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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 $$\sum_j \alpha(\beta s)\Big(\frac{diam (E_j)}{2}\Big)^{\beta s}\leq H^{\beta s}_\delta(A)+\epsilon \quad (*). $$

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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?

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Your argument is correct, and there is no way to improve the estimate. A simple example can be given in the context of abstract metric spaces: let $X $ be the metric space $ (A,d_\beta)$ where $d_\beta(x,y)=|x-y|^\beta$ is a metric. Let $f$ be the identity map from $A$ to $X$; it has the property $|f(x)-f(y)| = |x-y|^\beta$ for all $x,y$.

Unwinding the definition of $H^s$, one obtains

$$H^s(X) = \frac{\alpha(s) }{\alpha(\beta s) 2^{s-\beta s}} H^{\beta s}(A)$$

which shows that equality holds in your estimate.

The abstract metric space $ (A,d_\beta)$ can be realized as a subset of a sufficiently high-dimensional Euclidean space (i.e., it’s isometric to such a subset); this is a consequence of a theorem of Schoenberg on isometric embeddings into Euclidean spaces.

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