The Blaschke product here is taken over distinct points of the orbit; its existence is part of the hypothesis. It is not necessary to assert that every discrete orbit satisfies the Blaschke condition.

The modulus of a Blaschke factor is the pseudohyperbolic distance. Thus, away from the orbit,The logarithms converge absolutely at such a point, by the compact convergence estimate for the Blaschke product. Every $T\in G$ preserves the hyperbolic metric and permutes the orbit. Reindexing this convergent product yields $|B(Tz)|=|B(z)|$.

The functions $B\circ T$ and $B$ have the same simple zero set, since $T$ is a biholomorphism. Their quotient therefore extends holomorphically and without zeros across the orbit. It has modulus one everywhere. The open mapping theorem makes it a constant, denoted $\chi (T)$. FinallySince $B$ is not identically zero, cancellation givesThis is the automorphy character of an orbit Blaschke product. The equal-modulus argument is essential: equality of zero sets alone would not make the quotient of two bounded holomorphic functions constant.

A point fixed by $F^4$ satisfies $(2^4-1)x\in \mathbb{Z}$, so $x=j/15$ for $0\le j\le 14$. The points with period dividing two are $0,1/3,2/3$, corresponding to $j=0,5,10$. All other twelve points have least period four. Multiplication by two modulo fifteen groups them into the three cyclesThis exhausts the requested four-cycles. The displayed order is the dynamical order, and cyclic changes of the starting point describe the same cycle.

Reduce the polynomial modulo $p$, obtaining a nonzero degree-$d$ polynomial over the field ${\mathbb{F}}_p$. If $a$ is a root, polynomial division gives $\bar{f}(X)=(X-a)h(X)$. Any different root $b$ is a root of $h$, because $b-a\ne 0$ is invertible. Induction on degree therefore proves there are at most $d$ distinct roots modulo $p$.

Every nonzero residue is a root of $X^{p-1}-1$ by Fermat's little theorem. The difference between this monic polynomial and the product over all its $p-1$ nonzero roots has degree at most $p-2$, yet has $p-1$ roots. The root bound forces it to vanish identically over ${\mathbb{F}}_p$. Thus all coefficients of the corresponding integer polynomial are divisible by $p$.

For the requested failure modulo $p^2$, take $p=7$: $6!+1=721=7\cdot 103$, and $103$ is not divisible by seven. Hence the prime-square strengthening fails, even though some special primes do satisfy it.

The following unheaded requests concern the spherical measure, the wave solution and both uniqueness arguments. The missing part (i) above and this root Solution slot restore their coverage while retaining the existing part (ii) anchor.

For $k=|\xi |$, rotational symmetry of the spherical measure giveswith continuous value $4\pi t^2$ at $k=0$. The spatial Fourier transform of the wave equation gives ${\hat{u}}_{tt}+k^2\hat{u}=0$ with initial data zero and $\hat{g}$. Thus $\hat{u}=\sin (tk)\hat{g}/k$. The convolution theorem gives the Kirchhoff formulaIt is smooth for Schwartz data, solves the transformed equation, and has $u(0,x)=0$, $u_t(0,x)=g(x)$ by the second expression and its small-time limit.

To prove uniqueness for arbitrary $C^2$ solutions, take their difference $v$, with zero initial data, and set $e=(v_t^2+|\nabla v{|}^2)/2$, $p=-v_t\nabla v$. Direct differentiation gives $e_t+\nabla \cdot p=v_t(v_{tt}-\Delta v)=0$. For any centre $x_{\ast }$ and radius $R$, integrate over the shrinking ball $B(x_{\ast },R-t)$. Differentiating the moving-domain energy givessince $v_t{\partial }_{n}v\le (v_t^2+|\nabla v{|}^2)/2$. Initial energy is zero and energy is nonnegative, so it stays zero. In each such cone $v_t$ and $\nabla v$ vanish, and the initial value is zero, hence $v=0$. Choosing $R>T$ and arbitrary centre proves uniqueness at any time $T$, without imposing finite total energy or decay on the competing solution.

For the equation with the smooth positive time-independent potential $V$, use insteadThen ${\partial }_t{e}_V+\nabla \cdot p=v_t(v_{tt}-\Delta v+Vv)=0$. The same shrinking-ball estimate is nonpositive because $V{v}^2\ge 0$. Zero initial data therefore give $v=0$. The potential-wave initial-value problem has at most one $C^2$ solution, with no global energy-growth assumption required. Time independence of $V$ is what makes this particular local energy identity exact.

Communication is reflexive because $(P^0{)}_{ii}=1$, and symmetric by its definition. If $i\leftrightarrow j$ and $j\leftrightarrow k$, choose positive-probability paths $i\to j$ in $m$ steps and $j\to k$ in $n$ steps. The Chapman-Kolmogorov equation givesCombining the reverse paths gives a positive-probability path $k\to i$ as well. Thus communication is transitive and is an equivalence relation.

To transfer recurrence, first establish the recurrence criterion by return probabilities. Put $r_i={\mathbb{P}}_i(T_i^{+}<\infty )$ and let $N_i={\sum }_{n\ge 0}{1}_{\{X_n=i\}}$ under ${\mathbb{P}}_i$. After each return the Strong Markov property restarts the chain at $i$, so ${\mathbb{P}}_i(N_i\ge k+1)=r_i^k$. ThereforeFor $r_i<1$ this equals $(1-r_i{)}^{-1}$; for $r_i=1$ it is infinite. The expectation identity follows by summing the nonnegative indicators, so remains valid when the expectation is infinite. Consequently $i$ is a recurrent state exactly when its return-probability series diverges.

If $i\leftrightarrow j$, choose $a,b$ with $(P^a{)}_{ji}>0$ and $(P^b{)}_{ij}>0$. For every $n\ge 0$,Summing over $n$, recurrence of $i$ forces the return-probability series for $j$ to diverge. The proved criterion shows $j$ is recurrent, so recurrence is a property of the communicating class.

The sequence has uniform convergence to $f$ on $(0,1)$ precisely whenThe same $N$ must work throughout the interval; equivalently ${\sup }_{0<x<1}|f_n(x)-f(x)|\to 0$.

A real-valued function $f$ on an interval $I$ is uniformly continuous whenThe essential point in uniform continuity is that $\delta$ is independent of the locations of $x$ and $y$. Uniform continuity on the whole real line does not imply boundedness: $f(x)=x$ is Lipschitz continuous with constant $1$, hence uniformly continuous, but is unbounded.

For a steady inviscid incompressible fluid of constant density in uniform gravity $\mathbf{g}=-g{\mathbf{e}}_z$, the Euler equation for fluid motion and continuity equation areTaking the scalar product with $\mathbf{u}$ givesThus Bernoulli's quantity is constant along each streamline. For an irrotational connected flow it is constant throughout the flow; constancy across different streamlines is not assumed in general.

Since incompressibility makes $\nabla \cdot (\rho \mathbf{u}\otimes \mathbf{u})=\rho (\mathbf{u}\cdot \nabla )\mathbf{u}$, integrating over a fixed control volume and applying the divergence theorem yields the integral momentum balanceHere $\mathbf{n}$ is outward and the left side is the net outgoing momentum flux. Uniform atmospheric pressure has zero resultant on a closed control surface, so it may be subtracted from all pressures in the jet calculations.

Use the Fourier transform conventionDecay alone does not guarantee existence of the ordinary integrals, so impose the usual integrability and regularity assumptions. One sufficient version of the Fourier inversion theorem is that $f$ and $\tilde{f}$ are integrable and $f$ is continuous, in which case the inversion integral recovers $f$. Under the standard piecewise-smooth Dirichlet assumptions, inversion instead gives the midpoint of the two one-sided limits at a jump, with a suitable improper integral interpretation.

For integrable $f,g$, define $(f\ast g)(x)={\int }_{\mathbb{R}}f(x-y)g(y)dy$. The convolution theorem statesTo prove it, absolute integrability permits Fubini's theorem. Set $t=x-y$ in the inner integral:The absolute double integral is $\Vert f{\Vert }_1\Vert g{\Vert }_1$, justifying the interchange. The computations below assume $a,b>0$ and use the source PDF's factor $e^{ikx}$ in the final integral, omitted in the converted TeX.

The product topology on $X\times Y$ has as a basis the rectangles $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$. Equivalently, it is the coarsest topology making both coordinate projections continuous.

To show that the product of two Hausdorff spaces is Hausdorff, take distinct points $(x_1,y_1)$ and $(x_2,y_2)$. If $x_1\ne x_2$, choose disjoint open neighborhoods $U_1,U_2$ in $X$; the rectangles $U_1\times Y$ and $U_2\times Y$ separate the product points. If the first coordinates agree, the second coordinates differ, and the same argument uses disjoint open neighborhoods in $Y$. Thus every pair of distinct product points has disjoint open neighborhoods.

The three implications below prove the requested equivalence without assuming metrizability.

Allow an eigenvector $v\ne 0$ to have complex entries, since real matrices can initially have complex eigenvalues. Write $v^{\ast }$ for its conjugate transpose. Reality of $A$ gives $A^{\ast }=A^T$; symmetry then gives $A^{\ast }=A$. Thus $A$ is a Hermitian matrix. The scalar $v^{\ast }Av$ equals its complex conjugate, because $(v^{\ast }Av{)}^{\ast }=v^{\ast }{A}^{\ast }v=v^{\ast }Av$. If $Av=\lambda v$, thensince $v^{\ast }v>0$. This proves that all eigenvalues are real. The equality $A^{\ast }=A^T$ is precisely where reality was used; a complex symmetric matrix need not be Hermitian, as $iI$ illustrates.

If $Au=\lambda u$ and $Av=\mu v$, then $u^{\ast }Av=\mu u^{\ast }v$, but also $u^{\ast }Av=(Au{)}^{\ast }v=\overline{\lambda }{u}^{\ast }v=\lambda u^{\ast }v$. Consequently $(\mu -\lambda )u^{\ast }v=0$. Distinct eigenvalues force $u^{\ast }v=0$. For real eigenvectors this is ordinary Euclidean orthogonality. These arguments establish real symmetric spectral orthogonality without assuming a diagonalization in advance.

A real orthogonal matrix satisfies $P^{T}P=I$, equivalently $P^{-1}=P^T$. If $A$ is symmetric,so orthogonal similarity preserves symmetry. An arbitrary real invertible change of basis need not do so. For example,The final matrix is not symmetric. Orthogonal similarity preserves real symmetry; general similarity does not.

A subgroup $H$ is a normal subgroup of $G$ if $gH{g}^{-1}=H$ for every $g\in G$, equivalently if $gH=Hg$ for every $g$. In the symmetric group $S_3$, the subgroup $A_3=\{e,(123),(132)\}$ is normal: conjugation relabels a three-cycle as a three-cycle, so preserves this set. By contrast, $\{e,(12)\}$ is not normal because conjugating $(12)$ by $(23)$ gives $(13)$ outside the subgroup.

If $H$ is normal, define a product on its left cosets by $(gH)(kH)=(gk)H$. To check that this quotient group operation is well-defined, take different representatives $g{h}_1$ and $k{h}_2$, with $h_1,h_2\in H$. Their product isNormality puts the last two factors in $H$, so its coset is exactly $gkH$. Associativity follows from that in $G$, the identity coset is $H$, and the inverse of $gH$ is $g^{-1}H$. Thus all the group axioms hold, and the projection $g\mapsto gH$ is a group homomorphism.

For a real-valued function on a bounded regular region, apply the divergence theorem to $\varphi \nabla \varphi$. The product rule yields $\nabla \cdot (\varphi \nabla \varphi )=|\nabla \varphi {|}^2+\varphi {\nabla }^2\varphi$. The prescribed zero boundary values and Laplace equation therefore implyThe continuous nonnegative integrand must vanish everywhere. Hence $\varphi$ is constant on each connected component, and its boundary value makes each constant zero. This proves zero-boundary harmonic uniqueness. If complex-valued functions are allowed, apply the same argument separately to their real and imaginary parts. If ${\psi }_1,{\psi }_2$ solve the same Poisson equation and Dirichlet boundary data, their difference is harmonic with zero boundary value, so is zero. Thus there is at most one solution. An exterior unbounded region would require an additional condition at infinity; the preceding bounded-domain proof does not assert uniqueness without one.

For the radial calculation use Cartesian coordinates with $r=(x_i{x}_i{)}^{1/2}>0$. The chain rule gives ${\partial }_{i}r=x_i/r$ andDifferentiate again and sum the three coordinates:This derives the radial Laplacian directly in Cartesian coordinates. For ${\nabla }^2\psi =c$, set $u=r\psi$, so $u^{\prime \prime }=cr$. Integrating twice gives $u=c{r}^3/6+ar+b$, and consequentlyIf the solution must extend boundedly through the origin, $b=0$. The remaining polynomial is smooth there.

For the Cauchy-Schwarz inequality, take $\lambda =(\mathbf{x}\cdot \mathbf{y})/(\mathbf{y}\cdot \mathbf{y})$ and use positivity of the inner product:Multiplication by $\mathbf{y}\cdot \mathbf{y}>0$ provesEquality holds exactly when $|\mathbf{x}-\lambda \mathbf{y}{|}^2=0$. Positive definiteness of the inner product makes this equivalent to $\mathbf{x}=\lambda \mathbf{y}$. Conversely, substitution of a scalar multiple gives equality. Thus equality holds precisely for linearly dependent nonzero vectors.

For a bounded function, a partition of an interval is $P:0=x_0<x_1<\cdots <x_N=1$. Put $m_j={\inf }_{[x_{j-1},x_j]}f$ and $M_j={\sup }_{[x_{j-1},x_j]}f$. Its lower Darboux sum and upper Darboux sum areThe lower Darboux integral is ${\sup }_{P}L(f,P)$ and the upper Darboux integral is ${\inf }_{P}U(f,P)$. Riemann integrability means that these two real numbers coincide, and their common value is the Riemann integral.

Equivalently, the Riemann integrability criterion says that for every $\epsilon >0$ some partition satisfies $U(f,P)-L(f,P)<\epsilon$. To see equivalence, all lower sums are at most all upper sums: refine any two partitions to their common refinement and use Darboux sum refinement monotonicity. Thus a small gap forces the difference of the two integrals below $\epsilon$. Conversely, if both integrals equal $I$, choose a lower sum above $I-\epsilon /2$ and an upper sum below $I+\epsilon /2$; their common refinement has gap below $\epsilon$.

This also agrees with the tagged Riemann sum definition: there is $I$ such that every tagged sum on a partition of sufficiently small mesh is within any prescribed tolerance of $I$. Each tagged sum lies between the two Darboux sums. From a partition with small gap, an arbitrary small-mesh partition differs from its refinement at only the cells crossing the finitely many original division points; boundedness makes their total contribution tend to zero. Conversely, on a fine partition tags approximating each cell supremum or infimum force both Darboux sums close to the same tagged-sum limit.

Now suppose $f$ is continuous on $[0,1]$. It is bounded by the result of Question 11. It is also uniformly continuous: otherwise there are $x_n,y_n\in [0,1]$ with $|x_n-y_n|\to 0$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0>0$. The Bolzano-Weierstrass theorem gives a convergent subsequence of $x_n$; the corresponding $y_n$ have the same limit, contradicting continuity. Given $\epsilon >0$, this uniform continuity gives $\delta >0$ such that points less than $\delta$ apart have function values less than $\epsilon /2$ apart. A partition of mesh below $\delta$ then satisfiesHence every continuous function on the interval is Riemann integrable, with the required proof supplied by its oscillation estimates.

The mean value theorem states that a real-valued function continuous on $[a,b]$ and differentiable on $(a,b)$, with $a<b$, has a point $\xi \in (a,b)$ such thatFor the proof, subtract the secant line and put $h(t)=f(t)-f(a)-(t-a)[f(b)-f(a)]/(b-a)$. Then $h(a)=h(b)=0$. If $h$ is identically zero, every interior point works. Otherwise it has a positive maximum or a negative minimum. The extreme value theorem supplies that extremum, and its nonzero value makes its location interior. At an interior differentiable extremum, the difference quotients on the two sides have opposite weak signs, so their common limit is zero. Hence $h^{\prime }(\xi )=0$, proving the formula. This also gives the Rolle theorem argument rather than merely citing it.