I have to show that $0<\mathcal H_{\alpha }^\delta(\mathcal C)<\infty $ where $\mathcal C$ is the cantor set, $\alpha =\frac{\ln(2)}{\ln(3)}$ and $$\mathcal H_\alpha ^\delta(E)=\left\{\sum_{k=1}^\infty |U_i|^\alpha \mid E\subset \bigcup_{i=1}^\infty U_i, |U_i|<\delta\right\}$$ and $|A|=\sup\{|x-y|\mid x,y\in A\}$. So the fact $\mathcal H_\alpha ^\delta(\mathcal C)\leq 1$ is fine. I'm trying now to show that $\mathcal H_\alpha ^\delta(\mathcal C)\geq \frac{1}{2}$. My proof goes like:
Let $\{U_i\}$ a cover of $\mathcal C$. Since $\mathcal C$ is compact, we can suppose that $\{U_i\}$ is a finit collection of sub-interval of $[0,1]$. For all $i$, let $k$ s.t. $\frac{1}{3^{k+1}}\leq |U_i|<\frac{1}{3^k}$. Let recall that $\mathcal C=\bigcap_{i=0}^\infty C_i$ where $C_i$ is a set of $2^i$ intervals of length $3^{-i}$. Then $U_i$ intersect at most one interval of $C_k$. For $j\geq k$, $U_i$ intersect $2^{j-k}=2^j3^{-k\alpha }\leq 2^j3^\alpha |U_i|^\alpha $ interval of $C_j$. For $j$ big enough, since $\{U_i\}$ is finite, $3^{-(j+1)}\leq |U_i|$ for all $i$, and thus $$2^j\leq \sum_{i}2^j3^\alpha |U_i|^\alpha $$ and thus $\sum_{i}|U_i|^\alpha \geq 3^{-\alpha }=1/2$.
I just don't understand how we get $$2^j\leq \sum_{i}2^j3^\alpha |U_i|^\alpha.$$ I'm sure it's obvious, but I really don't see how we get this formula.
Let's say $I_{k,h}$ for $h=1,..,2^k$ every of $2^k$ subintervals of lenght $3^{-k}$ that make $\mathcal{C}_k$ so that $\mathcal{C}= \mathop\cap_{k \geq 0}\mathcal{C}_k.$
Let be $\{U_i\}_{i=1,...,t}$ a $\delta$-covering of $\mathcal{C}$ made up of closed intervals in $[0,1]$. For every $U_i$ let be $k_i$ the integer for which \begin{equation} \label{bounds da k} 3^{-(k_i+1)}\leq diam(U_i)\leq 3^{-k_i} \end{equation} so $U_i$ can intersect a unique of $I_{k_i,h}$ at most because they are distant at least $3^{-k_i}$ one from the other.
Let be now $j\in\mathcal{N}_0$ with$j\geq k_i$, $U_i$ intersects at most $2^{j-k_i}$ intervals $I_{j,h}$, this is indeed the max number of intervals $I_{j,h}$ in the same $I_{k_i,h}$. Being $s_0=ln_32$ so that $3^{s_0}=2$ we have \begin{equation} \label{minorazione 2^{j-k_i}} 2^{j-k_i}= 2^j 3^{-k_is_0}= 2^j 3^{-s_0k_i-s_0} 3^{s_0}= 2^j 3^{-s_0(k_i+1)} 3^{s_0}\leq 2^j 3^{s_0} diam(U_i)^{s_0} \qquad (1) \end{equation} where the last one comes from bounds given by the definition of $k_i$.
We choose now $j$ large enough so that $$ 3^{-(j+1)}\leq diam(U_i)\quad \text{for all $i=1,...,t$} $$ in other words $$j\geq k_i \quad \text{for all $i=1,...,t$}.$$ Remembering, by construction of $\mathcal{C}$, that $$I_{j,h}\cap C \neq \emptyset \qquad \text{for all $h=1,...,2^j$}$$ and because the union of family $\{U_i\}_{i=1,..,t}$ is a covering of $\mathcal{C}$ we have this family to intersect all intervals $I_{j,h}$ which are $2^j$ in number, in other words: $$ \mathop \cup_{i=1}^t U_i \cap I_{j,h}\neq \emptyset \qquad \text{for all $h=1,...,2^j$}.$$ But, for what said before, every $U_i$ can intersect at most $2^{j-k_i}$ of $I_{j,h}$, so that, calling $\eta_i$ the number of intervals $I_{j,h}$ intersected by $U_i$, we have \begin{equation} \label{eta_i minorazione} 2^{j-k_i} \geq \eta_i \quad \text{for all $i=1,...,t$}. \end{equation} and \begin{equation} \label{minorazione somma eta} \sum_{i=1}^t \eta_i \geq 2^j \end{equation} (a same $I_{j,h}$ of $2^j$ can be intersected by more than one $U_i$).
Therefore, using (1) and the above, we have $$ \sum_{i=1}^t 2^j 3^{s_0} diam(U_i)^{s_0}\geq \sum_{i=1}^t 2^{j-k_i} \geq \sum_{i=1}^t \eta_i \geq 2^j$$ so $$\sum_{i=1}^t diam(U_i)^{s_0} \geq \frac{1}{3^{s_0}}=\frac{1}{2}. $$