The Carleson–Hunt Time–Frequency Theorem: Examples

All four displayed intervals are half-open dyadic intervals. The area products are $1\cdot 1=1$ for s,u,v and $2\cdot (1\mathrm{/}2)=1$ for t. Moreover $[0,1)\subset [0,2)$ and $[0,1\mathrm{/}2)\subset [0,1)$, so $s\le t$ by the two defining inclusions. The reverse order fails because $1\in [0,2)$ but $1\mathrm{\notin }[0,1)$.

For u,v the frequency intervals agree, but neither spatial interval contains the other: $0\in I_u\setminus I_v$ and $1\in I_v\setminus I_u$. Thus both possible order relations fail, while every tile is comparable to itself.

Steps 1.1 and 1.2 verify the claimed comparable pair and the two failures for the incomparable pair, with midpoint and endpoint membership fixed by the half-open convention.

For $n\ge 0$ the minimum is $2^{-n}$, whereas for $n\le -1$ it is $2^n$. Finite geometric cancellation therefore gives, for integers $M\ge 1$ and $N\ge 0$, ${\sum }_{n=-M}^N\min (2^n,2^{-n})=(1-2^{-M})+(2-2^{-N})=3-2^{-M}-2^{-N}$. In particular the index zero is counted once and contributes one.

Every finite set of integer indices lies in an interval $[-M,N]$ of the above form, and all summands are nonnegative. Thus the supremum of finite subsums is bounded above by three by step 1.1. Taking $M=N\to \mathrm{\infty }$ in that same formula gives the lower bound three, since $2^{-M}\to 0$. The series is therefore three.

Put $Mf={\sup }_{N\ge 0}\mathrm{\mid }{S}_{N}f\mathrm{\mid }$. Each finite partial sum is measurable, hence this countable supremum is measurable. By F1 it equals infinity outside a null set on a space of measure one. For every positive integer k, the nonnegative simple function k on that conull set is bounded above by Mf, so the definition of the nonnegative integral gives $\int Mf\ge k$. Consequently $\mathrm{\parallel }Mf{\mathrm{\parallel }}_1=\mathrm{\infty }$.

Since $\mathrm{\parallel }f{\mathrm{\parallel }}_1<\mathrm{\infty }$ by F1, step 1.1 contradicts every proposed finite-constant inequality $\mathrm{\parallel }Mh{\mathrm{\parallel }}_1\le C\mathrm{\parallel }h{\mathrm{\parallel }}_1$ for all $h\in L^1$. It also contradicts every finite weak-(1,1) constant: $m\{Mf>t\}=1$ for all t, whereas $C\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{/}t<1$ for sufficiently large t. Thus neither assertion follows by adjoining the endpoint to the proposed positive-p range.