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The analytic Wiener algebra consists of the power series $f(z)={\sum }_{n\ge 0}{a}_n{z}^n$ with ${\sum }_{n\ge 0}|a_n|<\infty$, identified with their values on the unit circle. The norm is $\Vert f\Vert =\sum |a_n|$. Discrete convolution gives multiplication and a submultiplicative norm; completeness follows from the absolutely summable sequence space. Each series converges uniformly on the closed unit disk and is holomorphic in its interior.

Let a group $G$ of Möbius transformations of the unit disk have an orbit satisfying the Blaschke condition, counted once per distinct orbit point. For its Blaschke product $B$, invariance of the pseudohyperbolic distance and permutation of the orbit give $|B(Tz)|=|B(z)|$. The ratio $B(Tz)/B(z)$ extends across its zeros and has modulus one, so the open mapping theorem makes it a constant $\chi (T)$. Composition gives a group homomorphism $\chi :G\to \{\lambda :|\lambda |=1\}$. It is essential to compare moduli of the products: two arbitrary bounded holomorphic functions with identical zeros need not have constant quotient.

Knowing the radius of convergence does not settle convergence on its boundary. The three power seriesall have radius one, but behave differently on $|z|=1$.

For the geometric series, boundary terms all have complex modulus one, violating the term test for divergence. For the second series, absolute values give the divergent harmonic series at every boundary point. At $z=1$ the series itself is harmonic and diverges. At any other point on the unit circle, its geometric partial sums obeyThe Dirichlet test with the decreasing sequence $1/n\to 0$ therefore gives conditional convergence. In particular, $z=-1$ gives the negative of the alternating harmonic series, with sum $-\ln 2$. For the third series, the convergent p-series ${\sum }_{n\ge 1}1/n^2$ bounds the absolute values at every boundary point, giving absolute convergence by the comparison test for series.

At $z=-1$, the rational function $1/(1-z)$ is well defined and equals $1/2$, yet its geometric series diverges there. A formula for a function and a particular series representation can have different domains. Likewise, applying termwise differentiation of a power series preserves the radius but need not preserve the boundary convergence.

Normalize the autoregressive and moving-average polynomials to have constant coefficient one, cancel common factors, and require all their zeros to lie outside the closed unit disk. With positive innovation variance these conditions give a causal, invertible, minimal Gaussian process representation. To prove identifiability, compare two such models with the same time-series spectral density. The ratio of their scaled rational transfer functions and its reciprocal are analytic on a neighbourhood of the closed unit disk and have modulus one on its boundary. The maximum modulus principle makes the ratio constant; its positive value at zero forces equality of the innovation variances and equality of the rational functions. Coprimality and the constant-coefficient normalization then force equality of both polynomials. Without the root restrictions, reciprocal root reflection of an ARMA representation and a rescaling of the innovation variance can preserve the spectrum, so the unrestricted parameters are not identifiable.

For the power series ${\sum }_{n\ge 0}{a}_n(z-c{)}^n$, the Cauchy-Hadamard theorem giveswith $R=\infty$ when the limit superior is zero, and $R=0$ when it is infinite. Away from the centre, the root test explains the formula: the limit superior of the term roots at displacement $z-c$ is $|z-c|/R$, interpreted as zero for $R=\infty$ and infinite for $R=0$. Values below one give absolute convergence; values above one imply that the terms fail to tend to zero. A value of one leaves the boundary question open. At the centre the series converges regardless of $R$.

For $a_{2n}=1$ and $a_{2n+1}=0$, the coefficient roots alternate between one and zero, so their ordinary limit does not exist. Their limit superior is nevertheless one, giving $R=1$ for ${\sum }_{n\ge 0}{z}^{2n}=1/(1-z^2)$ inside the unit disk. A ratio of consecutive coefficients would be undefined at the zero coefficients; the root-limsup formula handles the gap directly.

In the first part of his 1892 study of functions given by Taylor series, Jacques Hadamard introduced an upper-limit formulation for sequences and used coefficient roots to determine the radius of convergence. Sections 4–6 separate finite positive, zero, and infinite root-growth cases; the later parts study singularities and continuation. See Hadamard, Essai sur l'étude des fonctions données par leur développement de Taylor, Journal de mathématiques pures et appliquées, fourth series 8 (1892), pp. 107–109. This original text explains the coefficient-growth question without implying that the study of power series began there.

This Möbius transformation maps the upper half-plane onto the unit disk. For $\mathrm{Im}z>0$, the identity $1-|C(z){|}^2=4\mathrm{Im}z/|z+i{|}^2$ puts the image inside the unit disk. Its inverse is $z=i(1+w)/(1-w)$, so the map is a conformal bijection. On the extended real boundary it maps onto the unit circle. It is the scalar complex-analytic version of a Cayley map, rather than the matrix parametrization used in Cayley transform.

Every algebra character of the analytic Wiener algebra is evaluation at a unique point $z$ of the closed unit disk. Indeed, for $u(w)=w$, continuity and multiplicativity give $z=\varphi (u)$, $|z|\le \Vert u\Vert =1$ and $\varphi (f)={\sum }_{n\ge 0}{a}_n{z}^n$. Conversely, this formula defines a continuous algebra character for every such $z$. The evaluation correspondence is a homeomorphism for the Gelfand topology, since the power series are uniformly convergent on the closed unit disk.

Let $F(\zeta ,t)$ map the unit disk conformally to a planar Darcy flow region, with $F(0,t)=0$ at a point source of flux $2\pi Q$. With zero velocity potential on the boundary, the logarithmic source potential becomes $\varphi =Q\log |\zeta |$. Differentiating the inverse map at fixed physical position gives ${\varphi }_t=-Q\mathrm{Re}(F_t/(\zeta F_{\zeta }))$, while $|\nabla \varphi {|}^2=Q^2/|\zeta F_{\zeta }{|}^2$. The boundary law ${\varphi }_t+|\nabla \varphi {|}^2=0$ therefore gives the displayed equation, for nonzero $Q$.

The vector $\zeta F_{\zeta }d\theta$ is outward normal times arclength on $\zeta =e^{i\theta }$. Consequently the equation gives the area growth law $d|D|/dt=2\pi Q$. For a circular map $F=R(t)\zeta$, it reduces to $R\dot{R}=Q$, and hence $R(t{)}^2=R(0{)}^2+2Qt$.

A convex set permits every intermediate mixture of two of its points. If two choices meet its constraints, continuously averaging them also meets those constraints. This geometric property makes many feasible regions easier to explore: there is no forbidden gap along a straight move between feasible choices.

Formally, a subset $C$ of a real vector space is convex whenThe parameter $t$ traverses the entire line segment, including both endpoints. The definition does not require a closed subset, a bounded set, or a nonempty interior. The empty set, the whole vector space, and every affine subspace are convex; an open unit disk and a closed unit disk are both convex.^[ref]

The triangle inequality verifies the disk example directly: if $\Vert x\Vert ,\Vert y\Vert \le 1$, then $\Vert (1-t)x+ty\Vert \le (1-t)\Vert x\Vert +t\Vert y\Vert \le 1$. For the open disk, the last inequality is strict. The unit circle fails the same test: $(1,0)$ and $(-1,0)$ belong to it, but their midpoint $(0,0)$ does not. A curved route along the unit circle connects the endpoints, so being path-connected is weaker than being a convex set.

The distinction between a convex set and a convex function matters. The former concerns where one may choose a point; the latter concerns how a value changes along line segments. Convex optimization combines both: averaging preserves feasibility, and a convex function obeys an upper bound by the corresponding average of endpoint values. A convex set alone does not make an arbitrary objective a convex function or eliminate its inferior local minima.

On the Dahlquist test equation, write the amplification polynomial as $D{\xi }^2-16z\xi -A$, where $D=z^2-7z+15$ and $A=z^2+7z+15$. The identity $|A{|}^2-|D{|}^2=28\mathrm{Re}z(|z{|}^2+15)$ shows that both roots of a polynomial can lie inside the open unit disk only if $\mathrm{Re}z<0$. But the complex quadratic Schur criterion would then require $32|\mathrm{Re}z|(|z{|}^2+15)<28|\mathrm{Re}z|(|z{|}^2+15)$, which is impossible. Thus the strict linear stability domain is empty. This concerns robust damping of all modes, rather than convergence on fixed time intervals or the bounded-root stability set of the symmetric two-derivative formula.

For an explicit linear multistep method, the highest-degree coefficient of $\rho (w)-z\sigma (w)$ is fixed while at least one lower coefficient grows unboundedly with negative real $z$. If all roots stayed in the unit disk, the elementary symmetric polynomials of those roots would keep every monic coefficient bounded. This contradiction proves that no nontrivial consistent explicit method is A-stable.

An exposed point is a point uniquely selected by maximizing a linear functional. In ${\mathbb{R}}^d$, a point $p$ of a convex set $C$ is exposed if some vector $c$ satisfiesGeometrically, a supporting hyperplane touches the convex set only at $p$. Every exposed point is an extreme point: a convex combination that attains the maximum can put positive weight only on maximizers, and there is only one.^[ref]

The unit disk calculation in extreme point exposes every point of the unit circle. For a corner $(a,b)$ of the square, $c=(a,b)$ exposes it: $ax+by$ attains two only at $(a,b)$. However, maximizing $x$ exposes no single point on the right edge, since all of them have the same objective value. A linear functional may have several optimal points, even when some of them are extreme.

If the derivative characteristic polynomial $\sigma$ of a linear multistep method has a simple root outside the unit disk and keeps its degree in the limit, then its absolute stability region is bounded. Otherwise a stable sequence with $|z|\to \infty$ would give amplification polynomials $\sigma -\rho /z$ tending coefficientwise to $\sigma$. A root near the exterior root would persist by continuity, contradicting the root condition for a multistep method. This argument excludes stability at every sufficiently large complex $z$, not only along the negative real axis.

An extreme point of a convex set is a point that cannot be made by genuinely mixing two different points of the set. For a convex set $C$ in a real vector space, the condition on $p\in C$ isThus $p$ lies on no open line segment between two distinct points of $C$. It is enough to exclude midpoint representations: if $p$ lies inside a line segment, two nearby points on opposite sides have midpoint $p$.^[ref]

The square $[-1,1{]}^2$ has just four extreme points, at $(\pm 1,\pm 1)$. In a convex combination giving a corner, each coordinate is already at its largest or smallest allowed value, forcing both input points to agree with that corner. In contrast, the boundary point $(1,0)$ is not extreme:It is the midpoint of two other points on the same edge. An interior point also admits a small line segment through it.

The unit disk $D=\{z\in {\mathbb{R}}^2:\Vert z{\Vert }_2\le 1\}$ has a different shape: every point on its unit circle is extreme. For a unit vector $p$, the Cauchy-Schwarz inequality givesand equality is possible only at $z=p$. If a convex combination equals $p$, applying this linear functional forces both ingredients to attain that equality. Both must therefore be $p$. Extremality need not look like a sharp corner.

These points help describe all allowable mixtures. Every $(x,y)\in [-1,1{]}^2$ is a convex combination of its four corners, with the weight on $(a,b)\in \{-1,1{\}}^2$ equal to $(1+ax)(1+by)/4$. For the unit disk, write $z=rp$, with $0\le r\le 1$ and $p$ a unit vector; at zero choose any $p$. ThenTwo extreme points suffice for that mixture. The Krein-Milman theorem gives a much broader reconstruction: a nonempty compact convex set in a locally convex space whose topology is Hausdorff is the closed convex hull of its extreme points.

Mark Krein and David Milman published their reconstruction theorem in 1940. Milman's colleagues describe that result and its subsequent use as a method of representing convex sets through their extreme points.^[ref]

The distinction appears at a smooth join between a straight segment and a curved boundary. Consider the convex hullThe upper tangent from $P$ meets the unit circle at $T=(1/2,\sqrt{3}/2)$. The supporting line isEvery point of the unit disk has $T\cdot z\le 1$, with equality only at $T$. Hence the convex face of $C$ where equality holds is exactly the line segment $TP$.

The point $T$ is extreme. If a nontrivial convex combination of two points of $C$ equals $T$, both must belong to the convex face $TP$. Since $T$ is an endpoint of that line segment, both must equal $T$.

Yet $T$ is not exposed. A nonzero linear functional maximizing at $T$ on $C$ must also maximize there on the included unit disk. Its vector must be a positive multiple of $T$, by the equality condition in the Cauchy-Schwarz inequality. It then gives $P$ the same value as $T$. No linear functional can select $T$ alone. Being an extreme point is a property of possible mixtures; being an exposed point adds the ability to distinguish it by one linear objective.

For $X_t=\alpha X_{t-1}+{\alpha }^2{X}_{t-2}+{\epsilon }_t$, let $\phi =(1+\sqrt{5})/2$. The autoregressive polynomial factors as $(1-\alpha \phi z)(1+\alpha z/\phi )$. Its roots lie outside the closed unit disk exactly when $|\alpha |<1/\phi$. In that causal region,The coefficients follow from ${\psi }_0=1,{\psi }_1=\alpha$ and ${\psi }_j=\alpha {\psi }_{j-1}+{\alpha }^2{\psi }_{j-2}$. They are absolutely summable in the stated region. Causality makes future white noise orthogonal to the observation past, giving forecasts $\alpha X_T+{\alpha }^2{X}_{T-1}$ and $2{\alpha }^2{X}_T+{\alpha }^3{X}_{T-1}$ at horizons one and two.

For absolutely summable linear-filter coefficients, the generating function is the displayed Laurent series, absolutely convergent on the unit circle. If the filter is one-sided, it is a power series analytic inside the unit disk. A two-sided filter need not have a nonempty larger annulus of convergence. Substituting the backshift operator gives the formal notation $Y=A(B)X$.

Stability of the partial differential equation requires a bound uniform over refining meshes and time steps on every fixed physical time interval. This is distinct from demanding an all-time bound for one fixed matrix. A fixed matrix is power bounded exactly when all eigenvalues are in the closed unit disk and its unit-modulus Jordan blocks have size one. A time-step family can nevertheless have harmless polynomial physical-time growth: $G_{\tau }=I+\tau N$ with $N^2=0$ gives $G_{\tau }^n=I+n\tau N$, uniformly bounded for $n\tau \le T$ despite its nontrivial unit-modulus Jordan block. Mesh dependence and the scaling of the step must therefore be kept in an eigenvalue argument.

With the positive logarithmic convention, a Green function with pole $p$ on a noncompact Riemann surface is positive and harmonic away from $p$, has local form $-\log |\zeta |$ plus a harmonic function in a coordinate $\zeta (p)=0$, and is minimal among positive functions with these properties. It is obtained, when finite, as the increasing limit of zero-boundary Green functions on an exhaustion by compact bordered subdomains. On the unit disk it isIts sign convention is $-\Delta G=2\pi {\delta }_p$ in a conformal coordinate, with the opposite overall sign from conventions using $\log |\zeta |$ as the singular part.

A harmonic majorant of a real function $u$ is a real harmonic function $h$ with $h\ge u$ pointwise. A least harmonic majorant lies below every other harmonic majorant. A continuous subharmonic function need not have one: $u(z)=(1-|z{|}^2{)}^{-1}$ on the unit disk has increasing circular means diverging to infinity, whereas any harmonic majorant would have the same finite mean on every centered circle.