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For a holomorphic scalar function on a neighborhood of a closed annulus, $M(r)$ is the maximum modulus on its radius-$r$ circle. The logarithm of its modulus is subharmonic; compare it with the harmonic function affine in $\log |z|$ which takes the logarithms of the two boundary maxima. The maximum principle for subharmonic functions proves the displayed bound. Zero boundary maxima are handled by adding a positive bound and taking its limit. The same bound holds for the norm of a Banach space-valued holomorphic function by applying scalar functionals and the Hahn-Banach theorem.

On a relatively compact coordinate disk $V$ in a Riemann surface, replace a continuous subharmonic function $u$ inside $V$ by the Poisson integral of its boundary values, leaving it unchanged outside. The resulting function $L_{V}u$ is continuous and subharmonic, equals $u$ off $V$, is harmonic on $V$, and satisfies $L_{V}u\ge u$. The maximum principle for subharmonic functions proves the inequality. The pasting assertion follows from harmonic comparison: a harmonic comparison function dominating the pasted function on the boundary of a test domain first dominates $u$, then dominates the disk lift on its part inside $V$.

The maximum principle for subharmonic functions here states that if $u\in C^2(D)\cap C(\overline{D})$ and $\Delta u=f\ge 0$ in $D$, thenIf $f\le 0$, apply this to $-u$ to obtain the corresponding minimum principle. In particular a harmonic function has both its maximum and minimum on the boundary. Without a sign condition on $f$, this conclusion need not hold.

First suppose $\Delta u>0$. By compactness and continuity, $u$ attains a maximum on $\overline{D}$. At an interior maximum, the first derivatives vanish and the Hessian matrix is negative semidefinite. Its trace is then nonpositive, contradicting $\Delta u>0$. Thus a maximum lies on the boundary.

For the non-strict case put $u_{ϵ}=u+ϵ(x^2+y^2)$, with $ϵ>0$. Then $\Delta u_{ϵ}=f+4ϵ>0$, so the strict case yields, for $z\in \overline{D}$,Let $ϵ\downarrow 0$. This gives $u(z)\le {\max }_{\partial D}u$, proving the theorem. The argument proves the weak maximum principle; it does not claim that every maximizer must be on the boundary when $u$ is constant.

For a real solution, $u^4\ge 0$, so $\Delta u\ge 0$. The maximum principle for subharmonic functions gives $u\le {\max }_{\partial D}u=0$. The additional nonnegativity condition gives $u\ge 0$. ThereforeThis function satisfies both the equation and boundary condition, so it is exactly the full set of nonnegative solutions. The sign restriction is used alongside the maximum principle, rather than treating the nonlinear equation as a linear uniqueness problem.

Use the Nevanlinna proximity function $m(r,f)=(2\pi {)}^{-1}\int {\log }^{+}|f(r{e}^{i\theta })|d\theta$ and the Nevanlinna integrated counting function $N(r,f)$ for poles, counting multiplicity. If the pole order at zero is ${\nu }_0$, thenBoundedness of the Nevanlinna characteristic refers to $r\uparrow 1$, equivalently to a uniform bound on $r_0\le r<1$ for any fixed $r_0>0$. This avoids the harmless origin normalization when ${\nu }_0>0$.

The Nevanlinna first main theorem states, for every fixed finite value $a$ and every nonconstant meromorphic function on the disk,For $a=\infty$, this is the definition of $T$. The error is bounded independently of $r$. Translation changes $T$ by $O(1)$, while Jensen's formula gives $T(r,h)-T(r,1/h)=\log |c|$ when $h(z)=c{z}^{\nu }(1+O(z))$ near zero; these observations explain the first theorem.

The Nevanlinna second main theorem in the unit disc states that for $q\ge 3$ distinct values $a_j$ on the Riemann sphere,as $r\uparrow 1$ outside an exceptional set $E$ satisfying ${\int }_{E}dr/(1-r)<\infty$. The truncated Nevanlinna counting function $\overline{N}$ counts each distinct preimage once. The boundary-distance error cannot simply be suppressed: on the disk it need not be $o(T)$. The disk error and exceptional-set estimates follow from the logarithmic-derivative estimates in mathweb.tifr.res.in/Documents/Publications/Lectures/tifr17.pdf .

Suppose first that $T(r,f)$ is bounded. Since $m(r,f)\ge 0$, its poles satisfy the Blaschke condition, counting their orders: let $r\uparrow 1$ in their integrated count and use $1-|b|\le -\log |b|$. Let $B$ be their Blaschke product, including the origin factor if necessary. Then $g=Bf$ extends to a holomorphic function on the whole disk, and $|B|\le 1$ givesHere the last bound follows from bounded $T$ and the uniform lower bound $N(r,f)\ge {\nu }_0\log r_0$ on $r\ge r_0$.

The nonnegative subharmonic function ${\log }^{+}|g|$ has a finite harmonic majorant. Indeed its Poisson integrals $H_R$ on expanding disks dominate it by the maximum principle for subharmonic functions, increase with $R$, and have $H_R(0)\le C$. The Harnack inequality for harmonic functions bounds them on every compact subdisk, so their increasing limit is a finite harmonic function $H\ge {\log }^{+}|g|$. Since the disk is simply connected, $H$ has a harmonic conjugate, giving a holomorphic function $A$ with $\mathrm{Re}A=H$. ConsequentlyThis proves the constructive half of the quotient characterization of bounded characteristic.

Conversely, if $f=u/v$ with $u,v$ bounded holomorphic functions and $v\not\equiv 0$, each pole of $f$ is a zero of $v$ of at least the same order. The elementary inequality ${\log }^{+}|u/v|\le {\log }^{+}|u|+{\log }^{+}|1/v|$ and the Nevanlinna first main theorem giveAny cancellation at an origin zero contributes only a bounded multiple of $\log r$ on $r\ge r_0$, which is included in this $O(1)$. The zero function is immediate; constant nonzero $v$ is handled directly. Thus bounded characteristic is exactly a quotient of two bounded analytic functions.

An unbounded example with bounded characteristic isIt is a ratio of the bounded holomorphic functions $1$ and $1-z$. More explicitly, Jensen's formula gives the mean of $\log |1-r{e}^{i\theta }|$ as zero, so $m(r,1/(1-z))=m(r,1-z)\le \log 2$, although $g(r)\to \infty$.

Finally, set $Y=(1+|g{|}^2{)}^{p/2}\ge 1$. Pointwise ${\log }^{+}|g|\le p^{-1}\log Y$. Jensen's inequality for the concave logarithm givesThere is no pole term because $g$ is holomorphic. Hence the stated integral bound implies bounded characteristic. This argument in fact works for every $p>0$.

First establish the norm estimate behind the spectral-radius three-circle inequality. If $g$ is a holomorphic Banach space-valued function on a neighborhood of an annulus $r_0\le |z|\le r_1$, then for $r_0<r<r_1$ and $\theta =\log (r/r_0)/\log (r_1/r_0)$,Here is the underlying Hadamard three-circle theorem argument. For any bounded linear functional $\ell$ of norm at most one, $\ell (g)$ is scalar holomorphic. Its logarithmic modulus is subharmonic. Compare it on the annulus with the harmonic function affine in $\log |z|$ whose boundary values are the logarithms of the two boundary norm bounds. The maximum principle for subharmonic functions gives the interpolated bound inside. Using slightly larger positive bounds and then taking limits handles zeros. The Hahn-Banach theorem identifies the norm with the supremum over these scalar functionals, proving the displayed Banach-valued estimate.

Apply this estimate to $g(z)=f(z{)}^{2^n}$ and take the $2^n$-th root. On either boundary circle the nonnegative continuous functionsdecrease, because submultiplicativity gives $q_{n+1}\le q_n$. The spectral radius formula gives their pointwise limit $r(f(z))$. Part (i) passes their boundary suprema to the limit. With $M_r={\sup }_{|z|=r}r(f(z))$ we obtain the general estimateIn particular, take $r_0=R^{-1}$, $r=1$, $r_1=R$, so $\theta =1/2$. Since $r(f(1))\le M_1$,This proves the requested inequality without assuming continuity of the spectral radius on the circles.

For the final spectral-radius Liouville theorem, suppose $r(f(z))\le C$ on the whole plane. The case $C=0$ is immediate. Center circles at an arbitrary $a$, and put $M_a(t)={\sup }_{|z-a|=t}r(f(z))$. For $0<\delta <r<R$, the same argument givesIf $M_a(\delta )=0$, this forces $M_a(r)=0$. Otherwise let $R\to \infty$ to get $M_a(r)\le M_a(\delta )$. Upper semicontinuity, proved in 2(a), gives $M_a(\delta )\le r(f(a))+\epsilon$ for all sufficiently small $\delta$. Thus every $z$ satisfies $r(f(z))\le r(f(a))$. The center $a$ is arbitrary, so reversing the two points gives equality. Hence the spectral radius is constant. The algebra-valued function itself need not be constant; $f(z)=zN$ with a nonzero square-zero matrix $N$ is a useful example.

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.

True. Continuity and compactness give a maximum of $\varphi$ on $\overline{\Omega }$. At an interior maximum the Hessian matrix is negative semidefinite, so its trace satisfies $\Delta \varphi \le 0$, contradicting $\Delta \varphi >0$. The maximum must therefore occur on $\partial \Omega$. This is the strict form of the maximum principle for subharmonic functions.

Suppose $u$ were positive somewhere. Since it tends to zero at infinity and to $-\infty$ at its vortex zero, it would attain a positive interior maximum away from the origin. At such a maximum the second-derivative test gives $\Delta u\le 0$, whereas the smooth Taubes equation givesThis contradiction with the maximum principle for subharmonic functions provesConsequently the Higgs magnitude of a vortex satisfies $|\varphi |=e^{u/2}\le 1$ everywhere.

Let $M={\max }_{S}u$ and suppose $u(z_0)=M$. The sub-mean inequality on every closed disk contained in $S$ givesEquality holds throughout. If $u$ were strictly below $M$ at one point of the circle, upper semicontinuity would make it uniformly below $M$ on a small arc, contradicting equality of the average. Thus $u=M$ on every sufficiently small circle centered at $z_0$, and hence throughout a neighborhood of $z_0$.

The set $\{u=M\}$ is therefore open. It is also closed because upper semicontinuity makes $\{u<M\}$ open. Since the domain is connected and $\{u=M\}$ is nonempty, it is all of $S$. This proves the maximum principle for subharmonic functions.

LetThe resolvent set is open, and the resolvent of an element is operator-valued holomorphic on each of its components. Fix $z_0$ in the resolvent set. For any $\delta >0$, choose unit vectors $x,y$ such thatThe scalar function $f(z)=⟨R(z)x,y⟩$ is holomorphic. Its modulus is subharmonic, soLetting $\delta \downarrow 0$ proves that the resolvent norm is subharmonic on the resolvent component containing $z_0$.

Use the convention that the reciprocal resolvent norm is zero on the spectrum. Suppose a bounded component $U$ ofcontained no spectral point. A spectral point in its boundary would belong to the same pseudospectral component, so $\overline{U}$ is contained in the resolvent set. On $\partial U$ one has $\Vert R(z)\Vert \le {ϵ}^{-1}$, whereas inside $U$ one has $\Vert R(z)\Vert >{ϵ}^{-1}$. Continuity on the compact set $\overline{U}$ makes the resolvent norm attain a maximum at an interior point. The subharmonic maximum principle would make it constant, contradicting its boundary values. Hence every bounded component of the pseudospectrum contains spectrum:

For a zero of order $N$ at the origin and no other zeros,Away from the zero, the flat Taubes equation isIf $u$ had a positive interior maximum, then the second-derivative test would give $\Delta u\le 0$ there, while $e^u-1>0$, a contradiction. The boundary values are $-\infty$ at the zero and $0$ at infinity, so the maximum principle gives $u\le 0$. Hence