Let the $s$-dimensional Hausdorff measure for a Borel set $F \subset \mathbb{R}^d$ be denoted as follows:
$$\mathcal{H}^s(F) := \lim_{\delta \to 0} \mathcal{H}^s_\delta(F) := \lim_{\delta \to 0} \inf \left\{ \sum_i |A_i|^s: \text{ $\{A_i\}_i$ is a $\delta$-cover of $F$ } \right\}, $$ where the infimum is taken over all the $\delta$-covers of $F$. Then the Hausdorff dimension of this set is defined as follows:
$$\dim_{H} F := \sup \{s>0; \> \mathcal{H}^s(F) = + \infty \} = \inf \{s>0 : \> \mathcal{H}^s(F) = 0 \}.$$
My question is as follows.
Let us assume that $\dim_{H} F = \alpha$. Can we prove an upper-bound of the following form? (for a fixed $\delta >0$ and bounded $F$) $$ \log \mathcal{H}^\alpha_\delta(F) \leq C (\alpha + c) \log \frac1{\delta},$$ for some $C,c > 0$.
I have the feeling that $c$ can be equal to $1$, but at this stage any $c$ should work for my purpose. On the other hand we can replace $\alpha$ with any $0<q < \alpha$ if it's going to be any easier. I would greatly appreciate if someone could help me on this issue.