Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 9 / 2

Use the Fourier transform convention $\hat{\nu }(\xi )=\int e^{-ix\cdot \xi }d\nu (x)$ throughout this question. The decay assumption implies square integrability, since polar coordinates giveNear zero the integrand is bounded by $r$, and at infinity it is bounded by $r^{-1-2ϵ}$. The positive $ϵ$ is what makes the latter integrable.

By the Plancherel theorem, there is a function $g\in L^2({\mathbb{R}}^2)$ whose Fourier transform is $\hat{\mu }$. It is the density of $\mu$ with respect to Lebesgue measure. To justify this step rather than assume a density, for every Schwartz function $\phi$, Fourier inversion givesThe finite measure and the locally integrable function thus define the same tempered distribution, so they agree as measures: $d\mu =gdx$. In particular $g\ge 0$ almost everywhere and $\int g=\mu ({\mathbb{R}}^2)<\infty$. This is the L2 density from a square-integrable Fourier transform principle.

For $f\in L^{\infty }(\mu )$, the density of $fd\mu$ is $fg$. The inequality $|f|\le \Vert f{\Vert }_{L^{\infty }(\mu )}$ holds wherever $g>0$, apart from a Lebesgue measure zero set, soA second application of the Plancherel theorem yields the explicit boundThe implicit constant in the requested estimate may depend on the measure and its Fourier-decay bound, but is independent of $f$.

Parametrize the unit circle by $\omega (\theta )=(\cos \theta ,\sin \theta )$, with $-\pi \le \theta <\pi$ and $d\sigma =d\theta /(2\pi )$. Extract the constant phase at $\omega (0)$:On the support, $|\theta |\le 2\delta /C$. The frequency rectangle and the elementary bounds $|1-\cos \theta |\le {\theta }^2/2$, $|\sin \theta |\le |\theta |$ implyChoose the fixed constant $C$ large enough that this is less than $\pi /3$. Every phase then has real part at least $1/2$. Nonnegativity of $\psi$ prevents cancellation after this phase rotation, while its central plateau gives $\int \psi d\sigma \ge \delta /(\pi C)$. Hence the Fourier transform satisfiesThis is the circle cap Fourier lower bound. No upper bound on the values of $\psi$ is needed here; the support, plateau and nonnegativity suffice. The long radial scale ${\delta }^{-2}$ comes from the quadratic term $\cos \theta -1$, whereas the transverse scale ${\delta }^{-1}$ comes from the linear term $\sin \theta$.

Write $R_j$ for the rectangles, $x_j$ for their centers and $e_j$ for their long-axis directions. For the finite exponent $p$ in the displayed estimate, take smooth rotated cap functions ${\psi }_j$, with $0\le {\psi }_j\le 1$, equal to one on angular distance at most $\delta /C$ from $e_j$ and supported within $2\delta /C$. Choose $C$ sufficiently large once and for all. The direction separation makes these cap supports disjoint, and $\int |{\psi }_j{|}^{p}d\sigma \lesssim \delta$.

DefineThe Fourier modulation and translation identity gives the second equality. Rotating the circle cap Fourier lower bound then gives $|h_j(x)|\gtrsim \delta$ on $R_j$. The half-side lengths of $R_j$ are no larger than the two frequency bounds used in part (b).

Let ${\epsilon }_j$ be independent Rademacher random variables. Because the input cap supports are disjoint, for every choice of signswhere $M=\#\mathcal{R}$. 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 ${\delta }^{-3}$, the restriction-to-rectangle overlap principle givesThe constants are independent of $\delta$, the centers and the collection. The finite-$p$ 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 $p\ge 4$: its output contributes at least ${\delta }^{p-3}$ to the $p$th-power norm, whereas its input contributes at most a constant times $\delta$.