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A Q-matrix satisfies $q_{ij}\ge 0$ for $i\ne j$ and $q_{ii}=-{\sum }_{j\ne i}{q}_{ij}$; on a countable state space the exit rate $-q_{ii}$ is required to be finite. For a finite-state continuous-time Markov chain with transition semigroup $P(t)$,The backward and forward Kolmogorov equations are respectively

On a probability space carrying an infinite IID sequence $(X_n)$ with law $m$, map $\omega$ to $(X_1(\omega ),X_2(\omega ),\dots )$. Its pushforward measure $\mu$ satisfiesfor every cylinder set. Cylinder sets form a pi-system generating $\sigma (\mathcal{C})$, so the sigma-finite uniqueness theorem for measures proves uniqueness. This is the countable product measure $m^{\otimes \mathbb{N}}$.

For a cylinder $A=A_1\times \cdots \times A_N\times \mathbb{R}\times \cdots$,The same generating-class argument extends this equality to every measurable set, so $\theta$ is a measure-preserving transformation.

If $A={\theta }^{-1}A$, then $A={\theta }^{-n}A$ for every $n$, so membership in $A$ is independent of the first $n$ coordinates. Thus $A$ belongs to the tail sigma-algebra. The zero-one law gives $\mu (A)\in \{0,1\}$, proving that the shift is ergodic.

Let $(X_n)$ be independent random variables and letbe their tail sigma-algebra. The Kolmogorov zero-one law states that every $A\in \mathcal{T}$ has probability zero or one.

Indeed, $A$ is independent of $\sigma (X_1,\dots ,X_n)$ for every $n$. The union of these finite-coordinate sigma-algebras generates $\sigma (X_1,X_2,\dots )$, and independence extended from generating pi-systems shows that $A$ is independent of that entire sigma-algebra. Since $A$ itself belongs to it,so $\mathbb{P}(A)\in \{0,1\}$.

No. If the image lies in a great circle, there is a fixed unit vector $a$ with $⟨N,a⟩=0$, so $a$ is tangent to $S$ everywhere. Since the ambient derivative of this constant vector is zero, the shape operator satisfies $S(a)=0$. One principal curvature is therefore zero. Minimality makes their sum zero, so both vanish and $S=0$. The Gauss map is then locally constant, and connectedness makes it constant on all of $S$, contradicting its image being a great circle.

A minimal surface can have a planar point, where $k_1=k_2=0$. There $dN=0$, which fails the stipulated conformality condition $\lambda >0$. Thus minimality need not make the Gauss map conformal everywhere; a plane is the simplest counterexample.

Conversely, conformality gives $|k_1|=|k_2|\ne 0$. It permits either $k_1=-k_2$, which is minimal, or $k_1=k_2\ne 0$, which is umbilical and nonminimal. On a round sphere, the Gauss map is conformal while both principal curvatures are equal and nonzero. Thus conformality alone does not imply minimality.

The Gauss map sends $p$ to the chosen unit normal $N(p)$. Differentiating $⟨N,N⟩=1$ shows that $d{N}_p$ maps $T_{p}S$ into itself. The shape operator $S_p=-d{N}_p$ is self-adjoint because its associated second fundamental form is symmetric. Its eigenvalues $k_1,k_2$ are the principal curvatures. A point is umbilical when $k_1=k_2$, and the surface is minimal when its mean curvature $H=(k_1+k_2)/2$ vanishes everywhere.

In an orthonormal principal basis, $d{N}_p$ has diagonal matrix $\mathrm{diag}(-k_1,-k_2)$. It is conformal exactly whenwhich is equivalent to $k_1^2=k_2^2\ne 0$. At a nonumbilical point this says $k_1=-k_2$, exactly the condition $H=0$. Hence, when there are no umbilical points, the surface is minimal if and only if its Gauss map is conformal.

If $f_0(z)=i\mathrm{Log}z$ is the projection from part (b), then any other Mercator projection $f$ gives an automorphism $f\circ f_0^{-1}$ of $E$ preserving its two ends. Part (c) therefore yieldsEvery map in this family plainly has the required limiting behaviour.

The map $h(z)=e^{-iz}$ is a conformal equivalence from $E$ to the plane slit along the nonpositive real axis. Conjugating $\varphi$ by $h$ gives an automorphism of the slit plane whose two prime ends at $0$ and $\infty$ are fixed. Taking the principal square root identifies the slit plane with the right half-plane; the automorphisms of a half-plane that fix $0$ and $\infty$ are precisely $w\mapsto aw$ with $a>0$. Undoing the square root changes only the positive scale, sofor some $a>0$. Since $E$ is connected and both sides use the same branch,Thus $\varphi$ is translation by a purely imaginary number.

Use the principal branch $\mathrm{Log}z=\log |z|+i\arg z$, with $-\pi <\arg z<\pi$. Thenis a conformal equivalence from the slit plane to the vertical strip. Its imaginary part tends to $-\infty$ as $z\to 0$ and to $+\infty$ as $z\to \infty$.

The uniformization theorem says that every simply connected Riemann surface is conformally equivalent to exactly one of the Riemann sphere, the complex plane, and the unit disc. A proper simply connected plane domain is noncompact, so it is not the sphere. It cannot be uniformized by the plane: after choosing a point outside the domain, a branch of a square root and then a Möbius transformation would produce a nonconstant bounded entire function, contradicting the Liouville theorem. It is therefore conformally equivalent to the disc, which is the Riemann mapping theorem.

Let $S$ be the left shift and defineShift invariance gives $\varphi (A_{p}x)=\varphi (x)$. The hypothesis says $A_{p}x$ converges uniformly in $n$, hence in the ${\ell }^{\infty }$ norm, to the constant sequence $y1$. By continuity,

Take $x_n=(-1{)}^n$. Its shift is $-x$, so linearity and shift invariance giveThus every such functional assigns the common valuealthough the sequence does not converge.

No. Shift invariance makes all $\varphi (e_n)$ equal. Positivity givesso $\varphi (e_n)=0$. If an ${\ell }^1$ sequence represented $\varphi$, then $y_n=\varphi (e_n)=0$ for every $n$, which would give $\varphi (1)=0$, a contradiction.

Let $m_N={\inf }_{n\ge N}{x}_n$ and $M_N={\sup }_{n\ge N}{x}_n$. Positivity and normalization giveShift invariance makes the middle term equal to $\varphi (x)$. Letting $N\to \infty$ yields

The polynomial $X^3+2X^2+1$ is irreducible by the rational-root test. The power basis $1,\theta ,{\theta }^2$ has discriminantFor the order $\mathbb{Z}[\theta ]\subseteq {\mathcal{O}}_L$,Since $59$ is squarefree, the index can only be one. Hence