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Work over the complex numbers. By Maschke's theorem, restriction to the normal subgroup $A$ is a direct sum of its isotypic components. For an irreducible $A$-type $W$, denote its isotypic component in $V$ by $V_W$. Normality makes $G$ permute these components, by conjugating the $A$-types. Any union of orbits spans a $G$-invariant subspace, so irreducibility of $V$ means that all occurring components form a single orbit.

If there is only one component, restriction is isotypic, as claimed. Otherwise take one component $V_W$ and its stabilizer $H=\{g:g{V}_W=V_W\}$, a proper subgroup. This is an irreducible representation of $H$: if $0\ne U⊊V_W$ were $H$-invariant, the direct sum of its translates over coset representatives for $G/H$ would be a nonzero proper $G$-invariant subspace of $V$. It is direct because the translates lie in distinct isotypic components, and proper because $\dim U<\dim V_W$.

The natural map from the induced representation sends $g\otimes v$ to $gv$. It identifies the direct-sum coset decomposition with all isotypic components of $V$, so is bijective and $G$-equivariant. ConsequentlyThis proves the dichotomy directly, including the irreducibility of the representation being induced.

For odd prime $n$, every subgroup is one of $\{1\}$, $⟨\sigma ⟩$, a reflection subgroup $H_j=⟨{\sigma }^j\tau ⟩$ for $0\le j<n$, or $G$. Indeed, a subgroup containing a nonidentity rotation contains all rotations; with any reflection it is then $G$. A subgroup with no nonidentity rotation contains at most one reflection, since the product of distinct reflections is a nonidentity rotation.

The Galois correspondence gives the following fixed fields. The trivial subgroup corresponds to $L=\mathbb{C}(z)$, the rotation subgroup to $\mathbb{C}(z^n)$, and $G$ to the base $\mathbb{C}(w)$. Since composition of the stated automorphisms gives $({\sigma }^j\tau )(z)={\zeta }_n^{-j}/z$, the reflection subgroup corresponds toThis rational invariant has degree two, hence generates the fixed field of its order-two subgroup, by the same degree argument as part (i). These fields have degrees respectively $2n,2,n,1$ over $\mathbb{C}(w)$.

An intermediate field is Galois over the base exactly when its corresponding subgroup is normal in $G$. The trivial subgroup, the rotations and $G$ are normal; no reflection subgroup is normal because $n$ is odd and greater than two. Hence the Galois intermediate extensions and groups areThe $n$ reflection fixed fields are not Galois over the base.

The automorphisms satisfy ${\sigma }^n={\tau }^2=1$ and $\tau \sigma \tau ={\sigma }^{-1}$. The $n$ rotations $z\mapsto {\zeta }_n^{j}z$ and $n$ inversion maps $z\mapsto {\zeta }_n^j/z$ are all distinct, so the dihedral group $G$ has order $2n$, including the case $n=1$ when it has order two. The invariantis fixed by both generators.

Use two standard field-theoretic results: the fixed-field theorem for a finite group of automorphisms gives $[L:L^G]=|G|$ and makes $L/L^G$ Galois with group $G$; and a nonconstant rational function $P(z)/Q(z)$ in lowest terms satisfies $[\mathbb{C}(z):\mathbb{C}(P/Q)]=\max (\mathrm{deg}P,\mathrm{deg}Q)$. Here $w=(z^{2n}+1)/z^n$ has coprime numerator and denominator and degree $2n$. Thus $[L:\mathbb{C}(w)]=[L:L^G]=2n$. Since $\mathbb{C}(w)\subseteq L^G$, the tower law proves the dihedral fixed field in one rational variable:

For a regular periodic energy orbit, differentiate the action integral with respect to its energy. On a branch $H(p,q)=E$, implicit differentiation gives $\partial p/\partial E=1/H_p=1/\dot{q}$. ThusThe integral follows the entire orbit, with its orientation; at turning points this formula is interpreted by the limiting integral from each branch. It is valid for a smooth family of nonseparatrix periodic orbits.

An adiabatic invariant means that the action variable remains constant to leading order when the external parameter changes on a timescale much longer than the orbital period. Denote that parameter by $\lambda$. With $H(0,\lambda )=0$, the displayed identity integrates toIf the parameter change decreases $T(J,\lambda )$ for every intervening action $J$, this integrand increases pointwise. At the adiabatically fixed positive $I$, the energy increases. This uses the phrase “with any $I$” as a statement about the family of orbits, not merely a period decrease at a single selected action; that weaker premise alone would not determine the integral.

With Hamilton's equations ${\dot{q}}_j=H_{p_j}$ and ${\dot{p}}_j=-H_{q_j}$, the chain rule givesFor the vortex problem, combine the two ordered terms for each unordered pair, writing $D_{ij}=(p_i-p_j{)}^2+(q_i-q_j{)}^2$ and $H=-(\kappa /2){\sum }_{i<j}\log D_{ij}$. Since $F_{q_j}=2q_j$, $F_{p_j}=2p_j$, its pairwise contribution to the Poisson bracket isEvery pair cancels, so $[F,H]=0$ and ${\sum }_j(q_j^2+p_j^2)$ is conserved along any collision-free solution. The logarithmic Hamiltonian is undefined at coinciding vortices, so the derivation applies on its natural domain.

For $h\lambda /2<1$, the implicit step map $z\mapsto y_n+(h/2)[f(t_n,y_n)+f(t_{n+1},z)]$ is a contraction, so the Banach fixed-point theorem supplies a unique next value. Smoothness of the exact solution gives the one-step defectuniformly on $[0,T]$, by the trapezoidal rule error estimate or a direct Taylor expansion. Let $e_n=y_n-y(t_n)$. Subtracting exact and numerical steps and applying the Lipschitz condition givesPut $r_h=(1+h\lambda /2)/(1-h\lambda /2)$. With $e_0=0$, iteration yields $|e_n|\le [C{h}^3/(1-h\lambda /2)](r_h^n-1)/(r_h-1)$. Since $r_h-1=h\lambda /(1-h\lambda /2)$, this is $C{h}^2(r_h^n-1)/\lambda$. For sufficiently small $h$, $\log r_h\le 2\lambda h$, so $r_h^n\le e^{2\lambda T}$ whenever $nh\le T$. ThereforeThis proves convergence with global order two directly from the defect bound and stability recurrence.

Convergence of a numerical method means that, for every fixed final time $T$ and admissible initial-value problem, the maximum grid error tends to zero:

For a simple stationary point at the endpoint $b$, only one half of the local quadratic neighborhood lies in the integration interval. The even Gaussian integrand therefore gives half the interior stationary phase contribution:

Use disjoint small neighborhoods of the finitely many nondegenerate stationary points $t_j$, apply the stationary phase formula in each, and integration by parts elsewhere. The leading contributions add:Interference between these oscillatory contributions can cancel the leading coefficient at particular values of $\lambda$.

Near the nondegenerate stationary point, write $g(t)=g(t_0)+g^{\prime \prime }(t_0)(t-t_0{)}^2/2+O((t-t_0{)}^3)$ and replace $f(t)$ by $f(t_0)$ to leading order. Scaling $t-t_0$ by ${\lambda }^{-1/2}$ reduces the local integral to the Fresnel integral, while the remaining nonstationary region contributes $O({\lambda }^{-1})$. The stationary phase result is

Fix $ϵ>0$. For each $N$, the set $Q_N={\bigcap }_{n,m\ge N}\{x:|f_n(x)-f_m(x)|\le ϵ\}$ is closed, because each $f_n-f_m$ is continuous. At every $x$, pointwise convergence makes $(f_n(x))$ a Cauchy sequence, so $\mathbb{R}={\bigcup }_{N\ge 1}{Q}_N$. The Baire category theorem supplies an $N_0$ for which $Q_{N_0}$ contains a nonempty open interval $I$. For $x\in I$ and $n\ge N_0$, pass $m\to \infty$ in $|f_n(x)-f_m(x)|\le ϵ$ to obtainThis local uniform Cauchy control for pointwise convergent continuous functions allows both $I$ and $N_0$ to depend on $ϵ$.

The closed-set version of the Baire category theorem says that if a nonempty complete metric space is a countable union of closed sets, at least one of those closed sets has nonempty interior. Equivalently, a nonempty complete metric space cannot be a countable union of closed nowhere dense sets.

Differentiate under the integral and use the evolution equation:Thus $\boxed{I_f[u(t)]\le I_f[u(s)](t>s>0)}$. This is a gradient flow of the periodic screened Poisson energy. Part (iii) gives convergence in the function and its first derivative, so $\boxed{I_f[u(t)]\to I_f[u_f]}$, the minimum from part (ii).

The Fourier coefficients obey ${\dot{u}}_n+(1+n^2)u_n=f_n$, soThe coefficient decay of $u_0,f$ and the additional exponential damping justify all derivatives for $t>0$ and recovery of the initial data. Thus this is a smooth periodic solution. Every transient mode decays, giving $\boxed{u(t,\cdot )\to u_f}$ uniformly, indeed with all spatial derivatives. The elliptic solution in part (i) is the attracting steady state of this forced heat equation.

Expand the quadratic functional and use periodic integration by parts:The first integral is zero because $-u_f^{\prime \prime }+u_f=f$. Thus the difference is $\frac{1}{2}\int ({\phi }^{\prime 2}+{\phi }^2)>0$ for every nonzero $\phi$. $u_f$ is the unique strict minimizer.

Write the Fourier series $f(x)={\sum }_{n\in \mathbb{Z}}{f}_n{e}^{inx}$, with $f_n=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }f(x)e^{-inx}dx$. Smoothness makes its coefficients decay faster than every inverse power. Settingtherefore gives a smooth periodic function, and termwise differentiation proves $-u_f^{\prime \prime }+u_f=f$. Conversely every solution must have these coefficients, proving uniqueness. Equivalently, a homogeneous solution has $\int (|u^{\prime }{|}^2+|u{|}^2)=0$ by integration by parts, so vanishes.

For this alternating arrival and service queue, set $\theta =\lambda (\mu +\beta )/[\mu (\lambda +\alpha )]$. Direct multiplication gives $(\theta ,\lambda /\mu )B=\theta (\theta ,\lambda /\mu )$, so the initial row is a left eigenvector and $({\pi }_{iC},{\pi }_{iW})={\theta }^{i-1}({\pi }_{1C},{\pi }_{1W})$. ConsequentlyThese nonnegative probabilities can be normalized exactly when $\theta <1$, equivalently $\mu \alpha >\lambda \beta$. Summing the geometric series givesWhen $\theta \ge 1$ no positive multiple of the invariant sequence is summable, so no stationary distribution exists. This rules out positive recurrence; absence of a stationary probability alone should not be used to distinguish null recurrence from transience.

The stationary balance equations at level zero are $(\lambda +\alpha ){\pi }_{0C}=\beta {\pi }_{0W}$ and $\beta {\pi }_{0W}=\alpha {\pi }_{0C}+\mu {\pi }_{1W}$. At every $i\ge 1$ they areThe zero-level equations give ${\pi }_{0C}=\beta {\pi }_{0W}/(\lambda +\alpha )$ and ${\pi }_{1W}=\beta \lambda {\pi }_{0W}/[\mu (\lambda +\alpha )]$. Substituting the latter in the level-one $C$ equation givesSolve the $W$ equation for ${\pi }_{i+1,W}$, then substitute it into the next $C$ equation. This yields the required row-vector recurrence with

Uniform integrability means ${\sup }_n\mathbb{E}[|X_n|1_{\{|X_n|>K\}}]\to 0$ as $K\to \infty$. In particular the first absolute moments are uniformly bounded. By the Portmanteau theorem, $\mathbb{E}|X|\le {\mathrm{liminf}}_n\mathbb{E}|X_n|<\infty$. Let $T_K(x)=\max (-K,\min (x,K))$, a bounded continuous truncation. Convergence in distribution gives $\mathbb{E}{T}_K(X_n)\to \mathbb{E}{T}_K(X)$. ThereforeThe first term tends to zero by uniform integrability, the second by integrability of $X$. Hence $\boxed{\mathbb{E}{X}_n\to \mathbb{E}X}$.

Convergence in probability means $\mathbb{P}(|X_n-X|>ϵ)\to 0$ for every $ϵ>0$. Convergence in distribution means $F_{X_n}(t)\to F_X(t)$ at every continuity point $t$ of $F_X$. The inclusions obtained by separating the event $|X_n-X|>ϵ$ giveTake lower and upper limits as $n\to \infty$, then $ϵ\downarrow 0$ at a continuity point. Both bounds tend to $F_X(t)$, proving that convergence in probability implies convergence in distribution.