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Write $x,y$ for the two coordinates. The PDF gives $\mu =e^{\pi i/3}$, a primitive sixth root of unity. The matrices satisfyThese relations reduce every word to ${\sigma }^j$ or $\tau {\sigma }^j$, $0\le j<6$. The six first matrices are diagonal and distinct, and the six second matrices are antidiagonal and distinct. Thus $G$ has exactly twelve elements; it is a binary dihedral group.

First calculate the polynomial invariant ring of $⟨\sigma ⟩$. A monomial $x^a{y}^b$ is invariant exactly when $a-b$ is divisible by $6$. Removing the smaller exponent shows that it is a product of powers ofThe sole relation is $AB=C^6$. Indeed, reduction by this relation leaves monomials $C^j{A}^a$ and $C^j{B}^b$ with $b>0$, whose images in $\mathbb{C}[x,y]$ have distinct exponent pairs. HenceOn these generators, $\tau$ interchanges $A,B$ and sends $C$ to $-C$. Put $S=A+B$ and $T=A-B$. Since $2$ is invertible, the preceding ring isEvery element has a unique form $P(S,C)+TQ(S,C)$. The induced involution fixes $S$ and negates both $C$ and $T$. Its invariants are therefore exactly the expressionsSetWe obtain the relationThe unique normal form above also proves that there are no further relations: $P_0(U,V)+W{Q}_0(U,V)=0$ forces both polynomials to vanish. Thus the polynomial invariant ring is

For completeness, the algebraic quotient by a finite group is the affine variety with this coordinate ring. Each element $h$ of $\mathbb{C}[x,y]$ satisfies the monic orbit polynomial ${\prod }_{g\in G}(Z-gh)$ with invariant coefficients, so the quotient map is a finite morphism. Invariants separate distinct finite orbits: interpolate a polynomial taking value $0$ on one orbit and $1$ on the other, and average it over $G$. Thus its fibres are precisely the orbits. The defining polynomial and resulting algebraic quotient by a finite group areUsing the converted TeX's fourth root would instead give a different group and a different binary dihedral invariant hypersurface; the sixth root from the original PDF is essential.

Suppose there were distinct singular points of an algebraic variety $P,Q$. Let $L$ be their joining projective line. The homogeneous restriction $F{|}_L$ has degree $3$. At a singular point of an algebraic variety of a plane algebraic curve, the local defining equation has no constant or linear term, so its restriction to any line through that point vanishes to order at least $2$, unless the restriction is identically zero. Therefore a nonzero $F{|}_L$ would have at least four zeros counted with multiplicity, two at $P$ and two at $Q$, which a degree-$3$ homogeneous polynomial on ${\mathbb{P}}^1$ cannot have.

Hence $F{|}_L=0$. After choosing coordinates with $L=V(z)$, this says that every monomial of $F$ contains $z$, so $z$ divides $F$. That contradicts the irreducible polynomial hypothesis on the cubic. ThusThis elementary version of the Bézout theorem argument works in every characteristic, without assuming that the partial derivatives of the cubic have their characteristic-zero form.

Choose real monic orthogonal polynomials $p_j$ of degree $j$, $j=0,1,\dots$, for $w(x)=x^a{e}^{-x}$ on $[0,\infty )$, and letExtend each ${\varphi }_j$ by zero to negative $x$. The resulting functions form an orthonormal set in $L^2(\mathbb{R},dx)$. Define the orthogonal polynomial projection kernelThe formula on the right is for nonnegative arguments; the kernel is zero when either argument is negative. With ${\Phi }_{ij}={\varphi }_{j-1}(x_i)$, the kernel evaluation matrix is $\Phi {\Phi }^T$. By the determinant multiplication identity and the preceding monic basis change,Thus initially ${\hat{c}}_m=c_m{\prod }_j{h}_j$.

Use the symmetric labelled joint probability density on the full orthant. Expanding the two copies of $\det \Phi$ into permutations and integrating all variables, orthonormality makes a term vanish unless the two permutations agree. Each of the $m!$ surviving terms integrates to $1$. HenceIf the eigenvalues are instead listed in increasing order on a single ordered chamber, that chamber's density is $m!$ times this symmetric density. We keep the full-orthant convention throughout.

An explicit choice is $p_j(x)=(-1{)}^{j}j!L_j^{(a)}(x)$ in terms of the Generalized Laguerre polynomials. To verify the normalization independently, their Rodrigues' formula givesIntegrating by parts $j$ times proves orthogonality to every lower-degree polynomial. The boundary terms vanish at zero and infinity for $a\ge 0$. Taking the other factor to be $p_j$ and using $p_j^{(j)}=j!$ gives $h_j=j!\Gamma (a+j+1)$, where $\Gamma$ is the Gamma function. The Laguerre projection kernel is thereforefor $x,y\ge 0$.

Let $V_m=\det [x_i^{j-1}{]}_{i,j=1}^m$. It is an alternating polynomial: exchanging two variables exchanges two rows of the Vandermonde matrix and reverses its determinant. Hence $V_m$ vanishes whenever $x_k=x_l$, so each factor $x_l-x_k$ divides it. The distinct linear factors are pairwise coprime polynomials, and their product therefore divides $V_m$. Both have total degree $0+1+\cdots +(m-1)=m(m-1)/2$, soThe coefficient of $x_m^{m-1}$ in the determinant is $V_{m-1}$, by expansion along its last row. The same coefficient in the product is ${\prod }_{k<l<m}(x_l-x_k)$. Thus $C_m=C_{m-1}$, and $C_1=1$. Consequently the Vandermonde determinant identity isThis is a polynomial identity, including repeated coordinates; no division by a possibly zero numerical Vandermonde product is needed.

On $T=V^{\otimes n}$, define the permutation and diagonal actions byThe inverse in the first formula gives a left action. Applying $g$ to every tensor factor and then reordering produces the same tensor as reordering and then applying $g$. Thus the two actions commute. The Schur algebra isIt is the commutant of an operator algebra of the permutation action.

To identify this commutant, use the multilinear isomorphismConjugation by a permutation reorders the factors on the right. Therefore the invariant subspace is the space of symmetric tensors of degree $n$ in $\mathrm{End}(V)$. It is spanned by $a^{\otimes n}$. Indeed, the polarization identityexpresses every symmetrized elementary tensor as a linear combination of pure powers. Those symmetrized elementary tensors span the invariant subspace in characteristic zero.

One may restrict $a$ to invertible endomorphisms without changing the span. For fixed $a$, the vector-valued polynomial $(a+tI{)}^{\otimes n}$ has degree at most $n$. Choose $n+1$ distinct values of $t$ away from the finitely many roots of $\det (a+tI)$. Polynomial interpolation expresses its constant term $a^{\otimes n}$ as a linear combination of those invertible tensor powers. ConsequentlyThe span is an algebra, since products are $(gh{)}^{\otimes n}$.

The image $A_T$ of $\mathbb{C}{S}_n$ is semisimple, as a quotient of a semisimple algebra. The double-centralizer theorem for semisimple operator algebras gives $A_T^{\prime \prime }=A_T$. Since the preceding computation says that $A_T^{\prime }$ is the span of the general linear action, the two actions are mutual commutants. More explicitly, semisimple module decomposition givesHere $D_{\lambda }(V)={\mathrm{Hom}}_{S_n}(S^{\lambda },T)$ is the multiplicity space, with its natural general linear action. The commutant algebra is the product of the full endomorphism algebras of these multiplicity spaces. Hence each nonzero $D_{\lambda }$ is irreducible, and distinct spaces have nonisomorphic general linear representations, because the operators $g^{\otimes n}$ span that commutant. A primitive $e_t$ picks a one-dimensional factor from $S^{\lambda }$, so $e_{t}T$ is an equivalent realization of $D_{\lambda }$ as a Schur module.

The range of shapes is exact. If $\lambda$ has more than $m$ rows, a column antisymmetrizes more than $m$ vectors and its action vanishes by ${\Lambda }^{m+1}V=0$. Conversely, for at most $m$ rows place the $i$th basis vector in every tensor position belonging to row $i$ of $t$. Row symmetrization multiplies this tensor by ${\prod }_i{\lambda }_i!$. Column antisymmetrization is nonzero, because all vectors within each column are distinct, and its different permutations give distinct basis tensors. Thus $h_{t}T\ne 0$. This proves the length bound for a Schur module and the stated Schur–Weyl duality. For $n=0$, the empty partition and $V^{\otimes 0}=\mathbb{C}$ give the trivial version.

For a partition $\nu$ with at most $m$ parts, let $D_{\nu }(V)$ be the Schur module constructed in Question 3, equivalently the image of a Young symmetrizer on $V^{\otimes |\nu |}$. For a weakly decreasing integer tuple $\lambda$, put $k=\max (0,-{\lambda }_m)$ and $\nu =\lambda +k(1,\dots ,1)$. Define the rational Schur module byHere ${\det }^{-k}$ is the one-dimensional representation $g\mapsto \det (g{)}^{-k}$. This is rational, and it is irreducible under the permitted irreducibility assumption; for nonnegative $\lambda$ it is the original polynomial Schur module. Larger shifts give the same module, as will also follow from the character formula below.

Write $s_r(x)={\sum }_{i=1}^m{x}_i^r$ and $p_{\alpha }(x)={\prod }_{r=1}^n{s}_r(x{)}^{{\alpha }_r}$. For a permutation $\sigma$ of the given cycle type, the trace of a permuted tensor power isIn a tensor basis, the trace contracts the matrix entries of $\xi$ around each cycle of $\sigma$; a cycle of length $r$ contributes $\mathrm{tr}({\xi }^r)$. This proof applies to nondiagonalizable endomorphisms as well. The eigenvalues give $\mathrm{tr}({\xi }^r)=s_r(x)$.

The Schur–Weyl duality decomposition, on which $\sigma$ and ${\xi }^{\otimes n}$ act on the two respective factors, gives the same trace asFor a partition, the character extends polynomially to all endomorphisms because the tensor-power action does. This extension is not asserted for determinant-twisted modules at singular matrices.

We now derive the alternant character formula for the general linear group. Put $\delta =(m-1,m-2,\dots ,0)$ andUse the permitted symmetric-group character result, the Frobenius alternant character formula:It concerns characters of $S_n$, rather than assuming the character formula we seek for the general linear group. Substitute the proved trace identity and setThen ${\chi }^{\lambda }(\alpha )={\sum }_{\mu }{c}_{\lambda \mu }{\chi }^{\mu }(\alpha )$ for every conjugacy class. Independence of the irreducible symmetric-group characters forces $c_{\lambda \mu }={\delta }_{\lambda \mu }$.

Each character ${\varphi }_{\mu }$ on diagonal matrices is a symmetric homogeneous polynomial of degree $n$, since conjugation by permutation matrices permutes its arguments. Thus $a_{\delta }{\varphi }_{\mu }$ is alternating of degree $n+\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. Every alternating polynomial of this degree has a unique expansion in alternants $a_{\lambda +\delta }$: a monomial with repeated exponents has zero coefficient, while each strictly decreasing nonnegative exponent vector is uniquely $\lambda +\delta$ for a partition $\lambda$ of $n$ with at most $m$ parts. Its coefficient at $x^{\lambda +\delta }$ is the coefficient of that alternant. The identities for $c_{\lambda \mu }$ therefore give $a_{\delta }{\varphi }_{\mu }=a_{\mu +\delta }$. We have derived the Weyl character formulaFor partitions the quotient is the Schur polynomial, with removable apparent singularities when eigenvalues coincide. For arbitrary dominant integer tuples, multiply the formula for $\nu$ by $(x_1\cdots x_m{)}^{-k}$; this shifts every numerator exponent by $-k$ and yields the same boxed formula. The variables must then be nonzero. The expression also shows independence of the shift used to define the determinant twist. Equality of characters identifies these irreducible modules: the group-algebra image on a direct sum of two irreducibles is finite-dimensional and semisimple, and its span of group operators detects the traces on every simple block.

Finally, the character of a dual representation evaluates the original character at $x_i^{-1}$. Reverse the numerator's columns after making this substitution. The exponents become $-{\lambda }_{m+1-j}+1-j$. Factoring $(x_1\cdots x_m{)}^{-(m-1)}$ converts these into ${\mu }_j+m-j$ for ${\mu }_j=-{\lambda }_{m+1-j}$. The denominator undergoes the identical column reversal and factor, so both signs and factors cancel. Hence ${\varphi }_{\lambda }(x^{-1})={\varphi }_{\mu }(x)$, andThis tuple is again weakly decreasing, so it is precisely the required dominant label.

The coordinate ring is $\mathbb{C}[W]=\mathrm{Sym}(W^{\ast })$. The induced action is contragredient on functions:It is a group action by degree-preserving algebra automorphisms, extending the dual action on linear polynomial functions. Its polynomial invariant ring isFor a monic polynomial $f(t)={\prod }_{i=1}^d(t-r_i)$, its polynomial discriminant is $\mathrm{disc}(f)={\prod }_{i<j}(r_i-r_j{)}^2$. This is symmetric in the roots, hence polynomial in the coefficients, and is zero exactly when a root is repeated.

For the alternating group action, let $e_1,\dots ,e_n$ be the elementary symmetric polynomials and put $\Delta ={\prod }_{i<j}(X_i-X_j)$. If $p$ is $A_n$-invariant and $\tau$ is any transposition, decomposeNormality and index two of $A_n$ show that $p_{+}$ is symmetric and $p_{-}$ transforms by the sign character of $S_n$. Any alternating polynomial vanishes when $X_i=X_j$, so every $X_i-X_j$ divides it. These pairwise nonassociate prime factors have product $\Delta$, so $p_{-}=\Delta q$ with $q$ symmetric. The Fundamental theorem of symmetric polynomials yieldsThe sum is direct, since a polynomial that is both symmetric and alternating is zero in characteristic zero. Also ${\Delta }^2=D(e_1,\dots ,e_n)$, where $D$ is the discriminant polynomial ofThus, more precisely than the requested quotient assertion,Surjectivity follows from the direct-sum expression. Divide any putative kernel element by the monic quadratic in $Z$; its remainder is $a(T)+Zb(T)$. The direct sum forces $a(e)=b(e)=0$, and algebraic independence of the $e_i$ gives $a=b=0$. Hence the displayed relation is the entire kernel. The argument also covers $n=2$, where $A_2$ is trivial.

Now let $G$ be finite with no nontrivial linear characters, and write $R=\mathbb{C}[W]$, $S=R^G$. The polynomial ring $R$ is a unique factorization domain. Factor a nonzero invariant $f$ in $R$; invariance permutes the associate classes of its irreducible factors and makes their exponents constant on each orbit. For an orbit $\mathcal{O}$, choose representatives and form its orbit product of polynomial factorsFor every $g$, $g\cdot q_{\mathcal{O}}=\theta (g)q_{\mathcal{O}}$ for a nonzero scalar $\theta (g)$. Applying two group elements proves that $\theta :G\to {\mathbb{C}}^{\times }$ is a homomorphism. The hypothesis forces $\theta =1$, so $q_{\mathcal{O}}\in S$.

This orbit product is prime in $S$. If it divides $ab$ with $a,b\in S$, one factor $p\in \mathcal{O}$ divides $a$ or $b$ in $R$. Invariance of that chosen polynomial makes every factor in the orbit divide it. Their product therefore divides it in $R$, and the quotient is invariant because numerator and denominator are invariant and cancellation is valid in the integral domain $R$. Thus $q_{\mathcal{O}}$ divides $a$ or $b$ in $S$. Every nonzero $f\in S$ is a scalar times a product of these prime orbit products, and the only units of $S$ are nonzero constants, as they are units in $R$. Therefore $\mathbb{C}[W{]}^G$ is a unique factorization domain. This gives a direct proof, without assuming a divisor-class-group theorem.

For a failure in characteristic zero, let the cyclic group of order two act on ${\mathbb{C}}^2$ by $-I$. Its coordinate invariants areEvery invariant monomial has even total degree: its exponents are either both even, or both odd, giving the indicated generators. Reducing powers of $b$ to at most one shows that $ac-b^2$ is the only relation, since $a^i{c}^j$ and $b{a}^i{c}^j$ map to distinct monomials. The three quadratic invariants are irreducible in $S$: each nonconstant invariant has degree at least two, so a product of two nonunits has degree at least four. They are pairwise nonassociate, butgives two different irreducible factorizations. Hence this invariant ring is not a unique factorization domain. The nontrivial sign character is precisely the kind of character excluded in the preceding theorem.

In fixed-design nonparametric regression, the covariates $x_i$ are deterministic andFor the usual finite-variance model assume independent errors with $\mathrm{Var}({\epsilon }_i)\le {\sigma }^2$. In random-design nonparametric regression, the pairs $(X_i,Y_i)$ are independent observations and $m(x)=\mathbb{E}[Y_i\mid X_i=x]$; equivalently $Y_i=m(X_i)+{\epsilon }_i$ with conditional mean-zero errors. Conditional variances are generally assumed bounded for risk bounds.

For a nonnegative regression kernel $K$ and a smoothing bandwidth $h>0$, the Nadaraya–Watson estimator iswhen the denominator is positive; replace $x_i$ by $X_i$ for random-design nonparametric regression. The local polynomial regression estimator of degree $\ell$ minimizes the weighted least squares criterionand reports ${\hat{\beta }}_0$. If the weighted Gram matrix is invertible, this minimizer is unique. The Nadaraya-Watson estimator is exactly local polynomial regression of degree zero. Higher degrees reproduce nonconstant polynomials and reduce boundary bias of an estimator.

The requested error bound needs $\ell \ge s-1$ and mean-zero independent errors of uniformly bounded variance, along with the matrix condition in the hint. The degree requirement is missing from the printed question. To see the problem, take $\ell =0$, $K=1_{[-1,1]}$, $m(t)=t$, zero errors, $x=0$, and $h=n^{-1/4}$. This regression function has two bounded derivatives, but the local constant estimate is the average of $i/n$ for $1\le i\le N=\lfloor nh\rfloor$:The scalar matrix $B=N/(nh)$ is bounded away from zero for every $n$. Nevertheless $(nh{)}^{-1/2}+h^2=n^{-3/8}+n^{-1/2}=o(h)$, so the printed conclusion with $s=2$ fails. A bounded-variance error hypothesis is also essential; otherwise an expectation bound need not even be finite.

Under the qualified hypotheses, use $U(t)=(1,t,\dots ,t^{\ell }/\ell !{)}^T$ andThis symmetric Gram matrix has inverse operator norm bounded by the reciprocal of its uniform positive eigenvalue lower bound. If $K$ is supported in $[-R,R]$, only $|x_i-x|\le Rh$ contribute. The equally spaced design and $nh\ge 1$ imply that there are at most $2Rnh+1\le Cnh$ such points. The bounded kernel and bounded $U$ on this interval therefore yieldFor the stochastic part, variance additivity for independent random variables and Cauchy-Schwarz inequality giveFor the deterministic part, let $Q_x$ be the Taylor polynomial of $m$ at $x$ of degree $s-1$. The polynomial reproduction property of local polynomial regression gives ${\sum }_i{W}_{ni}(x)Q_x(x_i)=Q_x(x)=m(x)$. By Taylor's theorem, $|m(x_i)-Q_x(x_i)|\le C|x_i-x{|}^s\le C{h}^s$ on the kernel window. ThusCombining the two parts provesThe argument also applies at the endpoints when the assumed matrix bound holds. With arbitrary degree, the generally valid smoothness term is $h^{\min (s,\ell +1)}$; extra symmetry can improve some interior biases but does not fix the general boundary counterexample.

A map $\Phi :S_D\to \mathbb{R}$ is Hadamard differentiable at $s_0$ if there is a continuous linear functional ${\dot{\Phi }}_{s_0}:S\to \mathbb{R}$ such that, whenever $t_j\to 0$ with $t_j\ne 0$, $a_j\to a$ in $S$, and $s_0+t_j{a}_j\in S_D$,For tangential Hadamard differentiability, the limit directions $a$ are restricted to a specified tangent subspace and the derivative is continuous linear on that subspace. The domain condition is indispensable.

Assume $X_n\in S_D$ almost surely and use the usual measurable, separably supported setting for norm-valued convergence in distribution, with the limit in the tangent space. The functional delta method givesTo justify this by Skorokhod representation theorem, a sufficient precise version is: for weakly convergent random elements in a separable complete metric space, one can construct copies of the whole sequence and its limit on one probability space which converge almost surely in the metric. More generally the separable-limit-support version of the representation theorem can be used. Apply it to $Z_n=r_n(X_n-s_0)$, obtaining ${\tilde{Z}}_n\to \tilde{X}$ almost surely. The variables $s_0+r_n^{-1}{\tilde{Z}}_n$ have the same laws as $X_n$ and lie in $S_D$ on a common probability-one event. The definition of Hadamard differentiability, with $t_n=1/r_n$, gives convergence almost surely of the transformed quotients to ${\dot{\Phi }}_{s_0}(\tilde{X})$, and hence the claimed convergence in distribution. If $X$ is a Gaussian measure, the limit is Gaussian because the derivative is linear. Without membership of $X_n$ in $S_D$, the expression $\Phi (X_n)$ in the question is not defined.

The final function space is $C_b(\mathbb{R})$ with supremum norm; it is not separable as a whole. This does not affect the deterministic differentiability proof below. Any statistical application of the preceding representation argument must use its separability or tightness hypotheses, rather than assuming that every normed space automatically satisfies them.

Fix admissible $F,G$ and set $T(F,G)={\int }_0^{u}G(x)F^{\prime }(x)dx$. The candidate derivative, defined even for continuous directions $a,b$ without derivatives, isThis follows formally from the product rule and integration by parts. We now establish the uniform-direction limit required by Hadamard differentiability.

Let $F_j=F+t_j{a}_j$ and $G_j=G+t_j{b}_j$ belong to the stated domain, with $a_j\to a$, $b_j\to b$ in supremum norm. The domain gives $\int |F_j^{\prime }|\le 1$ and $\int |F^{\prime }|\le 1$, while $F_j\to F$ uniformly. We first claim that, for every continuous $b$ on $[0,u]$,For a continuously differentiable test function $v$, integration by parts gives convergence directly from uniform convergence of $F_j$: both boundary terms and $\int v^{\prime }{F}_j$ converge. By the Weierstrass approximation theorem, approximate $b$ uniformly on $[0,u]$ by a polynomial $v$. The difference between the two integrals introduced by this approximation is at most $2\Vert b-v{\Vert }_{\infty }$, by the uniform bounded variation bounds. Letting the approximation error decrease proves the claim. This is weak convergence of bounded-variation integrators.

Now decompose the quotient in the order that retains the bounded-variation bound:The first term differs from $\int b{F}_j^{\prime }$ by at most $\Vert b_j-b{\Vert }_{\infty }$, so it tends to $\int b{F}^{\prime }$. For the second term, integration by parts yieldswhich tends to the corresponding expression with $a$. Hence the quotient tends to ${\dot{T}}_{F,G}(a,b)$ for every admissible perturbation sequence. Finally,This proves that the derivative is a continuous linear functional on the product space. Therefore $T$ is Hadamard differentiable at every point of the stated domain. The bounded-variation restriction is what controls the apparently dangerous interaction between uniformly small perturbations and possibly large perturbation derivatives; no convergence of $a_j^{\prime }$ or $b_j^{\prime }$ has been assumed.

Use the bit-query Boolean quantum oracle $O_x|i,b,z⟩=|i,b\oplus x_i,z⟩$, where $z$ denotes workspace. Initially every probability amplitude is independent of $x$, hence is a constant polynomial. An input-independent unitary gate only forms linear combinations of probability amplitudes, so it does not increase their polynomial degree.

If $a_{i,b,z}(x)$ is an amplitude before a query, its new value isOne query increases polynomial degree by at most one. Inductively, after $T$ queries each amplitude has polynomial degree at most $T$. The acceptance probability is a sum of squared absolute values of the accepting amplitudes, so it is a real polynomial of degree at most $2T$. Intermediate quantum measurements and classical adaptation can be retained coherently, or handled by summing unnormalized branch probabilities; either approach yields the same degree bound.

An exact algorithm has acceptance probability precisely $f(x)$ at every Boolean input. The multilinear reduction on the Boolean cube replaces positive powers of each $x_i$ by $x_i$, preserving these values without increasing polynomial degree. Therefore the polynomial method for quantum query lower bounds givesHere $\mathrm{deg}(f)$ is the degree of the unique multilinear polynomial representing the Boolean function, and $Q_E$ is its exact quantum query complexity. Uniqueness follows, for example, by evaluating successively on the indicator vectors of subsets: the value on a subset determines its coefficient once all smaller-subset coefficients are known.

Use $D_j=-i{\partial }_{x_j}$ and $⟨\xi ⟩=(1+|\xi {|}^2{)}^{1/2}$, so the Fourier transform of $P(D)u$ is $P(\xi )\hat{u}(\xi )$. Write $P_N$ for the homogeneous degree-$N$ part. An elliptic differential operator has $P_N(\xi )\ne 0$ for every real $\xi \ne 0$. By continuity on the unit sphere, $m={\min }_{|\omega |=1}|P_N(\omega )|>0$. The lower-degree terms are bounded by $C|\xi {|}^{N-1}$ for $|\xi |\ge 1$, whencefor sufficiently large $|\xi |$. This proves the high-frequency lower bound for an elliptic polynomial. Degree zero just means a nonzero constant and has no derivative gain.

With the Fourier convention in Question 3, the Sobolev space isThe condition includes that the weighted transform is represented by an $L^2$ function. For open $X$, the Local Sobolev space consists of $u\in {\mathcal{D}}^{\prime }(X)$ for which $\chi u$, extended by zero, belongs to $H^s({\mathbb{R}}^n)$ for every $\chi \in C_c^{\infty }(X)$. We use the elementary Sobolev multiplication by a smooth cutoff fact for every real $s$. It follows from Fourier convolution with the rapidly decreasing $\hat{\chi }$, the weighted inequality $⟨\xi {⟩}^s\lesssim ⟨\xi -\eta {⟩}^{|s|}⟨\eta {⟩}^s$, and the $L^1\ast L^2\to L^2$ convolution bound.

For a compactly supported distribution, continuity on test functions supported in a fixed compact neighborhood gives finite order: for some integer $M$,Insert a compact smooth cutoff function equal to one near the support, times $e^{-ix\cdot \xi }$. The resulting Fourier transform is smooth and satisfies $|\hat{u}(\xi )|\le C^{\prime }⟨\xi {⟩}^M$. The distribution is also tempered by this same finite-order bound. ThusIndeed the square of the weighted bound is integrable exactly when $2(t+M)<-n$. This is negative Sobolev regularity of a compactly supported distribution.

We next prove local regularity by a high-frequency reciprocal parametrix kernel, including its off-diagonal smoothing property. Choose $\psi \in C_c^{\infty }({\mathbb{R}}^n)$ equal to one on a ball containing all real zeros of $P$, and put $b(\xi )=(1-\psi (\xi ))/P(\xi )$, defining it smoothly as zero in the inner ball. Differentiation of the reciprocal and the elliptic lower bound giveThe Fourier multiplier $E=b(D)$ maps $H^s$ to $H^{s+N}$ by its zeroth-order bound, andThe kernel $K={\mathcal{F}}^{-1}b$ is smooth away from the origin: for $x\ne 0$, repeatedly integrate by parts using $e^{ix\cdot \xi }=(i|x{|}^2{)}^{-1}x\cdot {\partial }_{\xi }{e}^{ix\cdot \xi }$. After sufficiently many integrations, the differentiated symbol is integrable. For any desired $x$ derivative, repeat the argument with the additional polynomial ${\xi }^{\beta }$. A large-radius cutoff function justifies each step and its removal uniformly on compact sets away from zero. Thus convolution by $K$ carries a compactly supported distribution to a smooth function at points separated from its support. Also $R$ carries a compactly supported distribution to a Schwartz function, because $\psi \hat{u}$ is smooth with compact support.

For a target compact set in $X$, choose $\chi \in C_c^{\infty }(X)$ equal to one on a neighborhood of it. Then $\chi u$ and $[P(D),\chi ]u$ have compact support, and the commutator is supported where derivatives of $\chi$ occur, away from the target. The parametrix identity givesThe first term is in $H^{s+N}$ since $\chi P(D)u\in H^s$. The other two terms are smooth near the target by the proved kernel property. Since the target was arbitrary,This proof does not discard the cutoff function commutator; it places its support away from the set where regularity is sought.

For the final polynomial, select a multi-index ${\alpha }_0$ with $|{\alpha }_0|=N$ and ${\partial }^{{\alpha }_0}Q$ a nonzero constant. The derivative-ratio hypothesis immediately gives $|Q(\xi )|\ge c|\xi {|}^{\delta N}$ at large frequency. In particular $Q$ has no real zero there. For $N>0$, its degree bound also forces $\delta \le 1$. Differentiating $1/Q$ gives products of ratios ${\partial }^{\beta }Q/Q$, whose total derivative order is $|\alpha |$. Hence the corresponding high-frequency reciprocal satisfiesIts multiplier maps $H^s$ to $H^{s+\delta N}$. Its kernel is still smooth off zero: each frequency integration by parts now lowers the order by $\delta$, so more iterations may be needed, but $\delta >0$ supplies arbitrarily much decay. The same separated-support commutator identity applies without a change. The derivative-ratio Sobolev gain for a polynomial operator is thereforeSimply estimating the commutator by its differential order would lose this sharper gain; the off-diagonal kernel argument uses the full derivative-ratio hypothesis.