Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 7 (Harmonic Analysis)

A harmonic function on $\Omega$ is a twice continuously differentiable function satisfying $\Delta u=0$ there. For complex-valued functions this means that both its real part and its imaginary part are harmonic functions. The mean value property for harmonic functions characterizes harmonicity: a continuous function is harmonic exactly when its value at each point equals its average over every sphere, or equivalently every ball, whose closed ball lies in $\Omega$. In the forward direction, the derivative of the spherical average vanishes by the divergence theorem. Conversely, radial smoothing leaves a continuous function with the mean value property unchanged locally; it is therefore smooth. Expanding its small-ball average gives $u(x)+r^2\Delta u(x)/(2(d+3))+o(r^2)$ in dimension $d+1$, and hence $\Delta u=0$.

Write $M=\Vert u{\Vert }_{h^1}$ and $v_{d+1}$ for the volume of the unit ball in ${\mathbb{R}}^{d+1}$. We first obtain a pointwise estimate directly from the mean value property for harmonic functions, before using any boundary representation. For the ball of radius $r=t/2$ centered at $(x,t)$,Thus every upward shift of $u$ is a bounded harmonic function, with a continuous bounded boundary value.

The required map is the Poisson integralThe Poisson kernel for the upper half-space is positive, has integral one, and is harmonic as a function of $(x,t)$. Its normalization follows by polar coordinates and $r=\tan \theta$, which reduces the radial integral to ${\int }_0^{\pi /2}{\sin }^{d-1}\theta d\theta$; direct differentiation gives ${\Delta }_{x,t}{P}_t=0$. Differentiation under the integral on compact subsets proves that $P\mu$ is a harmonic function. Tonelli theorem givesHere the total variation norm of a measure is the norm on finite signed or complex Borel measures.

The Poisson kernel for the upper half-space is an approximate identity: for $\psi \in C_0({\mathbb{R}}^d)$, $P_t\ast \psi \to \psi$ uniformly. Indeed, uniform continuity controls translations smaller than a fixed radius, and the kernel mass outside that radius tends to zero. ConsequentlyThe dual characterization of the total variation norm of a measure, as the supremum over $\psi \in C_0$ of supremum norm at most one, now gives the reverse inequality $\Vert \mu {\Vert }_{\mathrm{TV}}\le \Vert P\mu {\Vert }_{h^1}$. Thus $P$ is a linear isometry and is injective.

For surjectivity, fix $s>0$ and put $u_s(x)=u(x,s)$. Bounded harmonic uniqueness in the upper half-space givesTo justify this uniqueness, subtract the Poisson integral of the bounded continuous boundary value from the upward-shifted harmonic function. Their difference is bounded and harmonic and has zero continuous boundary values. Its odd reflection across the boundary is an entire bounded harmonic function; the reflection is harmonic across the flat boundary by harmonic odd reflection across a hyperplane. The Liouville theorem for harmonic functions makes it constant, and oddness makes it zero. The initial pointwise estimate supplies the boundedness needed here. To see the reflection locally, take a ball centered on the hyperplane and extend the boundary data on its upper hemisphere oddly to its lower hemisphere. The harmonic solution on this ball is odd and zero on its equatorial disk. The maximum principle for harmonic functions identifies it with the given difference on the upper half-ball, proving that the reflected function is harmonic across the boundary.

Choose $s_j\downarrow 0$. The measures $u_{s_j}dx$ have total variation norm of a measure at most $M$. By the Banach-Alaoglu theorem, and separability of $C_0({\mathbb{R}}^d)$, a subsequence converges in the weak-star topology to a finite measure $\mu$ with $\Vert \mu {\Vert }_{\mathrm{TV}}\le M$. For fixed $(x,t)$ and $s_j<t$,The kernels in this display belong to $C_0$ and converge uniformly to $P_t(x-\cdot )$. The uniform measure norm bound therefore allows passage to the limit, giving $u(x,t)=P\mu (x,t)$. This proves the Poisson representation of harmonic h1 by finite measures, including onto-ness.

Finally $P_t(x)\le c_d{t}^{-d}$, so the Poisson representation of harmonic h1 by finite measures improves the preliminary pointwise constant:The last occurrence of $h^1({\mathbb{R}}^{d+1})$ in the printed question is understood as the harmonic Hardy space of the upper half-space defined earlier in that question.

For a positive Borel measure $\nu$, the centered Hardy-Littlewood maximal function and the uncentered maximal function of a finite measure areWe use open balls and allow infinite values. For signed or complex measures the definition uses their total variation measure. Since a centered ball is also an admissible uncentered ball, and every ball $B(c,r)$ containing $x$ lies in $B(x,2r)$,For functions, $m(f)$ means the centered Hardy-Littlewood maximal function of $|f|dx$.

The set $E_{\alpha }$ is the union of the open balls $B$ with $\nu (B)>\alpha |B|$. In particular, each witnessing ball lies entirely inside $E_{\alpha }$. If $K\subset E_{\alpha }$ is compact, take a finite witnessing cover of $K$. Choose its largest-radius ball, discard all balls meeting it, and repeat. The selected balls $B_j$ are disjoint. Every discarded ball has no larger radius than the selected ball that discarded it and is contained in its threefold dilation. HenceTaking the supremum over compact $K$ by inner regularity of Lebesgue measure proves the stronger, localized estimateIf the right side is infinite there is nothing to prove. Positivity of the measure is important: a signed version uses $|\nu |(E_{\alpha })$, not $\nu (E_{\alpha })$.

For the radial-kernel inequality, integrability implies $\varphi (r)\to 0$. For $0<s<\varphi (0)$, let $r(s)$ be the unique radius with $\varphi (r(s))=s$. The layer cake representation and Tonelli theorem giveThe last integral equals the total mass of $\varphi (|\cdot |)$, namely one. Taking the absolute value of the original convolution therefore proves the claim with constant exactly one. This is radial decreasing kernel domination by the maximal function.

An application is boundary convergence of the Poisson integral. Its normalized radial profile is $c_d(1+r^2{)}^{-(d+1)/2}$, so ${\sup }_{t>0}|P_t\ast f|\le m(f)$. For a continuous compactly supported $g$, the approximate identity property gives $P_t\ast g\to g$ pointwise. For arbitrary $f\in L^1$, put $h=f-g$; thenThe preceding weak type (1,1) estimate and the elementary integral bound for $|h|$ show that the set where this exceeds $2a$ has measure at most $(3^d+1)\Vert h{\Vert }_1/a$. Continuous compactly supported functions are dense in $L^1$, so this measure is zero. Thus the Poisson integrals of an integrable function converge to it almost everywhere; the same approximate identity also gives convergence in $L^1$.

We first construct a terminal integrable function without using the martingale convergence theorem. Put $G={\sup }_n|g_n|$ and $d\eta =Gdx$ on $(0,1{]}^d$. This is a finite positive measure. If $G=0$ almost everywhere the conclusion is immediate. Otherwise, for a bounded dyadic step function $h$ measurable with respect to ${\mathcal{F}}_n$ in the dyadic filtration, defineThe martingale property makes the definition independent of the choice of a sufficiently large $n$. Moreover $|\Lambda (h)|\le \int |h|d\eta$. Dyadic step functions are dense in $L^1(\eta )$: continuous functions on the closed unit cube are dense for this finite Borel measure (extend the measure by zero onto the omitted faces), and uniform continuity approximates each continuous function by step functions on fine dyadic partitions. Thus $\Lambda$ extends to a bounded linear functional on $L^1(\eta )$ of norm at most one. The duality of Lp spaces representation of this functional gives $H\in L^{\infty }(\eta )$ with $|H|\le 1$ and $\Lambda (h)=\int hHd\eta$. Set $g=HG$, taking $g=0$ where $G=0$. Then $g\in L^1(dx)$ and testing indicators of all dyadic atoms provesThis argument also works for complex-valued martingales, with a complex linear functional.

To prove convergence, choose a dyadic step function $h$ with $\Vert g-h{\Vert }_1$ arbitrarily small. For all $n$ beyond its dyadic level, $\mathbb{E}(h\mid {\mathcal{F}}_n)=h$. The L1 contraction of conditional expectation yieldsThis proves convergence in $L^1$. For convergence almost everywhere, Doob maximal inequality applied to the conditional expectations of $g-h$ givesLet the approximation error tend to zero, and then take positive rational $a$. Hence $g_n\to g$ both in $L^1$ and almost everywhere.

For the second assertion, write $M={\sup }_n\mathbb{E}|f_n|<\infty$ and take the given first-exit stopping time $\tau$. Each $g_n=f_{\tau \wedge n}$ is integrable, being a finite sum of restrictions of integrable variables. Its increments satisfyThe indicator is ${\mathcal{F}}_n$-measurable, so taking its conditional expectation proves that $(g_n)$ is a martingale.

The overshoot at $\tau$ need not be bounded by $N$. Instead, for a fixed $m$, on $\{\tau =j\}$ with $j\le m$ we have $|f_j|\le \mathbb{E}(|f_m|\mid {\mathcal{F}}_j)$. Integrating over this ${\mathcal{F}}_j$-measurable event, summing the disjoint events, and including $\{\tau >m\}$ givesThe variables $Z_m=|f_{\tau }|1_{\{\tau \le m\}}$ increase to $Z$, and $Z_m\le |g_m|$. The monotone convergence theorem gives $\mathbb{E}Z\le M$. Before stopping, the process has magnitude less than $N$; after stopping it remains at $f_{\tau }$. ConsequentlyThis is the integrable overshoot of a stopped L1-bounded martingale. The first part now proves that every such stopped process converges almost everywhere.

Finally, Doob maximal inequality gives $|\{{\sup }_n|f_n|\ge N\}|\le M/N$, by applying the finite-horizon estimate and then increasing the horizon. Thus ${\sup }_n|f_n|$ is finite almost everywhere. Outside the union of the null sets for all positive integer stopping thresholds, choose an integer $N$ larger than this supremum. No stopping occurs at that point, so $f_n=g_n$ for all $n$, and $f_n$ converges to a finite limit almost everywhere. We have not assumed, or concluded, $L^1$ convergence for this general $L^1$-bounded martingale.

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.

Put $M={\sup }_{y>0}{\int }_{\mathbb{R}}|f(x+iy)|dx$. The goal is to obtain an integrable non-tangential maximal function without assuming in advance that $f$ has an integrable boundary density. We use a square root to place the boundary data in $L^2$, where the Hardy-Littlewood maximal function is strongly bounded.

First recall this strong bound directly from Q2. For a nonnegative $h\in L^2(\mathbb{R})$ and $\lambda >0$, split $h=h{1}_{\{h>\lambda /2\}}+h{1}_{\{h\le \lambda /2\}}$. The second summand has centered maximal function at most $\lambda /2$; the first is in $L^1$. Q2's weak estimate, in dimension one, therefore givesThe layer cake representation and Tonelli theorem then implyThis is the needed Strong Lp bound for the Hardy-Littlewood maximal function, with an explicit adequate constant.

For $\epsilon >0$, let $f_{\epsilon }(z)=f(z+i\epsilon )$ and $h_{\epsilon }(x)=|f(x+i\epsilon ){|}^{1/2}$. Q1's elementary harmonic pointwise estimate makes $f_{\epsilon }$ bounded on the closed upper half-plane. Its boundary value is continuous, and $\Vert h_{\epsilon }{\Vert }_2^2\le M$. The function $|f_{\epsilon }{|}^{1/2}$ is subharmonic: away from zeros, for any $q>0$,and regularization, or the submean inequality at zeros, extends this to the whole domain. Bounded subharmonic comparison with the Poisson integral givesHere $h_{\epsilon }$ need not be in $L^1$. The Poisson integral is nevertheless defined, since $h_{\epsilon }\in L^2\cap L^{\infty }$ and $P_y\in L^1\cap L^2$, and it tends to its continuous boundary values. To justify comparison on this unbounded domain, the difference between the two sides is bounded above, subharmonic, and zero on the real boundary. On an upper half-disk of radius $R$, use the positive harmonic barrierOn its semicircular arc, $b_R(R{e}^{i\theta })=2/\sin \theta \ge 2$, while $b_R(z)\to 0$ for every fixed interior $z$ as $R\to \infty$. The maximum principle for subharmonic functions, with a fixed bound for the difference, proves the comparison. The barrier's positive blow-up at the arc endpoints causes no difficulty.

For $|x-x^{\prime }|\le y$, the elementary inequality $y^2+|x-s{|}^2\le 3(y^2+|x^{\prime }-s{|}^2)$ givesQ2's radial decreasing kernel domination by the maximal function extends to nonnegative $L^2$ inputs by applying it to increasing bounded compactly supported truncations. Hence, taking the supremum over the cone,For any fixed cone point, $f_{\epsilon }(z)\to f(z)$ as $\epsilon \downarrow 0$. Therefore $f^{\ast }(x)\le {\mathrm{liminf}}_{\epsilon \downarrow 0}{f}_{\epsilon }^{\ast }(x)$; use a sequence of positive shifts and the Fatou lemma to concludeThe precise numerical constant is unimportant. Suprema over countable dense sets of cone points give the same values, so the maximal functions used here are measurable.

This yields boundary absolute continuity of measures rather than assuming it. The holomorphic function $f$ is a complex-valued harmonic function, so Q1 represents it as $f(x+iy)=P_y\ast \mu (x)$ for a finite complex measure $\mu$, with $f(\cdot +iy)dx\to \mu$ weak-star. Since $|f(x+iy)|\le f^{\ast }(x)$, every compactly supported continuous $\psi$ satisfiesThe dual characterization of the total variation measure implies $|\mu |\le f^{\ast }dx$, so the Radon-Nikodym theorem gives $\mu =Fdx$ with $F\in L^1$. The approximate identity now gives $f(\cdot +iy)\to F$ in the L1 norm and almost everywhere. Thus the analytic Hardy space has an integrable boundary function, unlike the unrestricted harmonic $h^1$ space.

In the real-line form of the F. and M. Riesz theorem, start with a finite complex measure whose Fourier transform of a finite measure vanishes for $\xi <0$. Its Poisson integral is holomorphic: the transformed operators ${\partial }_y$ and $i{\partial }_x$ act by $-|\xi |$ and $-\xi$, which agree on that spectrum. These identities can first be read as distributional identities and then classically, since the Poisson integral is smooth. Its harmonic $h^1$ norm is finite by Q1, so the preceding argument makes the measure absolutely continuous. A one-sided Fourier spectrum forces a finite boundary measure to have a Lebesgue density.

For the customary circle form, normalize arc length by $dm=d\theta /(2\pi )$. If $\hat{\mu }(n)=\int e^{-in\theta }d\mu (\theta )=0$ for $n<0$, thenis analytic and $F(r{e}^{i\theta })$ is the Poisson integral on the unit disk of $\mu$. Positivity and unit mass of the Poisson kernel on the circle imply ${\sup }_r\int |F(r{e}^{i\theta })|dm\le \Vert \mu {\Vert }_{\mathrm{TV}}$. The same square-root argument on the disk gives an integrable radial maximum, as follows. For $0<\rho <1$, put $h_{\rho }(\theta )=|F(\rho e^{i\theta }){|}^{1/2}$. The maximum principle for subharmonic functions on the disk givesThe supremum of these Poisson integrals is bounded by a constant times the circular Hardy-Littlewood maximal function. Indeed, for $r\ge 1/2$ the circular kernel is bounded by a constant times $(1-r)/((1-r{)}^2+{\theta }^2)$ for $|\theta |\le \pi$; splitting into arcs of radii $2^j(1-r)$ gives a summable geometric series of arc averages. For $r<1/2$ the kernel is uniformly bounded, and the whole-circle average suffices. The covering proof and the truncation argument already used on the line give the strong $L^2$ maximal bound for arcs on the circle as well. ThusLet $\rho \uparrow 1$ and apply Fatou lemma to obtain an integrable radial maximum $R(\theta )={\sup }_{0<r<1}|F(r{e}^{i\theta })|$. The measures $F(r{e}^{i\theta })dm$ tend weak-star to $\mu$ and are dominated in magnitude by $Rdm$. The same continuous-test-function argument gives $|\mu |\le Rdm$. This proves the F. and M. Riesz theorem: a finite complex measure on the circle with all negative Fourier coefficients zero is absolutely continuous with respect to arc length. Reversing the Fourier convention interchanges positive and negative indices.