In what follows, $mathcal H^s(F)$ denotes the $s$-dimensional Hausdorff measure of $Fsubsetmathbb R^n$, and $dim_H F$ denotes the Hausdorff dimension of $Fsubsetmathbb R^n$.

Proposition$3.3(a)$:

Let $Fsubsetmathbb R^n$ and suppose that $f ∶ F → ℝ^m$ satisfies the Hölder condition $$|f (x) − f (y)| ⩽ c|x − y|^𝛼 quad (x,yin F)$$

Then $dim_H f(F) ⩽ (1∕𝛼)dim_H F$. In particular, if $f$ is a Lipschitz mapping, that is, if $𝛼 = 1$, then $dim_H f (F) ⩽ dim_H F$.

I think the proof given is incomplete, or I do not understand it well. Here it is:

If $s > dim_H F$, then by Proposition $3.1$, $mathcal H^{s/𝛼}(f (F)) ⩽ c^{s∕𝛼}mathcal H^s(F) = 0$,

implying that $dim_H f (F) ⩽ s/𝛼$ for all $s > dim_H F$. The conclusion for Lipschitz mappings is immediate on taking $𝛼 = 1$.

If $s > dim_H F$, I know that $mathcal H^s(F) = 0$. The conclusion from Proposition $3.1$ makes sense, and we get $dim_H f (F) ⩽ s/𝛼$ for all $s > dim_H F$. But, what about the cases $s = dim_H F$ and $s < dim_H F$? Do we not have to worry about those? Why? I’m not sure I understand how the author has proved the result, and I would appreciate any help.

For your reference, here is Proposition $3.1$:

Proposition$3.1$:

Let $F ⊂ ℝ^n$ and $f ∶ F → ℝ^m$ be a mapping such that $$|f (x) − f (y)| ⩽ c|x − y|^𝛼quad (x,yin F)$$

for constants $𝛼 > 0$ and $c > 0$. Then for each $s$, $$mathcal H^{s/𝛼}(f(F)) ⩽ c^{s/𝛼}mathcal H^s(F)$$

In particular, if $f$ is a Lipschitz mapping, that is, if $𝛼 = 1$, then $$mathcal H^s(f (F)) ⩽ c^smathcal H^s(F)$$