If $f$ is Lipschitz, then the following holds for the Hausdorff dimension: $$\dim_H f(A) \le \dim_H A.$$
Is the same true for the box counting dimension?
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3$\begingroup$ And what difference does this switch make in the proof you know? (I assume that the Lipschitz constant is uniform) $\endgroup$– fedjaCommented May 17, 2019 at 21:20
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Of course this is still true, you simply map the covering using $f$. Let $L$ denote the Lipschitz constant for $f$. For a given $\epsilon$, suppose that $A$ can be covered by $N_A(\epsilon)$ sets of diameter at most $\epsilon$, and $N_A(\epsilon)$ is minimal. Then mapping each set in the cover using $f$ we get a cover of $f(A)$ that yields $N_{f(A)}(L \epsilon) \le N_A(\epsilon)$. The upper box counting dimension of $A$ is the limsup of $(\log N_A(\epsilon))/(-\log \epsilon)$ as $\epsilon \to 0$. Now $$(\log N_{f(A)}(L \epsilon))/(-\log (L\epsilon)) \le (\log N_A(\epsilon))/(-\log \epsilon) \cdot (\log (L\epsilon))/ (\log \epsilon).$$ Taking limsup gives the statement for upper box counting dimension, the proof for lower dimension is similar.
answered May 22, 2019 at 6:24
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$\begingroup$ Yuval, could you please clarify why $N_{f(A)}(L\epsilon) \le N_{A}(\epsilon)$? I'm having trouble to see this in the case where $0<L<1$. $\endgroup$– anchovaCommented Jul 13, 2021 at 18:55
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$\begingroup$ If $A$ is covered by sets $\{S_j\}$ for $j \ge 1$ then $f(A)$ is covered by sets $\{f(S_j)\}$ for $j \ge 1$. Observe that diam$f(S) \le L$diam$(S)$ for any set $S$ by the definition of diameter. $\endgroup$ Commented Jul 13, 2021 at 21:08