2
$\begingroup$

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 $n$?

$\endgroup$
0

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.