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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.

A boundary limit taken while approaching within a cone that stays away from tangency. At zero in the complex upper half-plane, every fixed cone $|\mathrm{Re}z|\le A\mathrm{Im}z$ is allowed. A nontangential limit need not be a limit along all paths approaching the boundary.

// Destination: analysis.bigb

A nontangential limit at $\xi \in \partial \mathbb{D}$ restricts approach to a fixed Stolz region ${\Gamma }_A(\xi )=\{z\in \mathbb{D}:|z-\xi |<A(1-|z|)\}$, where $A>1$. Radial approach is a special case. The Fatou theorem for bounded holomorphic functions says that every $f$ in H-infinity space has a finite boundary value $f^{\ast }(\xi )$ for almost every $\xi$, with convergence through every fixed Stolz region and $|f^{\ast }|\le \Vert f{\Vert }_{\infty }$ almost everywhere.

Here is a proof outline. Write $f=u+iv$ with bounded real harmonic functions $u,v$. Take a sequence $R_j\uparrow 1$. The bounded functions $u(R_j\xi )$ have a subsequence converging in the weak-star topology of $L^{\infty }(\partial \mathbb{D})$ by the Banach-Alaoglu theorem and weak-star metrizability of the dual ball; its $L^1$ predual is separable. Denote the limit by $U$. For $|z|<R_j$, the Poisson integral on the unit disk givesThe Poisson kernels converge uniformly for this fixed interior point and the boundary functions stay uniformly bounded, so the weak-star topology permits passage to the limit:Repeat for $v$, obtaining $V$. Thus $f$ is the Poisson integral of $F=U+iV\in L^{\infty }$.

The Lebesgue differentiation theorem makes almost every boundary point a Lebesgue point of $U$ and $V$. The lemma that a Poisson integral converges nontangentially at Lebesgue points then gives $f(z)\to F({\xi }_0)$ in every fixed Stolz region. To see why, put $\delta =1-|z|$ and let $d(\xi ,{\xi }_0)$ be the shorter angular distance. In ${\Gamma }_A({\xi }_0)$ the kernel satisfiesSplit the circle into a central arc of radius comparable to $\delta$ and annuli with radii $2^k\delta$. For a Lebesgue point, averages of $|F-F({\xi }_0)|$ on sufficiently small arcs are arbitrarily small. The annular contributions are bounded by these averages times a summable geometric sequence $C_A{2}^{-k}$. On the remaining fixed distant arc the kernel is $O(\delta )$, so its contribution vanishes. This proves the claimed nontangential limit. The same Lebesgue points work for every integer aperture $A\ge 2$, hence for all finite apertures.

These boundary values recover the interior by the Poisson integral, so they determine $f$ uniquely. They satisfy $\Vert f^{\ast }{\Vert }_{L^{\infty }}=\Vert f{\Vert }_{\infty }$: one inequality follows by taking limits of the bounded interior values, and the other from positivity and unit integral of the Poisson kernel. The dominated convergence theorem also gives $f(r\xi )\to f^{\ast }(\xi )$ in every finite $L^p$ norm. Its negative Fourier coefficients vanish, because the Taylor series of $f$ has only nonnegative powers. This explains the boundary realization of analytic Hardy spaces.

The exceptional set cannot generally be removed. For example, $f(z)=\exp (i\mathrm{Log}(1-z))$ is bounded and holomorphic on the unit disc, using the holomorphic logarithm on the right half-plane: $|f(z)|=e^{-\arg (1-z)}$ lies between $e^{-\pi /2}$ and $e^{\pi /2}$. At the boundary point one its radial values $e^{i\log (1-r)}$ oscillate without a limit. Nor does the theorem assert convergence along arbitrary tangential paths or a continuous boundary extension. Its strength is almost-everywhere control on every fixed approach cone.

A normal family is a family such that every sequence has a subsequence converging locally uniformly in the spherical metric to a meromorphic limit, possibly the constant infinity. For a family uniformly bounded in modulus, the possible limits are finite holomorphic functions, and ordinary locally uniform convergence suffices. The limit need not take values in the open unit disc; a sequence of constants can approach its boundary.

Here is a direct compactness proof for disc-valued maps. For each compact subset $K\subset \Omega$, choose a slightly larger compact neighborhood contained in $\Omega$. The Cauchy estimate on a fixed-radius disk around each point bounds $|f^{\prime }|$ independently of $f$ because $|f|<1$. A finite covering of $K$ yields uniform boundedness and equicontinuity there. The Arzelà-Ascoli theorem, applied successively to a compact exhaustion and followed by a diagonal subsequence, gives a limit uniformly on every compact. The Cauchy integral formula passes to that limit on small circles, proving it is a holomorphic function. This proves the required normal family assertion rather than just invoking Montel theorem.

Every $g_n(z)=g(z/n)$ is a holomorphic function on the complex upper half-plane with values in the unit disc, so the same argument makes $(g_n)$ a normal family. If a subsequence converges locally uniformly to $h$, then for every $t>0$,The identity theorem for holomorphic functions gives $h\equiv \ell$. Since every convergent subsequence has this same limit, the whole sequence converges locally uniformly to $\ell$: otherwise a compact set and a subsequence with a fixed positive discrepancy would have a further locally uniformly convergent subsequence, a contradiction.

To obtain the nontangential limit, take any $z_j=x_j+i{y}_j\to 0$ with $|x_j|\le A{y}_j$ and $y_j>0$. Put $n_j=\lfloor 1/y_j\rfloor$. Then $n_j\to \infty$, $n_j{y}_j\to 1$, and the points ${\zeta }_j=n_j{z}_j$ lie, for large $j$, in a fixed compact rectangle inside the half-plane. ThereforeThis is the normal-family proof of angular boundary convergence; using integer rescalings requires this step because the points need not lie exactly on one fixed scaled ray.

For a tangential counterexample, useIf $z=x+iy$ with $y>0$, then $\mathrm{Re}(-i/z)=-y/(x^2+y^2)<0$, so $g$ is a holomorphic function and $|g|<1$. Also $g(iy)=e^{-1/y}\to 0$. ButTheir ratio $|\mathrm{Re}{z}_n|/\mathrm{Im}{z}_n=2\pi n^2$ tends to infinity. Thus the nontangential limit is zero while the unrestricted boundary limit does not exist.

A fixed-aperture approach region at a boundary point of the unit disc. Its points approach the boundary point without becoming tangent to the circle. Convergence inside every such fixed region is a nontangential limit; the corresponding supremum defines a non-tangential maximal function.