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Put $A=T^{\ast }T$. Then $A^{\ast }=T^{\ast }T$ and $(x,Ax)=\Vert Tx{\Vert }^2\ge 0$, with the inner product linear in its second argument as required by the later SVD formula. For nonreal $\lambda$, self-adjointness gives $\Vert (A-\lambda I)x\Vert \ge |\mathrm{Im}\lambda |\Vert x\Vert$. For $\lambda =-a<0$, positivity gives $\Vert (A+aI)x\Vert \ge a\Vert x\Vert$. These bounds give injectivity and closed range. The range is dense because its orthogonal complement is the kernel of the adjoint, to which the same bound applies. Thus the range is all of $H$ and the inverse is bounded in either case. ConsequentlyIf $T$ is compact, composition of the compact operator $T$ with bounded $T^{\ast }$ is compact, so $T^{\ast }T$ is compact too.

Let $w(\xi )=(1+|\xi {|}^2{)}^{s/2}$. The map $U:f\mapsto w\hat{f}$ is linear, and $\Vert U(f+g){\Vert }_2\le \Vert Uf{\Vert }_2+\Vert Ug{\Vert }_2$ proves the vector-subspace claim. The stated inner product is $\int w^2\hat{f}\overline{\hat{g}}$. By the Plancherel theorem it is positive definite.

More strongly, $U:H^s\to L^2$ is onto: for $h\in L^2$, $w^{-1}h\in L^2$, since $w\ge 1$, and $f={\mathcal{F}}^{-1}(w^{-1}h)$ satisfies $Uf=h$. Thus $U$ is an isometric linear bijection, and completeness of $L^2$ givesThis argument includes $s=0$ if zero is allowed in the printed ${\mathbb{R}}_{+}$ convention.

For $A\mathbf{u}=\lambda B\mathbf{u}$ and $A\mathbf{v}=\mu B\mathbf{v}$, symmetry givesIf $\lambda \ne \mu$, ${\mathbf{u}}^{T}B\mathbf{v}=0$: the eigenvectors are orthogonal in the inner product induced by $B$.

This is the Gram-Schmidt process applied degree by degree. Inductively, subtracting projections onto $p_0,\dots ,p_n$ makes $p_{n+1}$ orthogonal to every earlier polynomial. Since every subtracted polynomial has degree at most $n$, the leading coefficient of the monic $q_{n+1}$ remains one, so $p_{n+1}$ is a real monic orthogonal polynomial of degree exactly $n+1$.

Now take $q_{n+1}=x{p}_n$. For $k\le n-2$, symmetry of the inner product givesbecause $x{p}_k$ has degree at most $n-1$. Only the $p_n$ and $p_{n-1}$ projections remain, yielding the three-term recurrence for monic orthogonal polynomialswhereThe second identity uses $x{p}_{n-1}=p_n$ plus lower-degree terms.

For the symmetric Legendre inner product, parity gives ${\alpha }_n=0$. Starting with $p_0=1$, one obtains $p_1=x$, thenThus the first four monic Legendre polynomials are

Put $I={\int }_0^{1}f(t)dt$. If $I\ne 0$, take the inner product with the unit vector $u=I/\Vert I{\Vert }_2$. Thenby the Cauchy-Schwarz inequality; the case $I=0$ is immediate. This is the triangle inequality for a vector-valued integral.

Equality in the Euclidean argument requires $u\cdot f(t)=\Vert f(t){\Vert }_2$ for every $t$: the nonnegative continuous difference has integral zero. Equality in Cauchy-Schwarz inequality then says $f(t)=\lambda (t)u$ with $\lambda (t)\ge 0$. Conversely, if $f(t)=\lambda (t)v$ for one fixed $v\in {\mathbb{R}}^m$ and a continuous $\lambda \ge 0$, then for every norm,Thus these nonnegative-ray functions, including $f\equiv 0$, are exactly the required functions.

Use the coroot normalization ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$, with the pairing understood via the root system and its invariant inner product. A dominant integral weight, called a dominant weight here, is a functional $\lambda \in {\mathfrak{t}}^{\ast }$ such thatEquivalently, $\lambda ={\sum }_i{n}_i{\omega }_i$ with $n_i\in {\mathbb{Z}}_{\ge 0}$, where the fundamental weights satisfy $⟨{\omega }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$. If dominance and integrality are defined separately, dominance means the inequalities and membership in the weight lattice means integrality.

A highest-weight vector is a nonzero weight vector $v\in V_{\lambda }$, so $h\cdot v=\lambda (h)v$ for all $h\in \mathfrak{t}$, annihilated by every positive root space:The last condition is equivalently annihilation by the raising operators for the simple roots.

Choose the root basisIn the usual invariant inner product, each simple root has squared length $2$, while $({\alpha }_1,{\alpha }_2)=({\alpha }_1,{\alpha }_3)=-1$ and $({\alpha }_2,{\alpha }_3)=0$. The labeled Dynkin diagram is therefore the following chain, with a single bond between adjacent nodes:It is type $D_3$, equivalently the A3 root system. To check the root basis explicitly, the corresponding positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_1+{\alpha }_2,{\alpha }_1+{\alpha }_3,{\alpha }_1+{\alpha }_2+{\alpha }_3$; these and their negatives are all twelve roots.

Use the coroot ${\alpha }^{\vee }$ in the pairing. The original PDF retains this mark, which is lost in the local TeX transcription. Fix any positive root $\alpha$, and put $n=⟨\lambda ,{\alpha }^{\vee }⟩$. For a dominant integral weight, $n$ is a nonnegative integer; positive coroots are nonnegative integral combinations of simple coroots.

Let $v_{\lambda }$ be a highest-weight vector, and restrict to the sl2 subalgebra associated with a root $\alpha$. It is killed by its raising operator $e_{\alpha }$ and has $h_{\alpha }$ eigenvalue $n$. If $n\ge 1$, its lowering operator produces a nonzero vector: otherwise $[e_{\alpha },f_{\alpha }]v_{\lambda }=n{v}_{\lambda }$ would vanish. Therefore $\lambda -\alpha$ is a weight.

Choose a Weyl group invariant positive-definite inner product on the real weight space. All weights of a minuscule representation have the same norm, since they belong to $W\lambda$. ButIf $n\ge 2$, this is negative, a contradiction. Thus for every positive root,When $n=1$, the lower weight is precisely the Weyl reflection $s_{\alpha }\lambda =\lambda -\alpha$, consistent with the orbit condition.

We first prove the useful lemma that dominant root-lattice highest weights have zero weight. Write $Q=\mathbb{Z}\Phi$ for the root lattice, and let a dominant integral weight $\lambda \in Q$ be the highest weight of a finite-dimensional irreducible representation.

First, $\lambda$ is a nonnegative integral combination of simple roots. Indeed, write $\lambda ={\lambda }_{+}-{\lambda }_{-}$, separating its positive and negative coefficients in the root basis. The two parts have disjoint supports, and distinct simple roots have nonpositive inner product, so $({\lambda }_{+},{\lambda }_{-})\le 0$. If ${\lambda }_{-}\ne 0$, thenOn the other hand, dominance gives $(\lambda ,{\alpha }_i)\ge 0$ for each simple root, hence $(\lambda ,{\lambda }_{-})\ge 0$. This contradiction proves ${\lambda }_{-}=0$.

Now suppose a nonzero weight $\mu ={\sum }_i{m}_i{\alpha }_i$ has all $m_i\in {\mathbb{Z}}_{\ge 0}$. The identityshows that some $i$ satisfies $m_i>0$ and $⟨\mu ,{\alpha }_i^{\vee }⟩>0$. We use the standard sl2 Lie algebra fact that its lowering operator is injective on any positive $h$ eigenspace in a finite-dimensional representation. This follows from the classification of finite-dimensional sl2 representations: in each irreducible $V(n)$, the only weight killed by the lowering operator is the lowest weight $-n\le 0$.

Consequently a nonzero vector of weight $\mu$ lowers to a nonzero vector of weight $\mu -{\alpha }_i$. Its simple-root coefficients remain nonnegative and their sum decreases by one. Starting at $\lambda$, repeated lowering must therefore reach the zero weight. Notice that intermediate weights need not remain dominant; positivity of the chosen coroot pairing is enough at each step.

For a minuscule representation, every weight belongs to $W\lambda$, so the zero weight just obtained lies in this orbit. Every Weyl group element is invertible, and $w\lambda =0$ forces $\lambda =0$. This proves that minuscule weights in the root lattice are zero. Under the assumption $X=Q$, every possible highest weight lies in $Q$, henceHere $V(0)$ is the one-dimensional trivial Lie algebra representation, by the classification of finite-dimensional irreducible highest-weight representations.

The assertion about all $H_0^1(U)\cap H^k(U)$ is false as printed in the original PDF. Take $U=(0,\pi )$, $L=-d^2/d{x}^2$, and $u=x(\pi -x)$. This is a nonzero smooth Dirichlet function, but $Lu=2$ and $L^{2}u=0$. In particularAlready at $k=3$, $B[Lu,Lu]$ is not defined on its stipulated domain, because $Lu=2\notin H_0^1$. Extending $B$ to $H^1$ for this example would give zero, not an inner product. Additional boundary conditions are essential.

The corrected spaces are the Sobolev domains of powers of an elliptic Dirichlet operator. Let $A_D$ be the Dirichlet realization of an elliptic operator, and setIn particular $X_1=H_0^1$, $X_2=H^2\cap H_0^1$, and $X_3$ also requires $T(Lu)=0$. The formulas in the question define inner products on these spaces.

Here is the norm-equivalence proof on the corrected domains. The base cases are the $L^2$ inner product at $k=0$ and part (i) at $k=1$. Standard Dirichlet elliptic regularity, together with invertibility from the Lax-Milgram theorem, gives for every integer $r\ge 0$The first estimate applies when $Tu=0$; invertibility absorbs the usual lower-order $L^2$ term. Moreover, $A_D:X_{r+2}\to X_r$ is an isomorphism. To see surjectivity, solve $Lu=f$ with zero Dirichlet trace for $f\in X_r$; regularity gives $u\in H^{r+2}$, and $T(L^{j}u)=T(L^{j-1}f)=0$ for each further required trace. The formulas satisfyInduction therefore givesInjectivity of each iterate of $A_D$ gives positive definiteness. These spaces are $D(A_D^{k/2})$; the spectral characterization of elliptic Dirichlet domains in the next part makes the half-powers precise. On the paper's uncorrected spaces the claim is valid at $k=0,1,2$, but fails in general beyond them.

Formal self-adjointness gives symmetry of $B$, and strict positivity gives positive definiteness, so $B$ is an inner product. Bounded coefficients give $B[u,u]\le C\Vert u{\Vert }_{H^1}^2$. To prove the lower bound, uniform ellipticity givesIf no lower bound by the $H^1$ norm existed, choose $u_j\in H_0^1$ with $\Vert u_j{\Vert }_{H^1}=1$ and $B[u_j,u_j]\to 0$. By weak convergence in a Hilbert space and the Rellich-Kondrachov compactness theorem, a subsequence has $u_j\rightharpoonup u$ in $H_0^1$ and $u_j\to u$ in $L^2$. The principal energy has weak lower semicontinuity: it is the squared $L^2$ norm of the bounded multiplication operator $a^{1/2}Du$. The potential term converges because $c$ is bounded. HenceStrict positivity forces $u=0$. The preceding ellipticity estimate now gives $\Vert D{u}_j{\Vert }_2\to 0$, while $\Vert u_j{\Vert }_2\to 0$, contradicting $\Vert u_j{\Vert }_{H^1}=1$. Thus strict positivity implies coercivity for an elliptic Dirichlet form, andThe new norm is equivalent to the original norm, so the space remains complete.

At a point $p$, write $W=T_{p}L$ and take the orthogonal complement with respect to the Riemannian metric $g$. The compatible triple identity givesIf $L$ is a Lagrangian submanifold, then $\omega (u,v)=0$ for $u,v\in W$, so $JW\subset W^{\perp }$. Both have dimension $\frac{1}{2}\dim M$, hence equality.

Conversely, if $JW=W^{\perp }$, equality of dimensions gives $\dim W=\frac{1}{2}\dim M$, and $g(Ju,v)=0$ for $u,v\in W$ gives $\omega {|}_W=0$. ThereforeA positive-definite inner product has $W\cap W^{\perp }=\{0\}$. Hence $W\cap JW=\{0\}$, proving that every Lagrangian submanifold is a totally real submanifold for a compatible almost complex structure.

Work over $\mathbb{R}$ or $\mathbb{C}$, in finite dimension and characteristic zero. The Killing form of a Lie algebra is the trace form of its Adjoint representation:It is bilinear and symmetric by the cyclic property of the trace. It is an invariant bilinear form on a Lie algebra, since the Jacobi identity gives ${\mathrm{ad}}_{[x,y]}=[{\mathrm{ad}}_x,{\mathrm{ad}}_y]$ and henceEquivalently, $\kappa ([z,x],y)+\kappa (x,[z,y])=0$. A Lie algebra automorphism preserves the form, because it conjugates the adjoint matrices. Its radical of a bilinear form ${\mathfrak{r}}_{\kappa }=\{x:\kappa (x,\mathfrak{g})=0\}$ is an ideal of a Lie algebra by invariance. The center of a Lie algebra lies in this radical, so the form vanishes for an abelian algebra. On a direct sum of ideals the summands are orthogonal and the form restricts to their own Killing forms.

A short argument proves the requested implication without assuming the Cartan criterion for semisimplicity. Let $\mathfrak{a}$ be an abelian ideal of a Lie algebra, $x\in \mathfrak{a}$, and $y\in \mathfrak{g}$. The map ${\mathrm{ad}}_x$ takes $\mathfrak{g}$ into $\mathfrak{a}$ and vanishes on $\mathfrak{a}$, while ${\mathrm{ad}}_y$ preserves $\mathfrak{a}$. Therefore ${\mathrm{ad}}_x{\mathrm{ad}}_y$ has image in $\mathfrak{a}$ and is zero on $\mathfrak{a}$. In a basis extending a basis of $\mathfrak{a}$, both diagonal blocks are zero, so its trace vanishes. This proves that abelian ideals lie in the radical of the Killing form.

If the solvable radical $\mathfrak{r}$ were nonzero, its derived series of a Lie algebra would have a last nonzero term $\mathfrak{a}$. The Jacobi identity makes every derived term an ideal of $\mathfrak{g}$, and the last one is abelian. Thus $0\ne \mathfrak{a}\subseteq {\mathfrak{r}}_{\kappa }$, contradicting nondegeneracy. We obtainThe converse holds as well in characteristic zero; together these implications are the Cartan criterion for semisimplicity.

For a complex simple Lie algebra, that criterion gives nondegeneracy on $\mathfrak{g}$. Let $\mathfrak{h}$ be a Cartan subalgebra. Use the standard root-space decompositionFor $h,h^{\prime }\in \mathfrak{h}$ and a root vector $e_{\alpha }$, choose $t\in \mathfrak{h}$ with $\alpha (t)\ne 0$. Invariance givesThus $\mathfrak{h}$ is orthogonal to every nonzero root space. If $h\in \mathfrak{h}$ is also orthogonal to $\mathfrak{h}$, it is orthogonal to all of $\mathfrak{g}$ and hence is zero. This establishes nondegeneracy of the Killing form on a Cartan subalgebra:The same invariance calculation shows that $\kappa ({\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta })=0$ unless $\alpha +\beta =0$. Opposite root spaces are therefore paired nondegenerately. Tracing the adjoint action on the root-space decomposition yieldssince the root spaces of a complex semisimple algebra are one-dimensional and its adjoint action on $\mathfrak{h}$ is zero.

The relevant Euclidean subspace of a Cartan subalgebra iswhere the $h_{{\alpha }_i}$ are the simple coroots, normalized by ${\alpha }_i(h_{{\alpha }_i})=2$. The standard Euclidean property is that $\kappa$ is real and positive definite on this space. The displayed trace formula explains it: $\kappa (h,h)$ is a sum of real squares, and the roots span ${\mathfrak{h}}^{\ast }$, so all squares vanish only for $h=0$. Via the nondegenerate form, the roots can consequently be regarded as vectors in a real Euclidean normed vector space. The induced dual inner product makes root reflections orthogonal and allows root lengths and angles to be encoded by Cartan integers and the Dynkin diagram.

This positivity is on a specified real subspace; the complex Killing form is a bilinear form, not a positive Hermitian inner product. On a compact real form the Killing form is instead negative definite. For example the Killing form of the special linear Lie algebra gives $\kappa (X,Y)=4\mathrm{tr}(XY)$ for ${\mathfrak{sl}}_2$, so $\kappa (L_0,L_0)=8$ but $\kappa (i{L}_0,i{L}_0)=-8$.

Use anti-Hermitian representation matrices for the unitary representation, so $R(X{)}^{†}=-R(X)$. Absorb the gauge coupling into the connection, consistently with the transformation law in the PDF. The appropriate gauge covariant derivative isThis has a plus sign because the inhomogeneous part of the connection transformation is $-ϵ{\partial }_{\mu }X$. Its variation isThus gauge covariance of a scalar covariant derivative givesThe cancellation uses that $R$ is a Lie algebra representation, including its commutator identity.

The gauge field strength isIt is the curvature appearing in $[D_{\mu },D_{\nu }]\varphi =R(F_{\mu \nu })\varphi$. The index positions in the PDF are related by the constant Minkowski metric; lowering the index in its transformation law gives the convention used here.

Take signature $(+,-,-,-)$, let $B$ be a real symmetric gauge-invariant internal form, and let $⟨\cdot ,\cdot {⟩}_V$ be the positive Hermitian form preserved by the unitary matter representation. One consistent Lorentz transformation scalar Lagrangian density isHere $g_{\mathrm{YM}}>0$, $m^2\ge 0$, and $\lambda \ge 0$ give a simple stable potential. The coupling is in the kinetic normalization because it was absorbed into $A_{\mu }$; it is not inserted again into $D_{\mu }$. Taking a zero or another gauge-invariant potential is also possible. A positive nondegenerate $B$ gives the usual physical gauge kinetic energy, as for a compact gauge algebra. More generally the invariance calculation below applies to any symmetric invariant bilinear form on a Lie algebra, including the Killing form.

Under the infinitesimal gauge-field transformation law,Invariance and symmetry of $B$ give the cancellation. For matter, both $\varphi$ and $D_{\mu }\varphi$ transform by $R(X)$, so anti-Hermiticity givesfor either pairing used in the Lagrangian. Its potential is a function of the invariant scalar norm. Therefore $\boxed{{\delta }_X\mathcal{L}=0}$, establishing gauge invariance of this Yang-Mills theory coupled to a scalar. Contraction of spacetime indices with the Minkowski metric gives Lorentz invariance.

For the Adjoint representation, define the Lie algebra structure constants byInternal superscripts here are basis labels, as in the paper; spacetime indices alone are raised with the Minkowski metric. The adjoint covariant derivative and curvature have componentsThese formulas involve no extra factor of $i$, because the generators use the Lie-algebra convention rather than Hermitian physics generators.

To make the Killing-form sign explicit, write $B=s\kappa$ and $⟨\varphi ,\psi {⟩}_V=s{\kappa }^{ab}({\varphi }^a{)}^{\ast }{\psi }^b$. The literal contraction requested in the paper corresponds to $s=1$. For a compact real form of a complex semisimple gauge algebra, its actual positive inner product has $s=-1$, since the Killing form is negative definite. With a complex adjoint scalar, the fully expanded Killing-form Lagrangian for an adjoint scalar isAll repeated internal labels are summed; $s^2=1$ explains the absence of $s$ from the quartic term. Invariance is equivalently the component identity $f^{ca}{}_d{\kappa }^{db}+f^{cb}{}_d{\kappa }^{ad}=0$. For a real adjoint scalar, omit complex conjugation and insert $1/2$ in the scalar kinetic and quadratic mass terms; the same invariance proof applies.

The PDF's identification of the Killing form with an inner product thus requires a sign convention. It also requires nondegeneracy if it is to give ordinary kinetic terms: on an Abelian gauge factor the Killing form vanishes, so an additional invariant metric is needed there. These qualifications do not alter the covariance identities or the formal gauge invariance of the displayed Killing-form contractions.

For covariant quantization of the bosonic string,All other commutators between independent canonical variables vanish. The covariant Fock vacuum obeys ${\alpha }_k^m|0;p⟩=0$ for every $k>0$ and every $m$. Normal ordering moves these annihilating string oscillators to the right and givesThen ${\hat{N}}_{\mathrm{cov}}|0;p⟩=0$. Its commutator $[{\hat{N}}_{\mathrm{cov}},{\alpha }_{-k}^m]=k{\alpha }_{-k}^m$ grades a Fock state basis by $N=\sum k{n}_{km}$, although its inner product is indefinite because of the timelike oscillator.

The quantum Virasoro algebra has central charge $D$:The Virasoro central extension comes from commutators needed to order infinite oscillator sums; it is a quantum effect absent from the classical Poisson brackets. Replacing a classical Poisson bracket by a commutator cannot recover that term without a regularized ordering calculation.

If a state were annihilated by every nonzero ${\hat{L}}_j$, the $j=1,k=-1$ commutator would imply ${\hat{L}}_0|\psi ⟩=0$. The $j=2,k=-2$ commutator would then imply $(D/2)|\psi ⟩=0$. For $D>0$, only the zero vector is annihilated by all nonzero Virasoro constraints. This explains why only positive modes annihilate a physical string state.

The vacuum is a physical string state when $p^2=2\pi Ta$. At level one all states have the form $|A;p⟩=A_m{\alpha }_{-1}^m|0;p⟩$. Since $[{\hat{L}}_j,{\alpha }_k^m]=-k{\alpha }_{j+k}^m$, the positive-mode Virasoro constraints giveThus the complete level-one conditions and norm arewith the common vacuum normalization suppressed.

For $a>1$, choose spacelike momentum $p^m=(0,k,0,\dots )$, $k^2=2\pi T(a-1)$. A purely timelike polarization has $A\cdot p=0$ and norm $-1$. Hence a negative-norm physical string state exists when $a>1$.

For $a<1$, the mass-shell condition gives $M^2=2\pi T(1-a)>0$. In a rest frame, $A\cdot p=0$ forces $A_0=0$, leaving $D-1$ positive-norm vector-particle polarizations, those of a massive vector.

For $a=1$ and nonzero null momentum, $A\cdot p=0$ leaves a null direction $A_m\propto p_m$. The corresponding null string state is $p\cdot {\alpha }_{-1}|0;p⟩=\sqrt{\pi T}{\hat{L}}_{-1}|0;p⟩$. It is orthogonal to every physical string state because ${\hat{L}}_1$ annihilates them. Quotienting by this gauge redundancy, $A_m\sim A_m+\zeta p_m$, leaves $D-2$ positive vector-particle polarizations. Thus the level-one spectrum agrees with light-cone gauge in string theory at $a=1$. This level-one argument alone does not establish consistency or absence of negative norms at all higher levels.

Write the residual of an inverse problem as $r_k=K{u}_{\delta }^{(k)}-f^{\delta }$. The iteration can be rearranged asApplying $K$ gives $(I+\tau K{K}^{\ast })r_{k+1}=r_k$. Since $K{K}^{\ast }$ is a positive operator, its resolvent operator $(I+\tau K{K}^{\ast }{)}^{-1}$ has operator norm at most one. HenceMore explicitly, taking the inner product of the residual equation with $r_{k+1}$ givesThis directly proves the same monotonicity by the Cauchy-Schwarz inequality, including the case $r_{k+1}=0$. An unresolvable data component in $\mathcal{R}(K{)}^{\perp }$ remains unchanged.

An orthonormal wavelet is a function whose integer translates and dyadic dilatesform an orthonormal basis of the Hilbert space $L^2(\mathbb{R})$. Thus their pairwise inner products are ${\delta }_{j,j^{\prime }}{\delta }_{k,k^{\prime }}$, and their closed linear span is the whole space. Equivalently, every $f\in L^2(\mathbb{R})$ has the norm-convergent expansionThe factor $2^{j/2}$ preserves the $L^2$ norm under dilation.

Use the Fourier transform convention $\hat{\phi }(\xi )=\int \phi (x)e^{-ix\xi }dx$. Nesting and the orthonormal basis of $V_1$ give the scaling refinement equationHere $h_k=⟨\phi ,\sqrt{2}\phi (2\cdot -k)⟩$. Compact support makes these inner products zero for all but finitely many $k$, so $m$ is a trigonometric polynomial. Expanding the orthogonality of the integer translates of $\phi$ in the orthonormal basis of $V_1$ yieldsConsequentlyThis holds everywhere because $m$ is continuous. The invoked properties are nesting, dyadic dilation, orthonormal integer translates, and compact support; mere finite-energy refinement would not imply the quadrature mirror filter identity.