Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 8 (Harmonic Analysis)

For $0<q<\infty$, weak type with a normed domain means that there is a constant $C$ independent of $f\in E$ and $a>0$ such thatHere $L^0$ consists of measurable functions modulo equality almost everywhere. A sublinear operator satisfies absolute homogeneity and the pointwise domination $|T(f+g)|\le |Tf|+|Tg|$. In particular the definition controls the output's superlevel measures, not necessarily its strong $L^q$ norm. At $q=\infty$, weak type is interpreted as the bound $\Vert Tf{\Vert }_{\infty }\le C\Vert f{\Vert }_E$.

A useful theorem is dense approximation under a weak-type maximal bound. Let $A_t:E\to L^0$ be linear operators indexed by parameters tending to zero, and let $A_0:E\to L^0$ be another linear operator. Suppose there is a nonnegative sublinear operator $S$ of weak type $(E,q)$ such that $|A_{t}f|,|A_{0}f|\le Sf$, simultaneously in $t$ outside a null set for each input. Assume the suprema and limiting suprema used below are measurable. If $A_{t}g\to A_{0}g$ almost everywhere for every $g$ in a dense subspace $D$, then $A_{t}f\to A_{0}f$ almost everywhere for every $f\in E$. For a sequence of operators all suprema are automatically countable; continuously parametrized averages below also have measurable suprema.

To prove the theorem, fix $f$ and $g\in D$. Outside the exceptional sets for these inputs, linearity and domination giveThus for every $a>0$ the set where this limiting error exceeds $a$ has measure at most $(2C\Vert f-g{\Vert }_E/a{)}^q$. Choose a sequence $g_k\in D$ converging to $f$ in norm and take the union of their null exceptional sets. Letting $k\to \infty$ makes the bound zero. Taking countably many positive rational $a$ proves convergence almost everywhere. This proves the requested sufficient condition rather than merely invoking a convergence theorem.

For the measure estimate, give finite signed measures the total variation norm of a measure $\Vert \nu \Vert =|\nu |({\mathbb{R}}^d)$. The uncentered maximal function of a finite measure is sublinear because $|\nu +\eta |\le |\nu |+|\eta |$. Its set $E_a=\{m_u\nu >a\}$ is open: it is precisely the union of open balls whose variation mass exceeds $a$ times their volume.

Fix a compact subset $K\subset E_a$ and select a finite cover of $K$ by these high-density balls. Order that finite collection by decreasing radius. Keep a ball if it is disjoint from all previously retained balls. Every discarded ball intersects a retained ball of at least its radius and is contained in that retained ball's triple dilation. The retained balls $B_j$ are disjoint and their triples cover $K$. ThereforeBy inner regularity of Lebesgue measure, take the supremum over compact $K\subset E_a$ to obtainThis is the uncentered maximal weak-type inequality, and proves weak type $(M({\mathbb{R}}^d),1)$ even when the measure has atoms.

Finally work in dimension one. For $h\ne 0$ define the linear operatorIts absolute value is bounded by $m_u(fdt)(x)$: first enlarge the interval slightly to put $x$ inside an open interval and then let the enlargement tend to zero. The endpoint masses vanish because $fdt$ is absolutely continuous. Taking $A_{0}f=f$, both operators are dominated by $Sf=m_u(fdt)+|f|$. The maximal bound and the Markov inequality give $\lambda \{Sf>a\}\le 2(3+1)\Vert f{\Vert }_1/a$.

Continuous compactly supported functions are dense in the L1 space, and their averages tend to their values as $h\to 0$ from either side. For a fixed $x$, the interval integral varies continuously for $h\ne 0$, so the required suprema can be taken over rational $h$ and are measurable. Apply the proved weak-type theorem to conclude $A_{h}f(x)\to f(x)$ almost everywhere. Hence

The layer cake representation, followed by the assumed tail bound and Tonelli theorem, givesAll integrands are nonnegative, so this use of Tonelli theorem is valid before finiteness of the left-hand side is known. Since $q=p/(p-1)$, the result is

Write $F=f_2$. Split the integral in the assumed tail bound according to whether $g\le a/2$ or $g>a/2$:The finite-superlevel hypothesis permits subtraction of the first term. With $a=2b$, this becomesThus $F/2$ satisfies part (i), which proves finiteness and the preliminary estimateNow integrate the original tail bound against $p{a}^{p-2}$. The layer cake representation and Tonelli theorem givewhere the second step is Hölder's inequality. The preliminary estimate justified the finiteness needed here. If $\Vert F{\Vert }_p=0$ the result is immediate; otherwise divide by its $(p-1)$st power. This proves the Lp bound from a tail domination inequality:The printed strict signs cannot hold when $g=F=0$: both sides are zero. The universally valid formulation therefore has non-strict inequalities. If $\Vert g{\Vert }_p>0$, the sharper estimate is actually strict, as follows.

Suppose equality held and $\Vert F{\Vert }_p>0$. Equality in Hölder's inequality and the norm relation force $g=F/q$ almost everywhere; equality in the preceding integrated tail inequality forces equality in that tail inequality for almost every $a>0$. Put $H(a)=\mu \{F>a\}$ and $J(a)={\int }_a^{\infty }H(t)dt$. The layer-cake identity on $\{F>a\}$ gives ${\int }_{\{F>a\}}F=aH(a)+J(a)$, so equality impliesBecause $F\in L^p$ and $p>1$, $J$ is finite for positive $a$, locally absolutely continuous, and $J^{\prime }=-H$ almost everywhere. It follows that $J^{\prime }=-(p-1)J/a$, whose solutions on $(0,\infty )$ are $J(a)=c{a}^{1-p}$. If $c>0$, then $H(a)=c(p-1)a^{-p}$ almost everywhere, and $p{\int }_0^{\infty }{a}^{p-1}H(a)da$ diverges. If $c=0$, then $F=0$ almost everywhere. Both contradict the assumed nonzero finite equality case. If $F=0$ but $g\ne 0$, strictness is immediate.

Thus both requested strict estimates hold when $\Vert g{\Vert }_p>0$: the coarse constant is strictly larger than the sharp one since $q^{1/q}\le e^{1/e}<2$, so $q^p<2^{p}q$. The zero-data exception is the only qualification needed here.

Let $(X_k{)}_{k=0}^n$ be a nonnegative submartingale and assume $X_n\in L^p$. Set $X^{\ast }={\max }_{0\le k\le n}{X}_k$. For $a>0$, let $A_k$ be the event that $k$ is the first index with $X_k>a$. These events are disjoint, $A_k\in {\mathcal{F}}_k$, and their union is $\{X^{\ast }>a\}$.

The submartingale property gives $\mathbb{E}[X_n\mid {\mathcal{F}}_k]\ge X_k$. HenceThus part (ii) applies on the probability space with $F=X^{\ast }$ and $g=X_n$, giving the Doob Lp maximal inequalityThe same split at $a/2$ makes $X^{\ast }/2$ satisfy part (i), recovering the preliminary constant $2^{p}q$. Probability spaces automatically have finite superlevel measures. For a right-continuous nonnegative submartingale on $[0,T]$, apply these finite-time bounds to increasing finite grids containing $T$ and a dense set of times; right continuity and monotone convergence theorem extend the estimate to ${\sup }_{t\le T}{X}_t$.

Use the Fourier transform convention $\hat{h}(\xi )=\int h(x)e^{-2\pi ix\cdot \xi }dx$. The deleted-ball kernel need not be integrable at infinity, so first work with finite annuli. WriteWe will prove a uniform bound, convergence as $R\to \infty$, and then convergence as $\epsilon \downarrow 0$. This makes all Fourier-transform assertions precise for the annularly cancelling singular convolution kernel.

Fix $\xi \ne 0$ and put $y=\xi /(2|\xi {|}^2)$, so $e^{-2\pi iy\cdot \xi }=-1$. For $R>r\ge 4|y|$, let $\Omega =\{r<|x|<R\}$ and defineThe Hörmander integral kernel condition givesOn changing variables $u=x-y$, the phase factor shows $J_{r,R}=-{\int }_{\Omega -y}K(u)e^{-2\pi iu\cdot \xi }du$. The symmetric difference between $\Omega$ and $\Omega -y$ lies in the two shells whose radii are within $|y|$ of $r$ and $R$. On each such shell, the kernel's size bound gives, by polar integration,Thereforeand adding the two comparisons proves the oscillatory tail estimate for a singular kernel:The integral on the right is at most $B$. Moreover, for fixed $\xi$, it tends to zero as $r\to \infty$ because the translated difference is absolutely integrable outside $2|y|$. The shell term also tends to zero. Thus the far Fourier integral is Cauchy and converges as its outer radius tends to infinity.

Set $r_0=4|y|=2/|\xi |$. If $\epsilon <r_0$, annular cancellation allows subtraction of the constant phase on the near annulus:The same estimate applies to a shorter near annulus when $R<r_0$. The far annulus is bounded by $B/2+C_{d}A$, using the tail estimate; if $\epsilon \ge r_0$, apply that estimate with $r=\epsilon$. At $\xi =0$ the annular integral vanishes by cancellation. We have provedFor each fixed $\epsilon >0$, the size bound makes $K_{\epsilon }\in L^2$, and $K{1}_{\{\epsilon <|x|<R\}}\to K_{\epsilon }$ in $L^2$ as $R\to \infty$. By the Plancherel theorem, their Fourier transforms converge in $L^2$ to ${\hat{K}}_{\epsilon }$. The pointwise improper limits just established identify this transform, up to a null set, withFor example identification follows by selecting an almost-everywhere convergent subsequence from the $L^2$ convergence. HenceThe near-annulus integral with the constant phase subtracted converges absolutely at zero, since its bound is proportional to $|x{|}^{1-d}$. The far integral has already converged. Therefore $m_{\epsilon }(\xi )\to m(\xi )$ for every $\xi \ne 0$, with the same uniform bound; the value at zero may be set arbitrarily.

For $f\in L^2$, the convolution $K_{\epsilon }\ast f$ is absolutely convergent for each $x$ by Cauchy-Schwarz inequality. Its Fourier transform is $m_{\epsilon }\hat{f}$: first prove this for Schwartz functions and then approximate any $L^2$ input. The convolutions converge uniformly under this approximation by the same Cauchy-Schwarz bound, and the multiplier outputs converge in $L^2$ by their uniform operator bound, so these two definitions agree. Define the bounded Fourier multiplier operator $T$ by $\widehat{Tf}=m\hat{f}$. Dominated convergence theorem and the Plancherel theorem now give

For the final request, use a normalized bump function, $\int \varphi =1$, so that its dilates form an approximate identity. Multiplying Fourier transforms givesThese products are integrable by Cauchy-Schwarz inequality, since $T{\varphi }_{\epsilon },f\in L^2$, and also define $L^2$ functions because $m$ and $\hat{\varphi }$ are bounded. Thus the convolution identity is justified even though $T{\varphi }_{\epsilon }$ need not be in $L^1$:The allowed standard approximate identity results, applied to $Tf\in L^2$, prove convergence to $Tf$ in norm and almost everywhere. This is the mass normalization in mollified singular integrals; it does not require proving almost-everywhere convergence of the sharp truncations $T_{\epsilon }f$.

If “bump function” means merely smooth with compact support, the stated limit needs the unit-integral qualification. For arbitrary such $\varphi$, the limit is $(\int \varphi )Tf$ instead. This is also the almost-everywhere limit: at each Lebesgue point of $Tf$, subtract $(\int \varphi )Tf(x)$ inside the convolution and bound the remaining local average by the boundedness and compact support of $\varphi$. As a concrete counterexample without normalization, take $K(x)=1/(\pi x)$ in dimension one and $\varphi =2\rho$ for a unit-integral mollifier $\rho$. This kernel has the required size, odd annular cancellation, and translation-difference integral $\log (3)/\pi$ on $\{|x|>2|y|\}$ for every $y\ne 0$. Its limit operator is the nonzero Hilbert transform; by the Hilbert-transform Fourier multiplier, it is an $L^2$ isometry. For any nonzero $f\in L^2$ the mollified limit is $2Tf$, not $Tf$.

Let $Mf$ be the uncentered maximal function of $|f|$, defined by its averages over balls. First establish the strong $L^p$ estimate needed below. The uncentered maximal weak-type inequality gives the weak $L^1$ bound with $C_d=3^d$, and plainly $Mf\le \Vert f{\Vert }_{\infty }$ for bounded inputs. For $f\in L^p$, split it into its parts on $\{|f|>a/2\}$ and $\{|f|\le a/2\}$. The first part is in $L^1$, since $|f|1_{\{|f|>a/2\}}\le (a/2{)}^{1-p}|f{|}^p$, and the second part contributes at most $a/2$ to the maximal function. Sublinearity therefore givesIntegrating this bound with layer cake representation and Tonelli theorem yieldsThis proves the strong Lp bound from weak L1 and L-infinity bounds directly. In particular $Mf$ is finite almost everywhere.

Now split the positive Riesz potential of $|f|$ at a radius $r$. In the near region, decompose into annuli $2^{-j-1}r<|x-y|\le 2^{-j}r$. On the $j$th annulus,The geometric series converges because $\alpha >0$. In the far region, Hölder's inequality, with $p^{\prime }=p/(p-1)$, givesThe radial integral is finite exactly because $p<d/\alpha$. Combining these bounds, including the fixed Riesz normalization constant, givesFor $0<Mf(x)<\infty$, choose $r=(\Vert f{\Vert }_p/Mf(x){)}^{p/d}$. If $Mf(x)=0$, every ball average vanishes and $f=0$ almost everywhere, so the assertion is trivial. The zero-norm case is likewise trivial. The resulting Hedberg inequality isThis proves absolute convergence of the defining integral for almost every $x$, also for signed or complex $f$. Put $\theta =\alpha p/d\in (0,1)$. The proposed exponent satisfies $q=p/(1-\theta )$, so $(1-\theta )q=p$. Raising the pointwise estimate to the $q$th power and using the proved maximal bound givesThus the Hardy–Littlewood–Sobolev inequality follows from the estimates just derived, rather than being quoted in place of its proof.

For uniqueness of the target exponent, suppose such a uniform estimate holds with index $r$ instead. Choose a nonzero nonnegative bump function $f$ and let $f_t(x)=f(tx)$ for $t>0$. Homogeneity of the Riesz potential givesThe assumed estimate at $t=1$ makes the nonzero output's $L^r$ norm finite. For all $t>0$ it must therefore satisfywith $d/r=0$ when $r=\infty$. Letting $t$ tend to zero or infinity forces the power of $t$ to vanish:Hence no other target index is possible.

For the Sobolev embedding, take $d>1$ and $1<p<d$, so the preceding result is available with $\alpha =1$. For $u\in C_c^{\infty }({\mathbb{R}}^d)$, the vector field $(x-y)/|x-y{|}^d$ has zero divergence in $y$ away from $x$. Integrating by parts outside the ball $|x-y|\le \epsilon$, its inner boundary contributes $|{\mathbb{S}}^{d-1}|$ times the spherical average of $u$ about $x$. The outer boundary contributes zero because $u$ has compact support. Letting $\epsilon \downarrow 0$ provesThe absolute value is at most $C{I}_1(|\nabla u|)(x)$. ThusFor a general member of the first-order Sobolev space $W^{1,p}({\mathbb{R}}^d)=L_1^p({\mathbb{R}}^d)$, approximate in the Sobolev norm by compactly supported smooth functions. This density of smooth functions in a Sobolev space follows from cutoffs tending to one and subsequent mollification: the cutoff-gradient term tends to zero in $L^p$, and a mollifier approximates both the function and each weak derivative. The displayed estimate makes the approximants Cauchy in $L^{q_1}$. Their $L^p$ limit is the original function, so subsequences converging almost everywhere identify the $L^{q_1}$ limit with it. This proves continuous embedding. The range $1<p<d$ is essential for this consequence; the formula is not a general critical $p=d$ embedding into $L^{\infty }$.

For $d\ge 2$, write ${\hat{x}}_j$ for the $(d-1)$ coordinates other than $x_j$, and setBy Tonelli theorem, $\Vert S_j{\Vert }_1=\Vert s_j{\Vert }_1$. The key estimate is the functional Loomis-Whitney inequality:Here is a proof in every dimension. In dimension two, the left-hand side separates into a product of two one-dimensional integrals, so Tonelli theorem gives equality. For $d\ge 3$, first integrate $x_d$. Apply Hölder's inequality with $d-1$ factors to $S_1,\dots ,S_{d-1}$, which may depend on $x_d$, and leave the factor $S_d$ outside that integral. WriteEach $H_j$ depends on the $d-2$ coordinates other than $j,d$. With $x^{\prime }=(x_1,\dots ,x_{d-1})$, the full integral is at mostPut $G(x^{\prime })={\prod }_{j<d}{H}_j^{1/(d-2)}$. A second application of Hölder's inequality, with exponents $d-1$ and $(d-1)/(d-2)$, bounds this expression byBy induction in dimension $d-1$, $\int G\le {\prod }_{j<d}(\int H_j{)}^{1/(d-2)}$. Since $\int H_j=\int S_j$, substitution proves the claimed functional inequality. Nonnegative integrands justify all integrations; truncation gives the extended-valued version if necessary.

The pointwise hypotheses imply $f^d\le {\prod }_j{S}_j$, soApply the proved functional inequality and take the $(d-1)/d$ power to obtainFor the useful case all the right-hand norms are finite. If one of them is zero, its line integral vanishes for almost every transverse coordinate and forces $f=0$ almost everywhere; there is no division by that factor. In dimension one the corresponding direct assertion is $\Vert f{\Vert }_{\infty }\le \Vert s_1{\Vert }_1$.

For $u\in C_c^{\infty }({\mathbb{R}}^d)$, the fundamental theorem of calculus along the $j$th coordinate line givesTake $f=|u|$ and $s_j=|{\partial }_{j}u|$ in the introductory result. It follows thatThe last step is the arithmetic-geometric mean inequality. This bounds the target norm by the first-order Sobolev space norm. Approximate any $u\in W^{1,1}({\mathbb{R}}^d)=L_1^1({\mathbb{R}}^d)$ by compactly supported smooth functions, using cutoffs and mollification as in Question 4. The estimate applied to differences gives convergence in $L^{d/(d-1)}$, and subsequences of the simultaneous $L^1$ and target-norm convergence identify the two limits almost everywhere. Thus $W^{1,1}$ embeds continuously into $L^{d/(d-1)}$. The coordinate-derivative product bound also passes to the limit. For $d=1$, the line argument instead gives $\Vert u{\Vert }_{\infty }\le \Vert u^{\prime }{\Vert }_1$ and the embedding into $L^{\infty }$.

Each coordinate projection $S_j$ is compact, hence a Lebesgue measurable set, because it is the continuous image of the compact set $K$. Its $(d-1)$-dimensional measure is finite. Choose any nonnegative integrable function $\eta$ on $\mathbb{R}$ with integral one, and defineIf $x\in K$, every projection ${\hat{x}}_j$ belongs to $S_j$. If $x\notin K$, $f(x)=0$. Thus for every $j$, $f(x)\le 1_{S_j}({\hat{x}}_j)=\int s_j({\hat{x}}_j,t)dt$, so the introductory estimate applies. Tonelli theorem gives $\Vert s_j{\Vert }_1={\lambda }_{d-1}(S_j)$, andSubstitute these values and raise to the $d$th power:This proves the geometric Loomis--Whitney inequality, including zero-volume sets or zero-measure projections, without any division by their measures.