For Lemma 4.36, one applies Lemma 4.35 with d taken to be slightly larger than the upper bound in the second display, then no dissociated sets will occur.

]]>Hello. I’m reading your book and find chapter 4 an extremely fascinating introduction to Fourier analytic methods. Yet there are two points on which I wonder if you could kindly elaborate a little.

The first is exercise 4.2.3 (p. 159), about “dyadic decomposition” just below the display. I suppose, following the previous hints, one should split the function into a sum, with each summand ranging within a dilation of a dyadic interval. In this regard, the strategy is unable to cover all scenarios, for example when, arranging values in ascending order, we have the subsequent one is always more than twice (or N times) the preceding. In this case, each dyadic (or N-adic) interval can only contain one value, and the logarithm estimate degrades.

The second has to do with the “Fourier concentration lemma”, in p. 182. In the proof, the sentence before the second display confuses me. To be precise, if one directly uses lemma 4.35, then he merely can find a union of dissociated sets plus a cube, instead of the promised cube alone. If one incorporated the dissociated sets into the cube, then I truly wonder how he could deal with the dimension. In addition, I have taken a glance at [48], but only found the proof of the latter part.

Sincerely I look forward to your reply.

]]>*[Ah, you’re right, the error does not need to be so large for the upper bound, only for the lower bound. An erratum has been added – T.]*

I think there is a minor correction to add p.249 in the proof of Theorem 6.2. It says that “there are only elements of larger than ” but with the ordering, I think it should be . Then for all , and the following sum. Of course the result is the same though.

*[Correction added, thanks – T.]*