Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 9 2 c Solution Created 2026-10-03 Updated 2026-10-07
Write for the rectangles, for their centers and for their long-axis directions. For the finite exponent in the displayed estimate, take smooth rotated cap functions , with , equal to one on angular distance at most from and supported within . Choose sufficiently large once and for all. The direction separation makes these cap supports disjoint, and .
DefineThe Fourier modulation and translation identity gives the second equality. Rotating the circle cap Fourier lower bound then gives on . The half-side lengths of are no larger than the two frequency bounds used in part (b).
Let be independent Rademacher random variables. Because the input cap supports are disjoint, for every choice of signswhere . Apply the assumed Fourier extension estimate to the sum. Average over signs and use the Khintchine inequality pointwise, followed by the Tonelli theorem:There is no requirement that the spatial rectangles be disjoint; disjointness is used only for the input caps on the unit circle. Their spatial overlaps are precisely what the square function measures. The cap lower bounds now implySince each rectangle has area , the restriction-to-rectangle overlap principle givesThe constants are independent of , the centers and the collection. The finite- interpretation is the one for which the printed power integral is defined. A single cap also shows that the assumed diagonal Fourier extension estimate can hold only for : its output contributes at least to the th-power norm, whereas its input contributes at most a constant times .
Restriction-to-rectangle overlap principle 2026-10-07
Assume a finite diagonal Fourier extension estimate on the unit circle. For direction-separated rectangles of dimensions , choose disjoint frequency caps and translate their transforms to the rectangles by the Fourier modulation and translation identity. Randomize their signs. Disjointness gives an input th-power norm at most ; the Khintchine inequality converts the averaged output norm into the square function. The circle cap Fourier lower bound gives . Each rectangle has area , giving the displayed form. Input cap disjointness is compatible with arbitrary spatial rectangle overlap.