Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 7 / 4

The defining convolution integral is absolutely convergent for every $x$ when $e\in L^1+L^2$. Indeed, writing $e=a+b$ with $a\in L^1$ and $b\in L^2$, the Cauchy-Schwarz inequality givesDisjoint supports imply ${\sum }_n|e_n(y)|=|e(y)|$ almost everywhere. The Tonelli theorem therefore yieldsWe may exchange the sum and integral; the resulting numerical series is absolutely convergent. Its triangle inequality givesHere $e$ belongs to the stated operator domain, as is implicit when writing $T(e)$. Disjoint supports also ensure that each summand belongs to that domain.

Let $c$ be the center and $\ell$ the side length of $Q$. The zero integral of $f$ lets us subtract a constant kernel value:Choose $L=4\sqrt{d}$. For $x$ outside the concentric cube $\hat{Q}$ of side length $L\ell$, and for $y\in Q$,Thus the domain of integration in $x$ is contained in $\{|x-c|>2|y-c|\}$. After the translations $X=x-c$ and $Y=y-c$, the Hörmander integral kernel condition givesIf $Y=0$ the difference is zero and the same inequality holds. Apply the triangle inequality and Tonelli theorem to obtain the cancellation estimate outside a dilated cube:In particular the constant on the right is the same $C$ as in the hypothesis; only the dilation $L$ depends on the dimension.

We derive the required Calderón–Zygmund decomposition and apply the preceding two parts. Use the Fourier transform convention $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$, and put $A=\Vert \hat{K}{\Vert }_{\infty }$. The Plancherel theorem givesFor completeness this starts with smooth rapidly decreasing inputs, where $\widehat{K\ast f}=\hat{K}\hat{f}$, and extends by $L^2$ density. It agrees with the pointwise convolution in the question: if $f_j\to f$ in $L^2$, then $\Vert K\ast (f_j-f){\Vert }_{\infty }\le \Vert K{\Vert }_2\Vert f_j-f{\Vert }_2$. Thus there is no need to assume that $K$ itself is integrable.

Fix $f\in L^1({\mathbb{R}}^d)$, let $F=\Vert f{\Vert }_1$, and choose $\lambda >0$. Select the maximal dyadic cubes $Q$ for which $|Q{|}^{-1}{\int }_Q|f|>\lambda$. They exist above every qualifying cube because averages over increasingly large ancestors tend to zero. They are disjoint, their total volume is at most $F/\lambda$, and maximality givesOutside their union, dyadic differentiation gives $|f|\le \lambda$ almost everywhere. One can obtain this differentiation fact from the allowed martingale convergence theorem: on each fixed dyadic cube the averages are conditional expectations of the integrable restriction, their almost-everywhere limit exists, and approximation by dyadic step functions shows in $L^1$ that this limit is $f$.

Let $f_Q=|Q{|}^{-1}{\int }_{Q}f$, define $g=f$ outside the selected cubes and $g=f_Q$ on $Q$, and set $b_Q=(f-f_Q)1_Q$. This constructswith the estimatesThese follow by integrating separately on each selected cube; $|f_Q||Q|\le {\int }_Q|f|$ gives both the good-part norm bound and the last cancellation-error bound.

Let $\Omega ={\bigcup }_Q\hat{Q}$, with the dilation $L=4\sqrt{d}$ from part (ii). Then $|\Omega |\le L^{d}F/\lambda$. On its complement, part (i), followed by part (ii), givesFinally, $|Tf|>\lambda$ implies $|Tg|>\lambda /2$ or $|Tb|>\lambda /2$. The first alternative is controlled by the $L^2$ operator bound, and the second by the last integral outside $\Omega$. HenceThus $T$ has weak type $(1,1)$, with a constant depending only on the dimension, the bounded Fourier transform, and the Hörmander integral kernel condition.