# An integral estimate over rotations of the dyadic grid

I'm currently reading the paper Rectifiable Sets and the Traveling Salesman Problem (link) by Peter Jones (Invent. math. 102, 1-15 (1990)), and am having trouble understanding an integral estimate made by rotating the dyadic grid. The estimate is made in the appendix, section $5$, on page $14$ of the paper.

Here are the notations used in the paper:

• Given a dyadic square and a set $K$ in the complex plane, we define quantities $$\beta_K(Q) = \frac{\omega(Q)}{l(Q)}$$ where $l(Q)$ is the side-length and $\omega(Q)$ is the width of the smallest infinite strip containing $K \cap 3Q$ (so that $0 \le \beta \le 3$ measures the deviation from linearity of $K$ on the scale of $Q$).

• Now consider a Lipschitz curve $\Gamma$, parameterized as $\psi(\theta) = r(\theta) e^{i\theta}$, and define $$\Gamma^n_j = \Gamma \cap \{j 2^{-n + 1} \pi \le \theta \le (j + 1) 2^{-n + 1} \pi\}$$ That is, $\Gamma^n_j$ is the $j$-th part of the curve after it's divided into $2^n$ slices according to the angle.

• Again given an $n$ and $j$, $J^n_j$ is the line segment connecting the endpoints of $\Gamma^n_j$, and $$\tilde{\beta}(\Gamma^n_j) := 2^{-n} \sup_{z \in \Gamma^n_j} \operatorname{distance}(z, J^n_j)$$ (I believe that the definition should actually read $2^n$, in order to be consistent with notation and estimates used earlier in the paper).

Now it's shown in the paper that $$\sum_{n, j} \tilde{\beta}(\Gamma^n_j)^2 2^{-n} \lesssim 1$$

Then the author states

... then we may rotate the dyadic grid through $[0, 2\pi]$ to obtain quantities $\tilde{\beta}_{\theta}(\Gamma^n_j)$ and note that $$\sum_{l(Q) = 2^{-n-2}} \beta_{\Gamma}(Q)^2 l(Q) \le C \int_0^{2\pi} \left\{\sum_{j} \tilde{\beta}_{\theta}(\Gamma^n_j)^2 2^{-n} \right\} d\theta$$

where the sum is taken over all dyadic cubes and $C$ is some universal constant. It is this last estimate that I do not understand; any references to similar estimates with rotations of the dyadic grid, or an explanation of the estimate would be greatly appreciated.

In particular, we form $\tilde{\beta}_{\theta}(\Gamma^n_j)$ by looking at the $\tilde\beta$ corresponding to "rotated" $J$'s; that is, we form the line segment connecting the points at $j 2^{-n + 1} \pi + \theta$ and $(j + 1) 2^{-n + 1} \pi$, and the analogous $\Gamma^n_{j, \theta}$. It seems that the quantity $\tilde \beta_{\theta}(\Gamma^n_j)$ should be uniformly continuous in $\theta$, since the original curve is Lipschitz continuous - we cannot have huge variations in the deviation from linearity over small scales. Is this correct? If so, is the continuity also uniform in $n$?

## 1 Answer

I was able to complete this, and will post a solution in case anyone else finds it useful. Writing $J_{n, j, \theta}$ for the line segment corresponding to $J^n_j$ with everything rotated by angle $\theta$ (that is, the segment connecting $[j 2^{-n + 1} \pi + \theta, (j + 1) 2^{-n + 1} \pi + \theta]$, then all the estimates for $J^n_j$ still hold true.

Now whenever $J_{n, j, \theta}$ crosses $3Q$, we have that $\beta_{\theta}(\Gamma^n_j)$ is comparable to $\beta_{\Gamma}(Q)$. This happens with probability $1/4$, since we're summing over cubes exactly two levels down. The result then follows immediately.