Let $f\colon \mathbb R^d\to \mathbb R^k$ be a $\beta-$ Holder continuous function ($\beta \in (0,1)$) and $A\subset \mathbb R^d$. As for a Lipschitz function $g$ it holds that $H^s(g(A))\leq Lip(g)^s H^s(A)$, also for $f$ should hold a similar inequality. I want to know if what I've done is correct.
So, let $f$ be such that $|f(x)-f(y)|\leq L|x-y|^\beta$ for each $x,y$.
For each $\epsilon > 0$ there exists a covering $\{E_j\}_j$ of $A$ with $diam(E_j)\leq \delta$, such that $$\sum_j \alpha(\beta s)\Big(\frac{diam (E_j)}{2}\Big)^{\beta s}\leq H^{\beta s}_\delta(A)+\epsilon \quad (*). $$
Now $\{f(E_j)\}$ is a covering of $f(A)$ and $diam f(E_j)\leq L (diam(E_j))^\beta:=\delta'$, so $$H^s_{\delta'}(f(A))\leq \sum_j \alpha(s)\Big(\frac{diam f(E_j)}{2}\Big)^{s}\leq \sum_j \alpha(s) L^s \frac{(diam (E_j))^{\beta s}}{2^s}.$$
But $$\sum_j \alpha(s) L^s \frac{(diam (E_j))^{\beta s}}{2^s}=\frac{\alpha(s) L^s}{\alpha(\beta s) 2^{s-\beta s}}\sum_j \alpha(\beta s) \Big(\frac{(diam (E_j))}{2}\Big)^{\beta s}$$.
Now from $(*)$, we get that $$H^s_{\delta'}(f(A))\leq \frac{\alpha(s) L^s}{\alpha(\beta s) 2^{s-\beta s}} (H^{\beta s}_\delta(A)+\epsilon),$$ so $$H^s(f(A))\leq \frac{\alpha(s) L^s}{\alpha(\beta s) 2^{s-\beta s}} H^{\beta s}(A).$$
Is it correct? Can I get a better estimate?