Harmonic analysis

Harmonic analysis decomposes functions into elementary oscillations, generalizing Fourier series and the Fourier transform. On a locally compact Abelian group the elementary oscillations are continuous unitary characters, and Haar measure supplies translation-invariant integration.

A positive Borel measure near a domain boundary is a Carleson measure when its mass in boundary neighborhoods is bounded by a constant times their boundary measure. On the unit disc, this becomes the displayed bound for every Carleson box. The equivalent hyperbolic geodesic cap version follows from the geodesic-cap and Carleson-box comparison. The Carleson embedding theorem on the disk explains why this geometric growth condition controls integration of analytic Hardy spaces.

A positive Borel measure on the unit disc satisfies $\int |f|d\mu \le N\Vert f{\Vert }_1$ for every H1 space on the unit disk function exactly when it is a Carleson measure, with comparable optimal constants. The normalized squared Cauchy kernel tests necessity. For sufficiency, the boundary shadow of every point with $|f(z)|>t$ lies inside a component arc of $\{Nf>t\}$. The geodesic-cap and Carleson-box comparison bounds the measure of each shadow region by its component length, giving $\mu \{|f|>t\}\le C\Vert \mu {\Vert }_{\mathrm{Carl}}m\{Nf>t\}$. The layer cake representation and H1 non-tangential maximal inequality on the disk give the embedding. The broader kernel and covering mechanism is discussed in Terence Tao's harmonic-analysis notes, Exercise 2 after Lemma 4.1; the analytic endpoint here uses the fractional-power argument just given.

A boundary neighborhood over an arc $I$ of the unit circle, with radial depth equal to its normalized angular length $m(I)$. Changing that depth by any fixed positive factor gives an equivalent Carleson measure condition. Large arcs and central mass can be handled separately by finitely many fixed neighborhoods covering the unit disc.

This supremum function takes a supremum over a cone approaching the boundary point $x$; the positive number $a$ is its aperture. For an analytic upper-half-plane $H^1$ function, it belongs to $L^1$. To prove this without first assuming boundary absolute continuity, shift the function upwards, apply subharmonic Poisson comparison to its square root, and use the strong $L^2$ bound for the Hardy-Littlewood maximal function. Fatou lemma removes the shift.

For integrable boundary data $g$ on the unit circle, this follows from the Poisson kernel bound $C_A\delta /({\delta }^2+d^2)$ in a fixed Stolz region. The arc of radius $\delta$ and successive dyadic annuli contribute at most $C_A{2}^{-k}Mg(\xi )$, which sum to the displayed estimate. This links harmonic extension to the Hardy-Littlewood maximal function and its strong $L^p$ bounds for $p>1$.

For $0<p<\infty$, an analytic Hardy space controls the integral means of a holomorphic function as it approaches a domain boundary. On the disk, $H^p$ requires ${\sup }_{0<r<1}\int |F(r{e}^{i\theta }){|}^{p}d\theta <\infty$; on the upper half-plane it requires ${\sup }_{y>0}{\int }_{\mathbb{R}}|f(x+iy){|}^{p}dx<\infty$. For $p=\infty$ use a supremum norm. These analytic spaces differ from the harmonic Hardy space of the upper half-space; the Hardy space of the circle is the boundary Hilbert-space realization of the disk $H^2$ space.

This Hardy space consists of holomorphic functions with $\Vert f{\Vert }_1={\sup }_{r<1}{\int }_{\partial \mathbb{D}}|f(r\xi )|dm(\xi )<\infty$, where $m$ is normalized angular measure. The H1 non-tangential maximal inequality on the disk controls their approach to the boundary, and the Carleson embedding theorem on the disk characterizes measures against which all these functions remain integrable.

For an analytic H1 space on the unit disk function, its fixed-aperture non-tangential maximal function obeys the displayed strong $L^1$ bound. First replace $f$ by $f_R(z)=f(Rz)$, $R<1$. The fact that modulus powers of holomorphic functions are subharmonic gives $|f_R{|}^{1/2}\le P[|f(R\xi ){|}^{1/2}]$. Since a Poisson maximal function is dominated by the Hardy-Littlewood maximal function, the strong $L^2$ Hardy-Littlewood maximal inequality yields $\int N_A{f}_R\le C_A\int |f(R\xi )|\le C_A\Vert f{\Vert }_1$. Pointwise interior convergence and the Fatou lemma remove $R$. Analyticity permits this endpoint estimate; general $L^1$ boundary data have only a weak $L^1$ maximal bound.

For $a\in \mathbb{D}$, this holomorphic function has H1 space on the unit disk norm one: its boundary modulus is the unit-mass Poisson kernel on the circle. The radial integrals are bounded by that boundary integral by subharmonic comparison and tend to it by continuity. Choosing $a$ at depth comparable to an arc's length makes $|f_a|$ bounded below by the reciprocal length throughout its hyperbolic geodesic cap. It tests necessity in the Carleson embedding theorem on the disk.

This Hardy space consists of bounded holomorphic functions on the unit disc, with norm $\Vert f{\Vert }_{\infty }={\sup }_{z\in \mathbb{D}}|f(z)|$. Its almost-everywhere boundary realization follows from the Fatou theorem for bounded holomorphic functions.

A bounded holomorphic function on the unit disc has finite nontangential limits almost everywhere on its boundary. To outline a proof, bounded real and imaginary harmonic parts have $L^{\infty }$ boundary representations: apply the Banach-Alaoglu theorem to their circular traces and pass to the limit in the Poisson integral on the unit disk. Almost every boundary point is a Lebesgue point. The fact that a Poisson integral converges nontangentially at Lebesgue points gives the desired limits. Their $L^{\infty }$ norm equals the interior supremum norm, and the Poisson integral recovers the original function. Exceptional points can lack even radial limits; $\exp (i\mathrm{Log}(1-z))$ is a bounded example at the boundary point one.

If a finite complex Borel measure $\mu$ on the circle has Fourier coefficients $\hat{\mu }(n)=0$ for all $n<0$, then it is absolutely continuous with respect to arc length. The reflected Fourier convention gives the equivalent theorem with positive coefficients vanishing. The Poisson integral on the unit disk is analytic because of the one-sided Fourier spectrum. Subharmonic comparison for its square root and an $L^2$ maximal estimate produce an integrable radial maximum. This maximum dominates the weak-star boundary measure, forcing absolute continuity of measures. The same proof on the real line says that a finite measure with Fourier transform of a finite measure supported in $[0,\infty )$ has a Lebesgue density.

For $f\in L^1({\mathbb{R}}^d)$ and a positive threshold $\lambda$, take maximal dyadic cubes with mean of $|f|$ exceeding $\lambda$. They are disjoint, have total volume at most $\Vert f{\Vert }_1/\lambda$, and their means are at most $2^d\lambda$. Replace $f$ inside each selected cube by its mean to obtain $g$; write $b_Q=(f-f_Q)1_Q$. Then $\Vert g{\Vert }_{\infty }\le 2^d\lambda$, $\Vert g{\Vert }_1\le \Vert f{\Vert }_1$, $\int b_Q=0$, and ${\sum }_Q\Vert b_Q{\Vert }_1\le 2\Vert f{\Vert }_1$. The outside bound follows from the Lebesgue differentiation theorem. This separates a bounded part from localized cancellation errors.

A complex-valued function on a group is positive-definite when the displayed inequality holds for every finite choice of points and coefficients. This includes positive semidefinite, rather than necessarily strictly positive definite, Gram matrices. In particular $p(e)\ge 0$ and $|p(x)|\le p(e)$. On an Abelian group, every convolution square $u\ast u^{\ast }$ with $u\in C_c(G)$ is continuous and positive-definite: its quadratic form is the integral of $|{\sum }_i{z}_{i}u(t+x_i){|}^2$.

For an Abelian group, on finite formal sums of group points, define $⟨[x],[y]⟩=p(x{y}^{-1})$, with the first variable linear. Quotient by the null space of this positive semidefinite form and complete to a Hilbert space. Left translation is unitary, and $v=[e]$ generates a dense span of translated vectors. For continuous $p$, the equality $\Vert U_x[y]-[y]{\Vert }^2=2(p(e)-\mathrm{Re}p(x))$ gives strong continuity. This realizes $p$ as a matrix coefficient without assuming its integral representation.

For a locally compact Hausdorff Abelian group, a continuous positive-definite function $p$ is uniquely representable as $p(x)={\int }_{\hat{G}}\chi (x)d{\mu }_p(\chi )$ for a finite positive Radon measure on the Pontryagin dual group. Its total mass is $p(e)$. Conversely every such finite positive measure defines a continuous positive-definite function.

A positive-definite function $r$ on the discrete group $\mathbb{Z}$ is the sequence of Fourier coefficients of a unique finite positive Borel measure on the circle, with total mass $r(0)$. Positivity makes the finite sums ${\sum }_{|n|<N}(1-|n|/N)r(n)e^{-2\pi int}$ nonnegative. Their measures converge weakly to the required measure; uniqueness follows from density of trigonometric polynomials in continuous functions.

For a finite positive Borel measure on the circle, averaged squared Fourier coefficients tend to the sum of squares of its atomic masses. Express each squared coefficient as a double integral and average the circle character. The averaged character is bounded by one, tends to one on the diagonal and to zero elsewhere, so the dominated convergence theorem gives the diagonal mass. In particular an atomless measure has vanishing averaged absolute coefficients by the Cauchy-Schwarz inequality.

For finite sums of compactly supported convolution squares, $\hat{p},\hat{q}\ge 0$. Insert the Bochner theorem representation of $q$ into $p\ast q$ and apply the Fubini theorem: its representing measure is $\hat{p}{\mu }_q$. Reversing $p,q$ and using uniqueness gives the displayed identity. It makes the local quotients ${\mu }_p/\hat{p}$ consistent wherever $\hat{p}>0$.