Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 55 / 4

The Feigenbaum period-doubling map is quadratic-critical at zero: $g^{\prime }(0)=0$ and $g^{\prime \prime }(0)\ne 0$. Differentiating the renormalization equation once gives $g^{\prime }(x)=g^{\prime }(g(\beta x))g^{\prime }(\beta x)$. Differentiate again and set $x=0$:Cancellation of the nonzero second derivative therefore givesThe quadratic critical-point property matters. The normalization and functional equation alone, without specifying the Feigenbaum solution class, would not justify cancelling $g^{\prime \prime }(0)$.

Use the geometric properties of the Feigenbaum geometric potential: for some constants $0<b\le B<\infty$,Here ${\mathrm{var}}_r$ means variation between sequences agreeing in their first $r$ symbols. In particular $D={\sum }_{r\ge 1}{\mathrm{var}}_{r}U<\infty$. These are the standard bounds for the Feigenbaum geometric potential; the symbolic potential is not the singular expression $-\log |g^{\prime }|$ at a quadratic critical point.

First allow all $n$ spins and use a fixed remote tail $\eta$, writing $Z_n(\gamma ;\eta )$ for the weighted partition function. In a concatenated block of lengths $n$ and $m$, replacing the history before the second block by $\eta$ changes its energy by at most $D$: the $j$th summand has the same first $j$ symbols, so its difference is at most ${\mathrm{var}}_{j}U$. ThereforeFor $a_n=\log Z_n$, the almost-additive partition-function limit follows explicitly: $a_n+|\gamma |D$ is subadditive and $a_n-|\gamma |D$ superadditive, so Fekete's lemma gives the same finite limit for both after division by $n$. The finiteness follows from bounded $U$ and the $2^n$ words.

Changing the fixed remote tail also changes each full energy by at most $D$. The question fixes the first spin to one and sums the remaining $n-1$ spins; its partition function is $e^{\gamma U(1,0,0,\dots )}{Z}_{n-1}(\gamma ;(1,0,0,\dots ))$. The tail comparison and bounded first contribution show that its normalized logarithm has the same limit. Thus the requested $f(\gamma )$ exists and equals the topological pressure of $\gamma U$.

For any $\delta >0$, every length-$n$ energy lies between $-Bn$ and $-bn$, henceTaking normalized logarithms and limits givesSo $f$ is Lipschitz continuous and strictly decreasing. For each finite $n$, differentiation of the log-sum gives $f_n^{\prime \prime }(\gamma )=n^{-1}{\mathrm{Var}}_{n,\gamma }(H_n)\ge 0$. Thus $f_n$ is convex, and passing to its finite pointwise limit proves that $f$ is convex too.

At zero weight the prescribed sum has $2^{n-1}$ terms, so $f(0)=\log 2$. For $\gamma >0$ the preceding bounds give $\log 2-B\gamma \le f(\gamma )\le \log 2-b\gamma$. The function therefore becomes negative at sufficiently large positive $\gamma$. Continuity and strict decrease prove

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.