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Fourier transform on the infinite spatial lattice giveswith polynomial roots $G_{\pm }=iA\pm \sqrt{1-A^2}$. When $|A|$ is uniformly less than one, the polynomial roots have modulus one and a uniform positive separation. Diagonalizing the companion matrix then bounds all its powers independently of $\theta$ and $n$. Parseval's identity proves stability in the discrete $L^2$ norm for arbitrary square-summable starting perturbations at both levels.

At $\mu =1/2$ the coefficients giveConsequently $\mu =1/2$ is stable.

At $\mu =3/2$,The growing polynomial root at this phase has modulusTo turn this into a Cauchy problem linear instability proof, rather than rely on a plane wave of infinite norm, choose a small interval around $\pi /3$ on which the growing polynomial root has modulus at least $1+\delta$. Choose an initial Fourier transform amplitude that is a nonzero square-integrable function supported there, and set the level-one amplitude equal to that polynomial root times the level-zero amplitude. The recurrence then multiplies by the same growing polynomial root at every step on the support. Parseval identity gives $\Vert u^n{\Vert }_2\ge (1+\delta {)}^n\Vert u^0{\Vert }_2$, whereas both initial norms are bounded independently of $h$. Real data can be obtained by adding the conjugate interval around $-\pi /3$. Since $n=T/(\mu h)$ tends to infinity at fixed $T$, $\mu =3/2$ is unstable.

Statistical homogeneity makes the two-point correlation depend on the displacement $\mathbf{u}=\mathbf{r}-{\mathbf{r}}^{\prime }$. Substituting the Fourier transforms and changing variables from $(\mathbf{r},{\mathbf{r}}^{\prime })$ to $(\mathbf{u},{\mathbf{r}}^{\prime })$ givesThe last integral is a Dirac delta function, soThe plus sign occurs because neither Fourier transform has undergone complex conjugation. For a real magnetic field, the version with ${\hat{B}}_j({\mathbf{k}}^{\prime }{)}^{\ast }$ instead has $\delta (\mathbf{k}-{\mathbf{k}}^{\prime })$. Homogeneity determines dependence on the displacement vector; dependence only on its norm needs statistical isotropy as well.

A strongly continuous semigroup on the complex Banach space $E$ is a family $T_t\in \mathcal{L}(E)$ with $T_0=I$, $T_{s+t}=T_s{T}_t$ and $\Vert T_{t}f-f\Vert \to 0$ as $t\downarrow 0$ for every $f\in E$. Its semigroup generator is the linear operatorFor $f\in D(Z)$, boundedness of $T_t$ and the semigroup identity giveThus semigroup commutation with its unbounded generator meansThere is no assertion that $Z$ is defined on every vector. The orbit is differentiable on this domain, with $\frac{d}{dt}{T}_{t}f=T_{t}Zf$.

A contraction semigroup additionally satisfies $\Vert T_t\Vert \le 1$ for every $t\ge 0$. For $\mathrm{Re}\lambda >0$, define the Bochner integralStrong continuity gives a continuous integrand, and its norm is bounded by $e^{-t\mathrm{Re}\lambda }\Vert f\Vert$, so the integral exists andChanging variables in $T_h{L}_{\lambda }f$ yieldsAfter subtracting $L_{\lambda }f$ and dividing by $h$, the limit is $\lambda L_{\lambda }f-f$. ConsequentlyFor $f\in D(Z)$, integrate the derivative of $e^{-\lambda t}{T}_{t}f$ over $[0,\infty )$. Its limit at infinity is zero by the contraction bound, and its value at zero is $f$. ThereforeThus this is also the Laplace-transform formula for a semigroup resolvent, with $L_{\lambda }=(\lambda I-Z{)}^{-1}$.

To prove that $Z$ is a closed operator, suppose $f_n\in D(Z)$, $f_n\to f$ and $Z{f}_n\to g$. Fix one $\lambda$ with positive real part. The identity just proved and boundedness of $L_{\lambda }$ giveIts range lies in $D(Z)$, and applying $\lambda I-Z$ shows $Zf=g$. Thus the graph is closed. Linearity follows directly from linearity of the defining difference quotient, so $Z$ is a closed linear operator.

The closedness conclusion also holds without the contraction assumption. For a general strongly continuous semigroup, the Uniform boundedness principle gives a uniform operator bound on each compact time interval. Integrating the generator-orbit identity gives $T_t{f}_n-f_n={\int }_0^t{T}_{s}Z{f}_{n}ds$. Under $f_n\to f$ and $Z{f}_n\to g$, pass to the limit to obtain $T_{t}f-f={\int }_0^t{T}_{s}gds$. Divide by $t$ and let $t\downarrow 0$; strong continuity makes the right side converge to $g$, so $f\in D(Z)$ and $Zf=g$.

Use the joint energy conventionOn the generator domain it is $-⟨Lf,g⟩$. The symmetric Feller semigroup has symmetric two-point measure ${\nu }_s(dx,dy)=\mu (dx)P_s(x,dy)$, with both marginals equal to $\mu$. Expanding the product of increments gives the finite-time identityfor real functions. This supplies a proof that works for a general symmetric Markov kernel, without assuming a diffusion chain rule.

For $a\ge b\ge 0$, Cauchy-Schwarz inequality givesInterchanging $a,b$ gives the same inequality in the other order. Apply it with $a=f(x),b=f(y)$, integrate against ${\nu }_s$, and take $s\downarrow 0$. This proves the power inequality for symmetric Markov energiesInitially take bounded nonnegative functions in the form domain. On their bounded range, the power maps are Lipschitz; their finite-time squared-increment energies are therefore bounded by a constant times that of $f$. This puts the powers in the form domain and makes both joint energy limits well defined. Truncation gives the corresponding extended-energy interpretation when needed. In particular the proof applies to the bounded positive functions used below.

Normalize the logarithmic Sobolev inequality asStart with a bounded $f$ bounded away from zero, and put $u(t)=P_{t}f$, $q=q(t)$ and $F(t)=\int u^{q}d\mu$. Positivity and preservation of constants keep $u$ within the same positive bounds. Symmetry realizes $P_t$ as a self-adjoint contraction semigroup on $L^2(\mu )$, so for $t>0$ its orbit lies in the generator domain and ${\partial }_{t}u=Lu$; this justifies the following differentiation away from zero.

Differentiate both the evolving function and exponent:Since $\log \Vert u{\Vert }_q=q^{-1}\log F$, it follows thatApply the logarithmic Sobolev inequality to $g=u^{q/2}$, and then the joint energy inequality:ThereforeThe specified exponent satisfies $q(0)=2$ and $q^{\prime }=4(q-1)/c_{\mathrm{LS}}$. Thus the right side is zero and the evolving norm is nonincreasing. Integrating from a positive time $\delta$ and letting $\delta \downarrow 0$ givesHere the initial limit follows from strong $L^2$ continuity, boundedness of $f$ and $P_{\delta }f$, and $q(\delta )\to 2$.

For a bounded nonnegative $f$, apply the estimate to $f+ϵ$ and let $ϵ\downarrow 0$. For arbitrary $f\in L^2(\mu )$, truncate $|f|$ to bounded nonnegative functions. Their images converge in $L^2$ to $P_t|f|$; an almost-everywhere convergent subsequence and Fatou lemma give the same $L^{q(t)}$ estimate for that limit. Positivity of the Markov kernel gives $|P_{t}f|\le P_t|f|$, also for complex $f$, soThis proves that the logarithmic Sobolev inequality implies hypercontractivity, with the numerical exponent matched to the stated joint-energy normalization.

Write $d{\gamma }_1(x)=(2\pi {)}^{-1/2}{e}^{-x^2/2}dx$. On the dense polynomial algebra in $L^2({\gamma }_1)$, the Gaussian creation and annihilation operators areGaussian integration by parts gives $⟨a^{-}f,g⟩=⟨f,a^{+}g⟩$. Their closed realizations are adjoints. More explicitly, for the orthonormal basis $h_n={\mathrm{He}}_n/\sqrt{n!}$ of normalized Probabilists' Hermite polynomials,Thus the closed annihilation operator has domain $\{\sum c_n{h}_n:\sum n|c_n{|}^2<\infty \}$, and the closed creation operator has domain $\{\sum c_n{h}_n:\sum (n+1)|c_n{|}^2<\infty \}$. These describe the same set of $L^2$ functions, the Gaussian Sobolev space, though the two operators differ. Directly on polynomials, $[a^{-},a^{+}]=I$, and

Use the standard carré du champ operator convention $\Gamma (f,g)=\frac{1}{2}(L(fg)-fLg-gLf)$. The product rule for this semigroup generator givesso the requested squared-gradient and joint energy formulas, for real functions in the standard algebra, areFor complex functions the joint energy is sesquilinear, with $\overline{g^{\prime }}$ in the second factor. If “squared gradient” is defined without the factor $1/2$ in the product-rule expression, its value is instead $2f^{\prime }{g}^{\prime }$ and the joint energy is one half of its integral; the joint energy and logarithmic Sobolev constant used here remain as displayed.

For the Gaussian logarithmic Sobolev inequality, first take smooth bounded $f$ with bounded first and second derivatives, bounded away from zero. The Mehler formula for the Ornstein-Uhlenbeck semigroup isIt preserves ${\gamma }_1$ and differentiating in $x$ gives the Ornstein-Uhlenbeck gradient commutation identityPut $u_t=P_{t}f$. Differentiate its entropy functional. Since $\int u_{t}d{\gamma }_1=\int fd{\gamma }_1$ is constant, and Gaussian integration by parts applies,For such data, Mehler's formula and dominated convergence give $P_{t}f(x)\to \int fd{\gamma }_1$ as $t\to \infty$. The uniform positive upper and lower bounds justify integrating this limit inside the entropy, whose limit is zero. Hence

Apply the allowed weighted Cauchy-Schwarz inequality with $g=f^{\prime }$:After integration, invariance of ${\gamma }_1$ givesIntegrating in $t$ yieldsFor a real smooth compactly supported $g$, take $f=g^2+ϵ$. ThenLetting $ϵ\downarrow 0$ provesSmooth approximation and cutoff in the Gaussian Sobolev norm extend this to the full form domain. The entropy is lower semicontinuous under the resulting $L^2$ convergence: the square densities converge in $L^1$, their masses converge, and, along an almost-everywhere convergent subsequence, Fatou's lemma applies to $h\log h$ after adding the bound $1/e$. The same inequality for complex $g$ follows by applying it to $|g|$ and using the weak-derivative bound $|(|g|{)}^{\prime }|\le |g^{\prime }|$. Thus the constant is at most $2$.

For sharpness of the Gaussian logarithmic Sobolev constant, take $g_s(x)=e^{sx/2}$ with real $s\ne 0$. These functions belong to the Gaussian Sobolev space. The Gaussian moment-generating function givesThereforeThe entropy-to-energy ratio is exactly $2$, so no smaller constant can work. Combining this lower bound with the proved inequality gives

Write $g=(a,b)$ for the affine map $z\mapsto az+b$. Conjugation by $(a,0)$ sends translation by $b$ to translation by $ab$, so it permutes transitively the nontrivial characters of $U$. Since $U$ is normal, the induced character is zero outside $U$. On a translation by $b$ it is the sum of all conjugates of $\psi$, henceFor $b\ne 0$ the sum of the nontrivial additive characters is minus one, since the sum including the trivial character is zero. Its character norm isBy the irreducibility criterion for characters, $\rho ={\mathrm{Ind}}_U^G\psi$ is irreducible of dimension $p-1$. All choices of nontrivial $\psi$ give the same representation.

Two norms are Lipschitz equivalent norms if constants $c,C>0$ satisfy $c\Vert u{\Vert }_1\le \Vert u{\Vert }_2\le C\Vert u{\Vert }_1$ for every $u$. On the vector space of finitely supported sequences, the norms $\Vert x{\Vert }_1=\sum |x_i|$ and $\Vert x{\Vert }_{\infty }=\max |x_i|$ are not equivalent: a vector with $N$ entries equal to one has the two norms $N$ and one.

For dimension $d<\infty$, choose a basis and write $u=\sum x_i{e}_i$. Any norm satisfies $\Vert u\Vert \le \sum |x_i|\Vert e_i\Vert \le C|x{|}_2$. This also implies its continuity in the Euclidean coordinates, by $|\Vert u\Vert -\Vert v\Vert |\le \Vert u-v\Vert \le C|x-y{|}_2$. Its positive continuous values on the compact Euclidean unit sphere have a positive minimum $c$. Homogeneity gives $c|x{|}_2\le \Vert u\Vert$, proving equivalence with the Euclidean norm and hence between any two norms.

If a finite-dimensional subspace $U$ of a normed space contains a sequence converging in the ambient norm, its coefficient vectors are Cauchy by this equivalence. The coefficients converge in ${\mathbb{R}}^d$ or ${\mathbb{C}}^d$, and the reconstructed vector lies in $U$ and is the ambient limit. Thus finite-dimensional subspaces are closed.

A finite-dimensional normed space has nonempty open balls with compact closures by equivalence and Heine-Borel. Conversely let an open set $O$ have compact closure. Choose $w\in O$ and $r>0$ such that $B(w,r)\subset O$. Its closed ball of radius $r/2$ is a closed subset of the compact closure of $O$ and is therefore compact. Translation and scaling make the closed unit ball compact.

If the space were infinite-dimensional, construct unit vectors $x_j$ with $\mathrm{dist}(x_j,\mathrm{span}(x_1,\dots ,x_{j-1}))>1/2$. Here is the needed Riesz lemma: for a proper closed subspace $M$, choose $z\notin M$, put $d=\mathrm{dist}(z,M)>0$, and choose $m\in M$ with $\Vert z-m\Vert <2d$; then $(z-m)/\Vert z-m\Vert$ has distance greater than $1/2$ from $M$. The finite spans are closed by the earlier argument, so this constructs the sequence. Its members are separated by more than $1/2$, and have no convergent subsequence, contradicting compactness. Therefore the space is finite-dimensional exactly when it has a nonempty open set with compact closure.

The conjugacy classes are specified by cycle type, with sizes $1,6,3,8,6$ for $1,(12),(12)(34),(123),(1234)$. The trivial and sign representations give the first two rows. The permutation representation on four letters contains a trivial line; its orthogonal complement has character equal to the number of fixed letters minus one. Tensoring that representation by the sign gives another row. Finally the quotient $S_4/V_4\cong S_3$ supplies the two-dimensional standard representation. The resulting table isWith the class-size-weighted inner product, these characters have norm one and are pairwise orthogonal, so all are irreducible and inequivalent. Their squared dimensions sum to $1+1+9+9+4=24=|S_4|$, proving completeness.

The classes of $A_4$ are the identity, the three double transpositions, and two classes of four three-cycles. Its quotient by $V_4$ is cyclic of order three, giving characters $1,\chi ,{\chi }^2$, with $\chi$ taking values $1,1,\omega ,{\omega }^2$ on these classes. The restriction of $V$ has character $(3,-1,0,0)$ and norm $(9+3)/12=1$, so it is the three-dimensional irreducible representation $T$. These four representations are complete since $1+1+1+9=12$.

The sign is trivial on $A_4$, soFor the last equality the restricted character is $(2,2,-1,-1)$, the sum of the two indicated one-dimensional characters. Equality of characters proves the decomposition.

A subset is nowhere dense when its closure has empty interior. It is of first category, or meagre, when it is a countable union of nowhere-dense subsets; otherwise it is of second category.

The Baire category theorem says that in a complete metric space every countable intersection of open dense subsets is dense. To prove it, start inside an arbitrary nonempty open set and choose a closed ball of positive radius there and in the first dense open set. Inductively choose a closed ball inside the interior of the preceding ball and the next dense open set, with radius less than $2^{-n}$. The centres are Cauchy. Completeness gives a limit lying in all the nested closed balls, hence in all the dense open sets and the original open set. Thus a nonempty complete metric space is not meagre in itself.

For $p<r\le \infty$, put $E_m=\{x\in {\ell }^p:\Vert x{\Vert }_p\le m\}$, viewed inside ${\ell }^r$. Necessarily $p<\infty$. Each $E_m$ is closed: convergence in the ${\ell }^r$ norm gives coordinatewise convergence, and every finite partial sum of $\sum |x_i{|}^p$ is bounded by $m^p$ in the limit. Taking the supremum of partial sums proves the assertion.

It has empty interior. Given a point in it and any small ${\ell }^r$ radius, perturb a block of $N$ coordinates by entries of modulus $\delta N^{-1/r}$, choosing signs to avoid cancelling the existing coordinates. The ${\ell }^r$ distance is $\delta$, while the resulting ${\ell }^p$ norm is at least $\delta N^{1/p-1/r}$, exceeding $m$ for large $N$. For $r=\infty$, use entries of modulus $\delta$ on that block. Hence each $E_m$ is nowhere dense, andFor $r=p$, including $p=\infty$, the space is Banach and so second category in itself by Baire. There is no case $r>p$ when $p=\infty$.

Assume the consistency regularity implicit in the requested rate, for example $u\in C^4$ on the closed square. Bounded fourth derivatives and Taylor's theorem give a uniform residual ${\tau }_{ij}={\Delta }_{h}u(ih,jh)-f(ih,jh)$ with $|{\tau }_{ij}|\le C_0{h}^2$. Subtracting the exact grid values from the numerical equations gives ${\Delta }_{h}e=-\tau$, with zero boundary error.

Let $A=-h^2{\Delta }_h$ be the unscaled positive Dirichlet matrix. Its eigenvectors are products of discrete sine vectors, which form an orthogonal basis; direct substitution gives eigenvaluesSince $\sin (\pi h/2)\ge h$ for $0<h\le 1$, ${\lambda }_{11}\ge 8h^2$. Therefore $\Vert A^{-1}{\Vert }_2\le 1/(8h^2)$. The error equation is $Ae=h^2\tau$, and its residual has norm $\Vert \tau {\Vert }_2\le M{C}_0{h}^2\le C_{0}h$. Combining the bounds yieldsThis is the unweighted grid error for the five-point Poisson formula. The unweighted norm loses one power of $h$ because the grid has order $h^{-2}$ points; the area-weighted norm is second-order accurate. A regularity assumption is needed for the uniform truncation estimate and hence for this rate; the question does not specify it explicitly.