_m\|^m \mathcal{H}^m(A)$ for any $m$, whence $\mathcal{H}^m(A)\ge 1$ for any $m$, so in fact $\mathcal{H}^m(A)= \infty$ for any $m$.

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