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Use zero-based indices $0\le i<2^n$; the two index conventions in the printed question enumerate the same partition. Let $O_n(\gamma )$ be the sum of length powers over odd intervals and $H_n(\gamma )$ the sum over all intervals. For odd-address words, the given symbolic potential estimate has $c\le |{\Delta }_i^{(n)}|e^{-H_n^{\mathrm{pot}}(i)}\le C$, with constants independent of $n,i$. Raising these bounds to a fixed real power $\gamma$ and reversing the order when necessary gives finite positive constants $c_{\gamma },C_{\gamma }$ such thatTherefore $n^{-1}\log O_n(\gamma )\to f(\gamma )$. It remains to justify the even intervals, since the supplied estimate covers only half of the partition.

In the standard period-doubling partition, ${\Delta }_0^{(n)}={\beta }^n[-1,1]$ and ${\Delta }_i^{(n)}=g^i({\Delta }_0^{(n)})$. All odd-address intervals lie in the first image restrictive interval, a compact set away from the unique critical point zero. Hence there are uniform constants $0<d\le |g^{\prime }|\le D_1$ there. For an odd $i<2^n-1$, its image is the next even interval, and the mean value theorem gives $d|{\Delta }_i^{(n)}|\le |{\Delta }_{i+1}^{(n)}|\le D_1|{\Delta }_i^{(n)}|$.

For the wraparound pair the image of the last interval is a subinterval of ${\Delta }_0^{(n)}$, not necessarily the whole of it. Iterating the renormalization identity givesThus $|g({\Delta }_{2^n-1}^{(n)})|=|\beta {|}^n(1-\beta )$, while $|{\Delta }_0^{(n)}|=2|\beta {|}^n$. The image is the fixed fraction $(1-\beta )/2$ of the central interval. Combining this with the derivative bounds proves uniform comparability of the last odd interval and the central even interval as well. This establishes the odd-even comparison for Feigenbaum partition lengths for every pair, including wraparound.

Consequently, for each fixed real $\gamma$, the even sum is bounded above and below by constant multiples of $O_n(\gamma )$. The same is true of $H_n(\gamma )$, soFinally take $\gamma >{\gamma }_{\ast }$. Strict decrease gives $f(\gamma )<0$, so this length-power sum tends to zero exponentially. The cover mesh also tends to zero: negative bounded $U$ and the odd-interval estimate give $|{\Delta }_i^{(n)}|\le C{e}^{-bn}$ on odd intervals, and the odd-even comparison transfers this bound to all intervals. Since the intervals cover the Feigenbaum attractor, they are admissible arbitrarily fine covers in the definition of Hausdorff measure, and ${\mathcal{H}}^{\gamma }(F)=0$. Therefore the pressure upper bound for Hausdorff dimension yieldsThe covering argument supplies the requested upper bound; it does not assume a dimension equality or a lower bound not proved here.