(1-D)^{-1}$). Once $E_0^{(n)}$ has been picked, $G_n$ can again be defined as in \eqref{gn}, but with the shifted centers $E_0^{(n)}+m_n(k)$ taking the role of $m_n(k)$. This two step procedure (first choose $m_n(k)$'s, then shift by an appropriate $E_0^{(n)}$ to make $\widehat{f}_0 = 0$) can now be used to pick the $m_n(k)$'s and $E_0^{(n)}$ (inductively) for all $n\ge n_0$. Note that the construction ensures that $G_n\subset G_{n-1}$. We must still choose, for every $n\ge n_0$, a particular potential from the corresponding family $V_{\alpha_0}$ of finite gap potentials. Fortunately, this choice is easy: We fix once and for all a sufficiently large $N\in\mathbb N$ (where ``sufficiently large'' will be made precise at the end of the proof) and then simply take a $V_n$ that satisfies the conclusion of Theorem \ref{T3.2} for $L=L_n$. Note also that the assumptions of Theorem \ref{T3.2} on the location of the gaps (that is, $C_1 \le m_n(k)-E_0^{(n)}\le C_2$ for all $k=1,\ldots,g_n$) hold with $n$-independent constants $C_1,C_2>0$ because $E_0^{(n)}$ lies in a small interval centered at zero while all gaps are in $[1,2]$. Therefore, the constant $C$ from the statement of Theorem \ref{T3.2} is also independent of $n$. Finally, for $n(1+\delta)^{-1}$. Now routine estimates show that if $\delta>0$ was chosen sufficiently small, then \eqref{lsg} implies that \[ \int_{x_0^{(n)}}^{a_{n+1}} \left|f_n(x)\right|^2 \, dx \ge C_0 L_n = AC_0/l_n. \] This inequality contradicts \eqref{contra} if $A$ is sufficienctly large, so $T\subset S$, as claimed. The next step is to prove that $\dim T = D$. To this end, we introduce a Borel measure $\mu$ that reflects the self-similar scaling structure of $T$. More specifically, $\mu$ gives equal weight to the intervals of $G_n$ for every $n$: $\mu(I_n)= g_n^{-1}$ if $I_n$ is one of the intervals $[m_n(k)-l_n/2, m_n(k)+l_n/2]$. Moreover, we also demand that $\mu$ be supported by $T$: $\mu(\mathbb R \setminus T)=0$. It is not hard to show (for instance, by considering approximations $\mu_n$ supported by $G_n$) that there indeed exists a unique Borel (probability) measure $\mu$ satisfying these requirements. We will now establish the following property of the generalized derivatives of $\mu$: For every fixed $\gamma < D$, we have that \begin{equation} \label{deri} \lim_{\delta\to 0+} \sup_{|I|\le \delta} \frac{\mu(I)}{|I|^{\gamma}} =0 . \end{equation} The supremum is over all intervals $I\subset\mathbb R$ of length at most $\delta$. If \eqref{deri} holds, then, by general facts on Hausdorff measures \cite[Section 3.4, Theorem 67]{Rog}, $\mu$ gives zero weight to sets of dimension strictly less than $D$, and therefore $\dim T\ge D$, as desired. The converse inequality $\dim T\le D$ does not need explicit proof (although that would actually be easy to do) because we know that always $\dim S\le 2(1-\alpha)$ (this is the result whose optimality we are about to prove), and thus $\dim T\le\dim S\le D$ will follow automatically once we have established that $V(x)=O(x^{-\alpha})$. So let us prove \eqref{deri}: Fix $\gamma$, and let $I$ be an interval with $|I|\le\delta$, where $\delta>0$ is small. Then, define $n\in\mathbb N$ by requiring that $l_n<|I|\le l_{n-1}$. Clearly, $n$ is large if $\delta$ is small. We first treat the case when $|I| \le d_n$. Recall that $d_n$ is the minimal distance between adjacent gaps of $V_n$. So the above assumption implies that $I$ intersects at most two of the intervals that build up $G_n$. Each of these intervals has measure $g_n^{-1}$, hence \[ \frac{\mu(I)}{|I|^{\gamma}} \le \frac{2}{g_n |I|^{\gamma}} \le \frac{2}{g_n l_n^{\gamma}}. \] On the other hand, if $|I|> d_n$, then the number of subintervals of $G_n$ intersecting $I$ is $\le 3|I|/d_n$, thus in this case, \[ \frac{\mu(I)}{|I|^{\gamma}} \le \frac{3|I|^{1-\gamma}}{d_ng_n} \le 3\, \frac{g_{n-1}l_{n-1}}{d_ng_n}\, \frac{1}{g_{n-1}l_{n-1}^{\gamma}} . \] Now $g_nl_n^{\gamma} \gtrsim \exp( \sigma a^n)$, where $\sigma>0$ depends on $\gamma$, and $g_{n-1}l_{n-1}/(d_n g_n) \lesssim \exp(\epsilon_n a^n)$; indeed, relations of this type motivated our definition of $g_n$ and $d_n$. Since, as noted above, $n\to\infty$ as $\delta\to 0+$, \eqref{deri} now follows. It remains to show that $V$ satisfies the bound \eqref{hyp}. So, let $x\in (a_n,a_{n+1})$ with large $n$. Recall that $\widehat{f}_0=0$, where $f$ is the function from the trace formula for $V_n$. Theorem \ref{T3.2} therefore implies that \[ \left|V(x) \right| \le C \left[ g_n^{1/2} l_n \left(\ln (g_nL_n) \right)^{1/2} + g_nl_n (l_nd_n^{-1}\ln g_n)^{N+1} \right] \] for these $x$. On the other hand, \[ x \le a_{n+1} =\sum_{m=1}^n L_m \lesssim \sum_{m=1}^n l_m^{-1} \lesssim l_n^{-1}, \] and \eqref{hyp} indeed follows, provided we took \[ N+1\ge \frac{1-\alpha}{2\alpha - 1}\, \frac{a}{a-1}. \] (As expected, $N\to\infty$ as $\alpha\to 1/2+$.) $\square$ Actually, doing these final estimates carefully, we obtain a stronger bound of the form $x^{-\alpha-\epsilon_n/2}(\ln x)^{1/2}$. 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