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The region between two rays with a common vertex, with no upper bound on the modulus. In the displayed convention the vertex and boundary rays are excluded. A holomorphic logarithm with $\alpha <\arg z<\beta$ defines powers $z^p=e^{p\mathrm{Log}z}$ there. For real $p>0$ with $p(\beta -\alpha )\le 2\pi$, the power is a conformal bijection onto the sector $p\alpha <\arg w<p\beta$: it multiplies arguments by $p$ and sends moduli bijectively by $r\mapsto r^p$.

For a proper plane domain $D$, choose $a\in D$ and $w\notin D$ attaining $\delta (a)=\mathrm{dist}(a,\mathbb{C}\setminus D)$. The holomorphic functionhas exactly one zero in $D$, a simple zero at $a$. For $|z-w|>2\delta (a)$ its zero-free holomorphic logarithm satisfiesThus products of such factors can realize prescribed zeros approaching the boundary. If the zero has multiplicity $m$, choose the logarithmic error smaller by a factor $m$ before taking the $m$th power. On a compact subset of $D$, eventual uniform summability of these logarithms gives a nonvanishing product tail and a holomorphic function with precisely the requested zeros.

A branch of a multivalued analytic expression is a single-valued holomorphic function on a specified domain that consistently selects one of the expression's possible values. A branch cut is a convenient way to choose such a domain, rather than an intrinsic locus of singularities. On $D=\mathbb{C}\setminus (-\infty ,0]$, the principal logarithm supplies two square-root branchesFor $z=i$, these give $B_{+}(i)=e^{i\pi /4}=(1+i)/\sqrt{2}$ and $B_{-}(i)=-(1+i)/\sqrt{2}$. Their derivatives follow by differentiating $B_{\pm }^2=z$: $B_{\pm }^{\prime }=1/(2B_{\pm })$, so $B_{+}^{\prime }(i)=(1-i)/(2\sqrt{2})$. There are exactly two branches on $D$: for any other holomorphic square root $B$, the ratio $B/B_{+}$ is continuous, takes values in $\{1,-1\}$, and hence is constant on the connected domain.

The cut can move. Choosing $0<\theta <2\pi$ defines a holomorphic logarithm $L_{+}(z)=\log |z|+i\theta$ on $D_{+}=\mathbb{C}\setminus [0,\infty )$. At $-1$ this gives $L_{+}(-1)=i\pi$ and $\exp (L_{+}(-1)/2)=i$, both regular values despite $-1$ lying on the previous cut. In the complex upper half-plane, $L_{+}$ agrees with $\mathrm{Log}$; in the lower half-plane it equals $\mathrm{Log}+2\pi i$, so the corresponding square roots have opposite signs there. The overlap $D\cap D_{+}$ has two components, exactly why the identity theorem cannot force agreement between them. ^[ref]

For example, the limits of $B_{+}$ at $-r$, $r>0$, are $i\sqrt{r}$ from above and $-i\sqrt{r}$ from below. This jump records the chosen branch cut; each point $-r\ne 0$ still has a neighbourhood supporting a holomorphic square root. Moving the cut changes the bookkeeping, while the branch point at $0$ and its monodromy persist. Specifying a branch of a multivalued function is therefore essential before differentiating or performing contour integration with holomorphic square roots, complex logarithms or complex exponentiation.

If two nonzero entire functions have identical zeros and orders, their quotient extends to a nowhere-zero entire function. Its logarithmic derivative has a primitive on the simply connected plane, producing a holomorphic logarithm $h$ and the displayed relation. Any two choices differ by one constant integer multiple of $2\pi i$. On a multiply connected domain the logarithm can fail to exist globally.

Let $L>0$ and consider holomorphic functions $f$ from the unit disc into $e^{-L}<|w|<e^L$, normalized by $f(0)=1$. Their largest possible derivative modulus is $4L/\pi$. Lift $f$ to its holomorphic logarithm $h$ with $h(0)=0$, so $|\mathrm{Re}h|<L$. The function $\tan (\pi h/(4L))$ maps the disc into itself and fixes zero. The Schwarz lemma gives $|h^{\prime }(0)|\le 4L/\pi$. Equality holds precisely forHere $\arctan$ is its holomorphic branch vanishing at zero; $(4L/\pi )\arctan z$ is a conformal map from the disc onto the vertical strip. Thus the maximal derivative modulus is attained, with a circle of distinct extremizers rather than a unique map.

In the upper half-plane outside a circle of radius $R$ centered at $i{y}_0$, with $y_0>R$, set $b=\sqrt{y_0^2-R^2}$. The displayed harmonic function is zero on the real axis and equals $V_0$ on the circle. Indeed, the Apollonius circle with ratio $P>1$ has center $ib(P^2+1)/(P^2-1)$ and radius $2bP/(P^2-1)$; hence $\log P=\mathrm{arcosh}(y_0/R)$. The logarithmic singularity at $ib$ is inside the excluded circle, and the one at $-ib$ is below the real axis. The real part of a local holomorphic logarithm is harmonic even if its imaginary part is multivalued. The potential tends to zero at infinity. Finite boundary values alone do not impose a growth condition at infinity; uniqueness in the unbounded domain needs such an additional condition, for example boundedness with the prescribed limit.

A holomorphic logarithm of a nowhere-zero holomorphic function $f$ on a domain $U$ is a holomorphic function $g$ such that $e^{g(z)}=f(z)$ throughout $U$. Differentiating with the chain rule gives $g^{\prime }=f^{\prime }/f$. For $f(z)=z$, a holomorphic logarithm is a branch of the complex logarithm, and its derivative is $1/z$.

The principal logarithm is holomorphic on the slit complex plane $D=\mathbb{C}\setminus (-\infty ,0]$, withHere $\mathrm{Arg}$ is the chosen complex argument, and $\mathrm{Im}\mathrm{Log}z\in (-\pi ,\pi )$. One may conventionally assign a principal value on the negative real axis, but that axis is excluded from this holomorphic domain. ^[ref] For example, $1+i=\sqrt{2}{e}^{i\pi /4}$ gives $\mathrm{Log}(1+i)=\frac{1}{2}\log 2+i\pi /4$ and $(\mathrm{Log}{)}^{\prime }(1+i)=(1-i)/2$; all its other complex logarithms differ by $2\pi ik$, $k\in \mathbb{Z}$.

There is no holomorphic logarithm of $z$ on the punctured complex plane $\mathbb{C}\setminus \{0\}$. Such a function would be a holomorphic primitive of $1/z$, so its contour integral around every closed path would vanish. On the counterclockwise unit circle, however,This is the period obstruction to a holomorphic antiderivative, and also exactly the logarithm change in the monodromy diagram.

Every nowhere-zero holomorphic function $f$ on a simply connected domain $U$ does admit a holomorphic logarithm. Choose $a\in U$ and $b\in \mathbb{C}$ with $e^b=f(a)$, and defineThe Cauchy integral theorem makes this contour integral independent of the path in $U$. Then $g^{\prime }=f^{\prime }/f$ and $(f{e}^{-g}{)}^{\prime }=e^{-g}(f^{\prime }-f{g}^{\prime })=0$; its value at $a$ is $1$, proving $e^g=f$. Any two choices on a connected domain differ by a constant $2\pi ik$, since their difference is continuous and takes values in the discrete set $2\pi i\mathbb{Z}$. ^[ref]

For a computed example, take $f(z)=1+z$ on the unit disc, with $a=b=0$. Expanding $f^{\prime }/f=1/(1+z)$ as a geometric series and integrating givesThe construction also produces holomorphic square roots: $h=e^{g/2}$ satisfies $h^2=f$, so here $h(1/2)=\sqrt{3/2}$. More generally, $e^{\alpha g}$ gives a consistent branch of a multivalued function for complex exponentiation by $\alpha$. A holomorphic logarithm converts multiplicative expressions into additive ones while making the domain restrictions explicit.

A holomorphic square root of a nonvanishing holomorphic function $f$ is a holomorphic function $g$ satisfying $g^2=f$. It exists locally. On a connected open domain it exists globally exactly when the winding number of $f\circ \gamma$ around zero is even for every closed path $\gamma$ in the domain. Continuing a local square root once along such a path multiplies it by $(-1{)}^m$, where $m$ is that winding number. Thus the obstruction concerns parity, whereas a holomorphic logarithm needs every such winding number to be zero.

On $\mathbb{C}\setminus \{0\}$, the function $f(z)=z$ has no global holomorphic square root: its value winds once on the unit circle, and the lift shown under monodromy changes sign. In contrast, $f(z)=z^2$ winds twice there and has global square roots $g(z)=z$ and $g(z)=-z$. It still has no holomorphic logarithm on that domain. This distinguishes the two global continuation conditions.

Suppose $A\ne 0$ and $R>0$, and $H$ is an analytic function on a disk of radius strictly larger than $R$. The Taylor series at zero of $f(z)=A\log (1-z/R)+H(z)$ has radius of convergence $R$. Its coefficients are the sum of $-A/(n{R}^n)$ and the more rapidly decreasing coefficients of $H$. At $x=-R$, the logarithmic part is an alternating harmonic series and converges conditionally; at $x=R$ it diverges. The power series for $H$ converges absolutely at both endpoints and therefore cannot change either conclusion. The real convergence interval is $[-R,R)$.

For $f(z)=\log (2-e^z)$, take $R=\log 2$ and $H(z)=\log [(2-e^z)/(1-z/R)]$, with its branch normalized by $H(0)=0$. The quotient has nonzero removable value $2R$ at $z=R$. Its next zeros are at $R\pm 2\pi i$, so a holomorphic logarithm exists on a larger disk. This shows why the Taylor series converges at $x=-\log 2$ even though its singular positive endpoint is excluded.

Start at $a=1$ with complex logarithm germ of a holomorphic function $L(1)=0$ and holomorphic square root germ of a holomorphic function $S(1)=1$. On the punctured plane $D=\mathbb{C}\setminus \{0\}$, both admit analytic continuation along every path. To construct it locally, for any $c\ne 0$ choose a small open disc avoiding $0$, select a holomorphic logarithm there, and select its additive multiple of $2\pi i$ to agree with the arriving germ of a holomorphic function; exponentiating half this logarithm supplies the arriving holomorphic square root. A finite chain of these open discs follows any given path.

Compare two paths from $1$ to $i$, parametrized by $0\le t\le 1$:The first follows a counterclockwise quarter-circle; the second follows a clockwise three-quarter-circle. Following ${\gamma }_A$ and then path reversal of ${\gamma }_B$ makes one positive turn about $0$. Its winding number is $1$, so the two paths cannot be homotopic with endpoints fixed in the punctured plane.

The continued values along the chosen paths areAt their common endpoint,Both complex logarithm values exponentiate to $i$, and both holomorphic square root values square to $i$. These give different terminal germ of a holomorphic function representatives, consistently obtained from the same starting germ of a holomorphic function. The monodromy theorem applies, but does not equate these two non-homotopic paths.

The closed path $\lambda (t)=e^{2\pi it}$ instead returns to $1$, with continued values $L_{\lambda }(t)=2\pi it$ and $S_{\lambda }(t)=e^{\pi it}$. One counterclockwise turn changes the complex logarithm by $2\pi i$ and flips the holomorphic square root from $1$ to $-1$. Hence neither initial germ of a holomorphic function can come from a global holomorphic logarithm or a global holomorphic square root of $z$ on the punctured plane. Repeating a loop of winding number $n$ gives the changes $2\pi in$ and $(-1{)}^n$. ^[ref]

Now restrict to the slit domain $D_{\mathrm{slit}}=\mathbb{C}\setminus (-\infty ,0]$. The principal logarithm and $\exp (\mathrm{Log}z/2)$ realize the same initial germ of a holomorphic function for each expression globally. To see that this is a simply connected domain, polar form identifies this domain with $(0,\infty )\times (-\pi ,\pi )$, a convex set; every loop contracts there. The simply connected corollary of the monodromy theorem now forces equality along every pair of paths with matching endpoints. In particular, ${\gamma }_A$ and the segment $\sigma (t)=1-t+it$ both give $L(i)=\pi i/2$ and $S(i)=(1+i)/\sqrt{2}$. The explicit homotopy $H(s,t)=(1-s){\gamma }_A(t)+s\sigma (t)$ stays in the closed first quadrant away from $0$, hence in $D_{\mathrm{slit}}$.

The other route ${\gamma }_B$ crosses the removed ray at $t=2/3$ and is no longer admissible. In the punctured plane, that crossing is locally regular: analytic continuation passes through $-1$ smoothly, although evaluating the principal logarithm would reset its argument. The branch cut changes the allowed domain; it does not create additional branch points or poles. That distinction is the beauty of the obstruction: perfectly compatible local holomorphic functions can fail to make a global choice solely because of the allowed based loops. The branches of a multivalued function article develops how moving the branch cut changes the selected values.

If a compact spectrum of an element avoids zero and zero belongs to its unbounded complementary component, choose a simple arc from zero to infinity avoiding the Banach algebra spectrum. Its complement is a simply connected neighborhood of the Banach algebra spectrum, admitting a holomorphic logarithm. Apply holomorphic functional calculus to that logarithm and the composition rule for holomorphic functional calculus to the exponential to obtain the displayed identity. An invertible matrix over the complex numbers always satisfies the hypothesis because its finite Banach algebra spectrum cannot separate zero from infinity.

The logarithmic derivative of a differentiable function $f$ is $f^{\prime }/f$ wherever $f\ne 0$. For a nonvanishing holomorphic function, a local holomorphic logarithm differentiates to this expression even when no global logarithm exists. If a meromorphic function has local factorization $f(z)=(z-a{)}^{m}h(z)$ with $h(a)\ne 0$, differentiation gives $f^{\prime }/f=m/(z-a)+h^{\prime }/h$. Its residue is therefore $m$, positive for a zero and negative for a pole; this explains the multiplicity count in the argument principle. For $\Re s>1$, the Euler product for the Riemann zeta function gives

The quotient $q(w)=\varphi (w)/w$ extends to a holomorphic function at zero with value one. Since $\varphi$ is injective and vanishes at zero, $q$ has no zero in the unit disc. A holomorphic logarithm $\ell$ exists there, with $\ell (0)=0$. Define the odd square-root transform of a normalized univalent function byIt satisfies $h(z{)}^2=\varphi (z^2)$, $h(0)=0$, $h^{\prime }(0)=1$ and $h(-z)=-h(z)$. To check injectivity, $h(z)=h(w)$ gives $\varphi (z^2)=\varphi (w^2)$, hence $z^2=w^2$. If $w=-z$, oddness gives $h(z)=0$, so $z=w=0$. Thus the two arguments always coincide, and $h\in \mathcal{S}$. Choosing the square root by its derivative at zero removes its sign ambiguity.

Fix a positive integer $q$ and an integer $a$ coprime to $q$. We use Dirichlet characters modulo $q$, extended by zero on integers not coprime to $q$, and write ${\chi }_0$ for the principal Dirichlet character. For a nonprincipal Dirichlet character, the sum over one complete period is zero. Its partial sums $A_{\chi }(u)={\sum }_{n\le u}\chi (n)$ are therefore $O(q)$. Partial summation givesThe integral converges locally uniformly there, so these Dirichlet L-functions are holomorphic at one. On the other hand,has a simple pole at one, with residue $\phi (q)/q$.

The assumed nonvanishing of a nonprincipal Dirichlet L-function at one for real Dirichlet characters also excludes vanishing for nonreal ones. To prove this, form the product over all Dirichlet charactersUse the logarithm supplied by the absolutely convergent Euler product. The Orthogonality of Dirichlet characters yieldsThus $\mathcal{F}(\sigma )\ge 1$. If a nonreal Dirichlet character $\chi$ had $L(1,\chi )=0$, its distinct conjugate $\overline{\chi }$ would also have a zero there, since $L(1,\overline{\chi })=\overline{L(1,\chi )}$. These two zeros would outweigh the single pole of the principal factor; all other factors are bounded near one. It would follow that $\mathcal{F}(\sigma )\to 0$ as $\sigma \downarrow 1$, contradicting the lower bound. Together with the assumption for real Dirichlet characters, this proves $L(1,\chi )\ne 0$ for every nonprincipal Dirichlet character.

The first-degree part of the Euler product logarithm givesThe omitted terms are bounded uniformly, since ${\sum }_p{\sum }_{m\ge 2}(m{p}^{m\sigma }{)}^{-1}\ll {\sum }_p{p}^{-2}<\infty$. For each nonprincipal factor, its logarithm remains bounded: choose a holomorphic logarithm near the now established nonzero value $L(1,\chi )$; the Euler-product branch differs from it by a fixed integral multiple of $2\pi i$ on a sufficiently small real interval. For the principal factor,Applying Orthogonality of Dirichlet characters again givesThe right side diverges as $\sigma \downarrow 1$, so the arithmetic progression contains infinitely many primes. This proves Dirichlet's theorem on arithmetic progressions: every reduced residue class modulo $q$ contains infinitely many primes. The argument also covers $q=1$ and $q=2$, when there need not be a nonreal character.

For the sharper statement about reciprocal primes in a fixed arithmetic progression, use the given prime-counting function asymptotic with $q$ fixed:The second equivalence follows by integration by parts in the logarithmic integral function. Partial summation yieldsFor any $\epsilon >0$, choose a fixed $T$ so large that the relative error in $\pi (t;q,a)$ is at most $\epsilon$ for $t\ge T$. The initial integral is a constant, while the remaining integral lies betweenThe boundary term is $O_q(1/\log x)$. Divide by $\log \log x$, let $x\to \infty$, and then let $\epsilon \downarrow 0$. We obtainOnly a relative asymptotic is deduced from the given information; an additive $O(1)$ error, or uniformity for a growing modulus $q$, would require additional estimates.

Give $\mathcal{O}(U)$ the compact-open topology. Choose a finite polygonal cycle $\Gamma$ in $U\setminus {\sigma }_A(x)$ whose winding number is one on the Banach algebra spectrum and zero outside $U$. One can take the oriented boundary of a finite union of small squares surrounding the compact Banach algebra spectrum and contained in $U$. DefineThe integral is an algebra-norm integral of a continuous function on finitely many segments. Cauchy's theorem, applied after any bounded linear functional on $A$, makes it independent of the cycle. Moreoverso it is continuous for the compact-open topology.

Here is a direct verification of the algebra properties and uniqueness using the allowed Runge theorem. The version required is that every holomorphic function on an open plane set is a locally uniform limit of rational functions with poles outside that set, allowing a polynomial part. A rational function $r=p/q$ of this kind has the unambiguous value $r(x)=p(x)q(x{)}^{-1}$, since the roots of $q$ avoid the Banach algebra spectrum. These rational values form a unital algebra homomorphism.

They agree with the contour integral. First $\frac{1}{2\pi i}{\int }_{\Gamma }(z1-x{)}^{-1}dz=1$, by replacing the cycle with a large circle and integrating its Neumann expansion. The identity $z^{j}1-x^j=(z1-x){\sum }_{k=0}^{j-1}{z}^{j-1-k}{x}^k$ then proves the assertion for polynomials. For $a\notin U$, the resolvent identity givesThe cycle has winding number zero at $a$, so integration gives $(x-a1{)}^{-1}$. This identity holds on a neighborhood of $a$ disjoint from the cycle and the Banach algebra spectrum. Differentiating with respect to $a$ gives the analogous result for every power $(z-a{)}^{-k}$, and partial fractions give it for all the rational functions under consideration.

If $r_n\to f$ and $s_n\to g$ locally uniformly, the contour estimate makes their evaluated sequences converge. Since $r_n{s}_n\to fg$ locally uniformly and multiplication in $A$ is continuous, their limits satisfy ${\Theta }_x(fg)={\Theta }_x(f){\Theta }_x(g)$. Also ${\Theta }_x(1)=1$ and ${\Theta }_x(Z)=x$. Conversely any continuous unital complex-algebra homomorphism sending $Z$ to $x$ must send $Z-a$ to $x-a1$ and its reciprocal to the inverse. It therefore agrees on rational functions and then, by Runge density and continuity, on all of $\mathcal{O}(U)$. This proves existence and uniqueness of the holomorphic functional calculus.

For its spectral mapping property, first note compatibility on a smaller open spectral neighborhood $W\subseteq U$: composing restriction to $W$ with the calculus there gives the calculus on $U$ by uniqueness. If $\mu \notin g({\sigma }_A(x))$, choose such a $W$ where $\mu -g$ never vanishes. Thenwith the right side evaluated on $W$. Conversely, for $\lambda \in {\sigma }_A(x)$, write $g(z)-g(\lambda )=(z-\lambda )q(z)$ with $q$ holomorphic on $U$. The evaluated factors commute. If their product were invertible, both factors would be invertible, contradicting $\lambda \in {\sigma }_A(x)$. ThusPullback $h\mapsto h\circ g$ is continuous from $\mathcal{O}(V)$ to $\mathcal{O}(U)$ because the image under $g$ of a compact set is compact. The composite $h\mapsto {\Theta }_x(h\circ g)$ is a continuous unital homomorphism taking the coordinate function to $y={\Theta }_x(g)$. Uniqueness of the calculus at $y$ provesFor the exponential conclusion, let $K={\sigma }_A(x)$. There is a simple polygonal arc from zero through its unbounded complementary component to outside a disk containing $K$, continued by a ray to infinity and avoiding $K$. Loops in an initial polygonal path can be removed to make the arc simple. The complement of this slit is a simply connected open neighborhood of $K$ avoiding zero, and hence admits a holomorphic logarithm $\ell$. Set $y={\Theta }_x(\ell )$. The composition rule givesThe calculus of the entire exponential agrees with the usual algebra exponential because its power-series partial sums converge locally uniformly. Finally an invertible matrix over the complex numbers has a finite Banach algebra spectrum avoiding zero. Removing finitely many plane points leaves a connected set, so the hypothesis always holds: every invertible matrix over the complex numbers is an exponential of a matrix over the complex numbers.

Fix a modulus $q$ and a residue class $a$ with $(a,q)=1$. The Dirichlet characters modulo $q$ are the one-dimensional characters of a representation of $(\mathbb{Z}/q\mathbb{Z}{)}^{\times }$, extended by zero on nonunits. Let ${\chi }_0$ be the principal Dirichlet character.

For every nonprincipal Dirichlet character the partial sums ${\sum }_{n\le u}\chi (n)$ are bounded: the sum over a full period is zero, and a remaining incomplete period has bounded length. Partial summation therefore continues its Dirichlet L-function holomorphically to $\Re s>0$ byThis shows in particular that it has no pole at one. By contrasthas a simple pole there. We use the stipulated nonvanishing $L(1,\chi )\ne 0$ for every real nonprincipal Dirichlet character, and establish the remaining nonvanishing as follows.

For real $\sigma >1$, take the product over all Dirichlet characters. The Euler product logarithms and Orthogonality of Dirichlet characters giveThus this product is real and at least one. If a nonreal Dirichlet character had $L(1,\chi )=0$, complex conjugation would also give $L(1,\overline{\chi })=0$. These are distinct factors. Two zeros and only one possible simple pole force the whole product to tend to zero as $\sigma \downarrow 1$, contradicting the lower bound. Hence all nonprincipal Dirichlet characters have $L(1,\chi )\ne 0$, under the given real-character assumption.

For $\sigma >1$, absolute convergence of each Euler product giveswith bounded remainder as $\sigma \downarrow 1$: the terms of prime-power exponent at least two have total modulus at most ${\sum }_{n\ge 2}1/(n(n-1))$. Since a nonprincipal $L$ defines a holomorphic function that is nonzero near one, its Euler-product logarithm differs from a local holomorphic logarithm by a fixed integer multiple of $2\pi i$ on a short real interval. It is therefore bounded there. For the principal Dirichlet character the pole instead givesApply the Dirichlet character orthogonality relation at each prime. The prime-character sum near one yieldsThis tends to infinity. A finite set of primes would have a bounded sum as $\sigma \downarrow 1$, so every coprime residue class contains infinitely many primes. This proves the Dirichlet theorem on primes in arithmetic progressions from the specified assumption. In fact it proves divergence of the corresponding reciprocal-prime series and gives Dirichlet density $1/\phi (q)$.

Finally, for real $s>1$ each prime satisfiesThe integrands are nonnegative, so Tonelli theorem permits summation inside the integral. At a fixed $t$, exactly $\pi (t,q,a)$ primes from the residue class have entered. ThereforeBoth sides are finite, since they are bounded respectively by ${\sum }_{n\ge 2}{n}^{-s}$ and $s{\int }_1^{\infty }{t}^{-s}dt$. The values of a counting function at its isolated jump endpoints do not change this integral.

Put $\alpha =-\log |c|>0$. Since $f$ has no zeros and the unit disc is simply connected, it has a holomorphic logarithm. Thus $F=-\log f$ is holomorphic with $\mathrm{Re}F>0$ and $\mathrm{Re}F(0)=\alpha$. The Cayley transform between the half-plane and disk givesThe Schwarz lemma yields $|{\varphi }^{\prime }(0)|\le 1$, so the Carathéodory derivative bound for the right half-plane gives $|F^{\prime }(0)|\le 2\alpha$. Since $f^{\prime }=-F^{\prime }f$, the sharp answer isEquality in the Schwarz lemma holds exactly when $\varphi (z)=\omega z$, $|\omega |=1$. Solving for $F$ and exponentiating gives exactly the extremizersTheir logarithmic real part is $\alpha \mathrm{Re}[(1+\omega z)/(1-\omega z)]>0$, so each is indeed a zero-free map into the unit disc with the prescribed value at zero.

For a fixed $w\in \mathbb{D}$, put $r=|w|$. Apply the Harnack inequality on the unit disk to the positive harmonic function $\mathrm{Re}F$ and then exponentiate its negative. The zero-free Schwarz lemma givesFor $w\ne 0$ both endpoints are attained: take $\omega w=r$ for the lower endpoint and $\omega w=-r$ for the upper endpoint in the displayed extremizers. At zero both are $|c|$. If the question is interpreted as allowing $w$ to vary as well, the sharp infimum and supremum over all admissible pairs $(f,w)$ are respectively $0$ and $1$, approached as $r\uparrow 1$; neither value is attained inside the unit disc. An individual function need not have these global extremal values: the constant function $f=c$ is a simple example.

The zero-free quotient $g$ from solution has a holomorphic logarithm, so $\log |g|$ is harmonic. Its radial mean is the constant $\log |g(0)|$. Subtract the radial logarithmic means of $f$ and $B$. The first tends to zero by the absolute logarithmic hypothesis; the second tends to zero by solution. It follows thatSince $|g|\le 1$ and $|g(0)|=1$, the maximum modulus principle makes $g$ a constant of modulus one. Absorb that constant into the unimodular factor of $B$. Therefore $f$ itself is a Blaschke product. This proves the vanishing absolute logarithmic mean characterization of Blaschke products; conversely a Blaschke product satisfies the absolute version too, since $\log |B|\le 0$.

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.

The disc lies in the upper-right half-plane and avoids zero and the negative real axis. On the slit plane take the holomorphic logarithm $\mathrm{Log}z=\log |z|+i\arg z$ with $-\pi <\arg z<\pi$. Its derivative is $1/z$, and $e^{\mathrm{Log}z}=z$. Therefore the holomorphic square rootis analytic and satisfies $f^2=z$. One can also construct it without a global branch theorem: fix either scalar square root $a$ of $3+3i$ and expand $a(1+(z-3-3i)/(3+3i){)}^{1/2}$ by its convergent binomial series. The series converges on the entire disc because $|z-3-3i|/|3+3i|<1/\sqrt{18}$. Either sign gives an admissible choice.