# Hausdorff measure and projections

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.

• 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 – Anton Petrunin Nov 10 '13 at 8:42
• 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. $\pi_V(x_1, x_2, \dots) = (x_1, \dots, x_m)$. – Axiom Nov 10 '13 at 17:41
• 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? – Pietro Majer Nov 10 '13 at 21:24

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 $m$.