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For $k>0$, put $q=\sqrt{k^2+l^2}$ and define the Squire transformationThe two-dimensional Orr-Sommerfeld equation then has the same transverse operator because $k^{\prime 2}=q^2$, and the same viscous coefficient because $k^{\prime }{R}^{\prime }=kR$. Its boundary conditions on $v$ and $v^{\prime }$ are unchanged. Thus it has the same eigenfunction and the same phase-speed eigenvalue.

If the original disturbance grows, $\mathrm{Im}c>0$, so the transformed disturbance also grows. Since $R^{\prime }\le R$, every unstable three-dimensional mode maps to an unstable two-dimensional mode at a no larger Reynolds number. It follows that the minimum critical Reynolds number is attained among two-dimensional disturbances. This is the Squire theorem for temporal normal-mode onset, not a statement that three-dimensional transient amplification is impossible. The case $k=0$ is treated separately without dividing by $k$; its streamwise-uniform viscous normal modes decay, although coupling can produce transient growth.

At $c_L=\beta$, the half-space decay exponent $p$ tends to zero and the dispersion relation requires $qh=n\pi$. These are the Love-wave cutoff frequencies,An overtone exists strictly above its cutoff. At the cutoff its half-space field is constant in depth rather than decaying, so it is a limiting nonlocalized state, not a trapped finite-energy eigenfunction. The $n=0$ fundamental has cutoff zero and exists for every positive frequency. Each fixed branch approaches $c_L={\beta }_0$ from above at high frequency. The half-space penetration depth $1/p$ diverges on approach to cutoff.

Choose field $a$ to be $({\hat{\mathbf{u}}}^a{)}^{\ast }$ at frequency $-{\omega }_a$, and field $b$ to be ${\hat{\mathbf{u}}}^b$ at ${\omega }_b$. Zero body forces and zero boundary tractions make all source and surface terms vanish. ThereforeFor distinct squared frequencies this proves weighted orthogonality,Assume positive mass density bounded above and below and finite-energy fields in the ordinary bounded domain. The weighted space $H=L^2(\mathcal{D},\rho dV{)}^3$ is a separable Hilbert space. Select one unit eigenfunction for each distinct squared frequency. They form an orthonormal set. Let $\{d_j\}$ be a countable dense subset of $H$; assign to each eigenfunction a point $d_j$ within distance $1/3$. Two orthogonal unit eigenfunctions have distance $\sqrt{2}$, so the same $d_j$ cannot be assigned to both. This gives an injection into a countable set.

Hence there are at most countably many squared eigenfrequencies, and at most twice that many signed frequencies. Multiplicity at one frequency does not change the argument. A zero frequency associated with rigid translations or rotations is included. This proves the countability of elastic eigenfrequencies directly; proving discreteness or finite multiplicity would require additional spectral regularity arguments.

If two harmonic solutions have the same body force and prescribed boundary traction, their difference has no body force and is traction-free. A nonzero difference would therefore be a free eigenfunction at that frequency. For any frequency outside the countable exceptional set found above, the difference must vanish, so the forced boundary-value solution, when it exists, is unique. At an eigenfrequency, a free eigenfunction can be added to any existing solution, and a resonant forcing may fail to admit a steady harmonic response. This distinguishes uniqueness from existence.

For the full time-dependent problem, let $\mathbf{w}$ be the difference between two fields with identical forces, tractions and initial displacement field and velocity. It obeys $\rho {\mathbf{w}}_{tt}=\nabla \cdot \mathbf{\sigma }(\mathbf{w})$, zero boundary traction, and zero initial data. Its energy uniqueness for traction-driven elasticity follows by direct integration:Multiply the equation by ${\mathbf{w}}_t$ and integrate by parts. Major symmetry turns the strain term into the derivative of elastic energy, giving $E^{\prime }(t)={\int }_{\partial \mathcal{D}}{\mathbf{w}}_t\cdot \mathbf{t}(\mathbf{w})dS=0$. Stable elasticity makes $E\ge 0$, and $E(0)=0$ gives ${\mathbf{w}}_t=0$. Thus $\mathbf{w}$ equals its initial value zero. Even rigid zero-strain motions cannot survive these initial conditions.

The PDF correctly specifies zero initial displacement field and zero initial velocity; the TeX duplicates displacement field in the second condition. The complete causal displacement field is unique, including when its Fourier representation contains resonant modal contributions fixed by the initial data.

Set $y=e^{-x/2}v$, the Dirichlet gauge transform for constant drift. Direct differentiation givesThe homogeneous endpoint conditions on $y$ become $v(0)=v(1)=0$. Thus the eigenvalue equation is $v^{\prime \prime }=(\lambda +1/4)v$. To exclude other branches without assuming the eigenvalue real, multiply by $\bar{v}$ and integrate by parts:for a nonzero eigenfunction. Therefore $\lambda +1/4=-k^2$ with $k>0$. The boundary conditions in $v=A\sin (kx)+B\cos (kx)$ give $B=0$ and $k=n\pi$, $n\ge 1$. HenceEquivalently, $L[y]=\lambda y$ is the Sturm-Liouville form $(e^x{y}^{\prime }{)}^{\prime }=\lambda e^{x}y$, with positive weight $e^x$. Its orthogonality relation isIt is the weighted inner product, not the unweighted one, that makes these eigenfunctions orthogonal.

Expand $x{e}^{-x/2}={\sum }_n{b}_n{y}_n$ with the coefficients from part a. Completeness and the nonzero eigenvalues give the Sturm-Liouville eigenfunction expansionThe coefficients are $O(n^{-3})$, so the function series converges uniformly and vanishes at both endpoints. Its first derivative also converges uniformly. Applying the transformed operator in the square-integrable sense gives the part-a expansion of $x$; equivalently $v^{\prime \prime }-v/4=x$ on the interior. Integrating this continuous right-hand side gives the classical twice-differentiable solution there. No nonzero homogeneous solution satisfies both endpoints, so this solution is unique. The mismatch of the forcing's sine series at a single endpoint does not affect the interior equation or these boundary conditions.

The orbital angular momentum operators are $\mathbf{L}=\mathbf{x}\times \mathbf{p}$, soThe canonical commutation relations are $[x_i,x_j]=[p_i,p_j]=0$ and $[x_i,p_j]=i\hslash {\delta }_{ij}$. On a common domain of smooth wavefunctions, the product commutator identity givesContract this with ${ϵ}_{iab}{ϵ}_{jcd}$. The first term contributes $i\hslash ({\delta }_{ij}\mathbf{x}\cdot \mathbf{p}-x_j{p}_i)$ and the second contributes $i\hslash (x_i{p}_j-{\delta }_{ij}\mathbf{x}\cdot \mathbf{p})$. Their sum isThis verifies the angular momentum commutation relations explicitly. Similarly,

In the position representation $\mathbf{p}=-i\hslash \nabla$. For a radial wavefunction, $\nabla {\Psi }_0={\Psi }_0^{\prime }(r)\mathbf{x}/r$, parallel to $\mathbf{x}$, soWrite ${\Phi }_i=h(r)x_i$ with $h(r)={\Psi }_0^{\prime }(r)/r$. The operator $L_3=-i\hslash (x_1{\partial }_2-x_2{\partial }_1)$ annihilates the radial factor. It gives $L_3{x}_1=i\hslash x_2$, $L_3{x}_2=-i\hslash x_1$, and $L_3{x}_3=0$, henceThus these components, when nonzero, are the specified eigenfunctions with eigenvalues $0,+\hslash ,-\hslash$. This is an example of angular momentum of a radial function times a linear polynomial.

On the torus ${\mathbb{T}}^n=(\mathbb{R}/2\pi \mathbb{Z}{)}^n$, choose normalized Haar measure, so that $e_k(x)=e^{ik\cdot x}$, $k\in {\mathbb{Z}}^n$, is an orthonormal basis of $L^2$. A periodic Sobolev space is defined, for any real $s$, byFor negative $s$, the coefficients describe distributions rather than necessarily functions. For a nonnegative integer $s$, the weight is comparable to ${\sum }_{|\alpha |\le s}|k^{\alpha }{|}^2$, so this norm is equivalent to the sum of the squared $L^2$ norms of all weak derivatives through order $s$. Fourier truncation proves that smooth trigonometric polynomials are dense. The $L^2$ pairing identifies the continuous dual of $H^s$ with $H^{-s}$, giving Sobolev duality.

Two Fourier estimates explain the usefulness of these spaces. If $s>m+n/2$, the Cauchy-Schwarz inequality givesUniform convergence of the differentiated Fourier series proves the Sobolev embedding $H^s\hookrightarrow C^m$. If $\delta >0$, a bounded subset of $H^{s+\delta }$ has uniformly small high-frequency tails in $H^s$:Finite-dimensional Fourier truncation therefore proves the compact embedding $H^{s+\delta }\hookrightarrow H^s$, the periodic version of Rellich-Kondrachov compactness theorem. In addition, for $s>n/2$ multiplication is continuous $H^s\times H^s\to H^s$, making it a Sobolev algebra. Indeed the weighted coefficient of a product is bounded by the sum of the convolutions of $⟨k{⟩}^s|\hat{u}(k)|$ with $|\hat{v}(k)|$ and of $|\hat{u}(k)|$ with $⟨k{⟩}^s|\hat{v}(k)|$; Cauchy-Schwarz puts unweighted coefficients in ${\ell }^1$, and the ${\ell }^1\ast {\ell }^2\to {\ell }^2$ case of Young's convolution inequality finishes the estimate.

For eigenfunction expansions, distinguish ellipticity from self-adjointness. An arbitrary non-self-adjoint elliptic operator need not have an orthonormal basis of eigenfunctions. A useful definite setting iswith smooth real periodic coefficients, $a_{ij}=a_{ji}$, real $V$, and uniform ellipticityTake the self-adjoint realization in $L^2$ with domain $H^2$. Its construction and spectral properties can be seen from the associated form on $H^1$,For a sufficiently large constant $C$, $b_C=b+C(\cdot ,\cdot )$ is a positive coercive form: $b_C(u,u)\ge c_1\Vert u{\Vert }_{H^1}^2$, and $L+C\ge I$. The Lax-Milgram theorem supplies the unique weak solution of $(L+C)u=f$ for $f\in H^{-1}$; the form is Hermitian, which gives symmetry of its inverse on $L^2$.

The regularity input is the periodic elliptic estimateHere is why an elliptic operator gains two derivatives. With coefficients frozen at a point, its principal symbol $\sum a_{ij}{\xi }_i{\xi }_j$ dominates $c|\xi {|}^2$, and the estimate follows by multiplying Fourier coefficients by that symbol. On a sufficiently small coordinate patch, the difference of the leading coefficients from these constants has arbitrarily small supremum norm. In the integer-order differentiated estimate, this absorbs the highest derivatives into the left side; derivatives falling on coefficients contribute only lower-order terms. A partition of unity reduces the variable-coefficient problem to such patches. Its commutators with $L$ have order one, controlled bywhich follows by splitting low and high Fourier modes. Absorbing these terms gives the displayed estimate. For a weak $H^1$ solution, the same localization with difference quotients first bounds its second weak derivatives for $f\in L^2$; differentiating and repeating then gives $f\in H^s\Rightarrow u\in H^{s+2}$ for nonnegative integer $s$. Interpolation and duality extend the scale to real orders. In particular $(L+C{)}^{-1}:L^2\to H^2$ is bounded.

Let $R=(L+C{)}^{-1}$. Compactness of $H^2\hookrightarrow L^2$ makes $R$ a compact operator. Symmetry of the form makes $R$ self-adjoint, and positivity of $L+C$ makes $R$ positive and injective. Its range is dense because $\ker R^{\ast }=\ker R=0$. The spectral theorem for compact self-adjoint operators consequently gives an orthonormal basis ${\varphi }_j$ withElliptic regularity bootstraps every ${\varphi }_j$ to a smooth function. Each eigenvalue has finite multiplicity. Thus a self-adjoint uniformly elliptic second-order operator on the torus has a complete smooth orthonormal eigenfunction expansion with discrete eigenvalues tending to infinity.

Writing $u_j=(u,{\varphi }_j)$, every $u\in L^2$ has $u={\sum }_j{u}_j{\varphi }_j$ in $L^2$, with $\Vert u{\Vert }_2^2={\sum }_j|u_j{|}^2$. Iterating the elliptic estimate shows that the graph norm of $(L+C{)}^m$ is equivalent to the $H^{2m}$ norm. Interpolation and duality give the more general spectral descriptionThe comparison constants depend on the operator and $s$. In particular smooth functions have coefficients decaying faster than every power of ${\lambda }_j+C$, and their expansions converge in every Sobolev norm and hence in every $C^m$ norm.

This gives a practical way to solve elliptic equations. The equation $Lu=f$ is solvable if and only if $f_j=0$ whenever ${\lambda }_j=0$; then one may take $u_j=f_j/{\lambda }_j$ on the other eigenspaces and add an arbitrary vector in $\ker L$. The Sobolev description proves the two-derivative gain and the Fredholm alternative, rather than leaving them as formal series manipulations. Similarly the solution of the heat equation ${\partial }_{t}u+Lu=0$ isFor every $t>0$ the exponential decay at high eigenvalues places this solution in every Sobolev space, hence it is smooth.

For the Laplacian $-\Delta$ the eigenfunctions are precisely $e^{ik\cdot x}$ with eigenvalues $|k{|}^2$. The number with $|k{|}^2\le R$ is the number of lattice points in a radius-$\sqrt{R}$ ball, asymptotic to its volume by comparison with unit cubes. For the general self-adjoint operator above, bounds on its form above and below by positive multiples of $\Vert \nabla u{\Vert }_2^2$ plus constants, together with the Courant–Fischer min-max principle, imply ${\lambda }_j+C\asymp j^{2/n}$. Thus the growth reflects dimension and the two-derivative order. Compactness of the torus is essential: on ${\mathbb{R}}^n$, translated bumps prevent the compact embedding, and the Laplacian instead has continuous spectrum described by the Fourier transform.

We construct the two domains explicitly from seven congruent triangles. Let a scalene reference triangle have sides $a,b,c$ opposite its three labelled vertices. Choose its three side lengths in a small open neighbourhood of $(1,1,1)$, with a fixed strict ordering. Thus the three angles are distinct and lie between $\pi /4$ and $\pi /2$. Each copy keeps these side labels.

For the first assembly glue the following tile pairs, reflecting across the indicated side:Each coloured adjacency graph is a tree with central tile $7$ and three two-tile arms. Place tile $7$ in the plane and obtain every other tile by the prescribed reflections. At the equilateral reference shape the copies have disjoint interiors and form a polygonal disk; the same holds throughout a sufficiently small neighbourhood, since the only contacts between tiles are the prescribed edges and their incident vertices. In particular each assembly is a bounded connected planar domain.

To prove equality of spectra, index tiles by the seven nonzero vectors $x\in {\mathbb{F}}_2^3$, using $x=(x_1,x_2,x_3)$ with label $x_1+2x_2+4x_3$. LetThese involutions act on points by $x\mapsto s_{s}x$ and on nonzero covectors by $y\mapsto s_s^{-T}y=s_s^{T}y$. Their swaps are exactly the two columns of the table. Write $P_s,Q_s$ for the corresponding permutation matrices. The Fano plane incidence matrixsatisfies $B{P}_s=Q_{s}B$, since incidence is preserved by a linear transformation and its inverse transpose. Each point lies on three lines, while two different points have one common line. Thereforewhere $J$ is the all-ones matrix, proving that $B$ is invertible.

For the Dirichlet boundary condition, a glued side has a positive swap and an exposed side has diagonal entry $-1$. Both trees have the same bipartition, with sign matrixTheir signed reflection matrices are $M_s=-D{P}_{s}D$ and $N_s=-D{Q}_{s}D$. On a glued edge its endpoint signs are opposite, so the off-diagonal entries are $+1$; on an exposed edge the entry is $-1$. Consequently the invertible matrix $C=DBD$ satisfies

We spell out why this gives isospectrality. Pull each function back from the seven tiles to the reference triangle. Its seven boundary values along side $s$ form a vector $u_s$ satisfying $M_s{u}_s=u_s$: this imposes equal values at glued edges and zero values at exposed edges. Outward normal derivatives of a smooth eigenfunction satisfy $M_s{v}_s=-v_s$, giving opposite normal derivatives on glued edges. The identities above preserve both conditions under $u\mapsto Cu$. Since $C$ is constant, it also commutes with the scalar Laplacian on each triangle.

One can include all weak eigenfunctions without any corner regularity issue by orthogonalization of a transplantation matrix. The matrix $C^{T}C$ commutes with the symmetric $M_s$, so $O=C(C^{T}C{)}^{-1/2}$ is orthogonal and still intertwines $M_s,N_s$. On the disjoint tiles, $O$ preserves the $L^2$ norm and the sum of the Dirichlet energies; it maps the trace-matching form domain $H_0^1({\Omega }_P)$ bijectively to $H_0^1({\Omega }_L)$. Thus it is a unitary equivalence of the Dirichlet Laplacians, proving equality of every eigenvalue with its multiplicity.

Finally, the only reentrant vertices of either polygon are the three vertices of tile $7$, with interior angles four times the corresponding reference-triangle angles. All other boundary angles are less than $\pi$. These three distinct angles force any hypothetical isometry to map each labelled central vertex to its counterpart. An isometry between connected open planar domains is a restriction of a rigid motion: its derivative is orthogonal, and preservation of the flat Levi-Civita connection makes that derivative constant. After aligning the two central triangles, this rigid motion fixes three noncollinear points and is the identity. But the two domains do not coincide: the tile beyond side $a$ of the central triangle has its next attachment along $b$ in the point assembly and along $c$ in the line assembly. These occupy different open regions. This contradiction proves nonisometry.

Therefore the three freely varying side lengths give a three-parameter family of nonisometric Dirichlet-isospectral bounded planar domains. The parameters include scale; no congruence normalization removes it.

Use the streamfunction convention $u={\psi }_z$, $w=-{\psi }_x$, and take $v=0$. With $J(A,B)=A_x{B}_z-A_z{B}_x$, the material derivative is $D={\partial }_t-J(\psi ,\cdot )$. This convention differs from the horizontal quasi-geostrophic streamfunction convention used below. The nonlinear Boussinesq equations areThe $y$ component of vorticity is $\zeta =u_z-w_x={\nabla }^2\psi$. Differentiate the first momentum equation in $z$ and subtract the $x$ derivative of the second. The pressure terms cancel; the derivatives of the advecting velocity cancel by $u_x+w_z=0$. Thus $D\zeta =-{\sigma }_x$, orConservation of total buoyancy, with background gradient $N^2(z)$, givesBoth equations retain the nonlinear advection terms.

Write $\psi =\bar{\psi }(z)+\varphi$, with ${\bar{\psi }}_z=U(z)=\bar{u}(z)$, and take the background buoyancy anomaly to be zero. The linear equations for $(\varphi ,\sigma )$ areFor a normal mode $\varphi =\hat{\psi }(z)e^{ik(x-ct)}$, the second equation gives $\hat{\sigma }=N^2\hat{\psi }/(U-c)$. Substitute this into the first to obtain the Taylor–Goldstein equationHere $k\ne 0$ and the calculation is initially away from critical levels $U=c$. The first term is the restoring effect of stable buoyancy on vertically displaced parcels. The second comes from advection of the background vorticity gradient $U^{\prime \prime }$: a displacement creates a vorticity anomaly, whose induced velocity acts back on the displacement. This is the shear-flow counterpart of a Rossby wave restoring mechanism; its sign can also permit instability, so it is not invariably stabilizing.

Rigid boundaries imply $w=-{\varphi }_x=0$, hence $\hat{\psi }(0)=\hat{\psi }(\pi H)=0$. For a real, regular normal mode, multiply the Taylor–Goldstein equation by $\hat{\psi }$, integrate, and integrate its second derivative by parts. The endpoint term vanishes, leavingFor a smooth real dispersion branch, differentiate this identity with respect to $k$. The stated stationarity of $I$ removes the term involving $d\hat{\psi }/dk$, so $2k=I_{c}dc/dk$. Direct differentiation at fixed $\hat{\psi }$ givesSince the frequency is $kc$, the group velocity is $d(kc)/dk$, and thereforeThis is variational group velocity of a Taylor-Goldstein mode. Singular critical levels or a vanishing denominator require separate limiting analysis; no stationarity proof is needed here.

For the sinusoidal profiles, choose $c=0$. In the interior,The profile ratios extend to the endpoints in this reduced equation. Thus $n=1$ always gives a nonzero real $k$, with $k_1^2=N_0^2/U_0^2$. For this normal mode,The denominator in the group velocity expression is $2H(2N_0^2/U_0^3+1/(U_0{H}^2))$. It follows thatAlthough $U$ vanishes at the endpoints, this weighted integral is finite; the vanishing of ${\hat{\psi }}_1$ supplies the required cancellation. The formula is the limiting derivative approached from the regular real branch with $c<0$, before any interior critical level is encountered.

If instead $N=N_0$ and $U=U_0$ are constant, still take $c=0$. The same vertical eigenfunctions satisfy the Taylor–Goldstein equation, now withThere is no value of $n\ge 1$ guaranteed to give real $k$ for arbitrary positive parameters: if $N_{0}H<U_0$, all these squared wavenumbers are negative. A nonzero real $k$ exists exactly when $N_{0}H>U_0$; equality gives only the degenerate $n=1$, $k=0$ limit. For a real branch the same group velocity calculation gives $c_g=k_n^2{U}_0^3/N_0^2$.

To test exact nonlinear validity, take a real normal mode $\varphi =A\sin (nz/H)\cos (kx)$ and its corresponding buoyancy perturbation $\sigma =a(z)\varphi$, where $a=N^2/(U-c)$. Becausethe quadratic vorticity self-advection is zero: $J(\varphi ,{\nabla }^2\varphi )=0$. The quadratic buoyancy self-advection, however, isThis nonlinear exactness test for a stratified streamfunction mode distinguishes the cases. For constant $N,U$, $a=N_0^2/U_0$ is constant, so both quadratic terms vanish. The background-mode terms already satisfy the linear equations, hence the constant-profile modes are exact finite-amplitude solutions of the ideal Boussinesq equations. Pressure can be recovered from the momentum equations: the vorticity equation supplies their compatibility condition in this simply connected strip.

For the sinusoidal profiles, $a=(N_0^2/U_0)\sin (z/H)$ has a nonzero derivative in the interior. For every nonzero $A$ and nonzero real $k$, $a^{\prime }\varphi {\varphi }_x$ is not identically zero; it produces a second horizontal harmonic that is absent from the proposed solution. Thus none of these nontrivial sinusoidal-profile modes is an exact finite-amplitude solution. A $k=0$ field is merely an added horizontal shear with no vertical motion and is a degenerate exception to the wave calculation, not a propagating member of it.

Use the nonnegative Laplacian convention $\tilde{\Delta }=-{\sum }_{j=1}^{d+1}{\partial }_j^2$ and $\Delta =-\mathrm{div}\mathrm{grad}$ on the unit sphere. In polar coordinates the Euclidean Riemannian metric is $d{r}^2+r^2{g}_{S^d}$, its volume form is $r^{d}drd{V}_{S^d}$, and therefore the Laplacian in polar coordinates isConsequently the required restriction identity isFor the opposite sign convention all eigenvalues below change sign, and the restriction identity changes accordingly.

Let ${\mathcal{P}}_n$ be the vector space of degree-$n$ homogeneous polynomials in $d+1$ variables, and let ${\mathcal{H}}_n$ be its subspace of harmonic polynomials. Write $H(r\omega )=r^{n}Y(\omega )$ for $H\in {\mathcal{H}}_n$. Substitution in the Laplacian identity givesThus each restricted harmonic polynomial is a spherical harmonic with eigenvalue ${\lambda }_n=n(n+d-1)$.

To count these spherical harmonics, give polynomials the Fischer inner product $⟨x^{\alpha },x^{\beta }{⟩}_F=\alpha !{\delta }_{\alpha \beta }$. Multiplication by $x_j$ is adjoint to ${\partial }_j$, so multiplication by $r^2$ is adjoint to the ordinary signed Laplacian ${\sum }_j{\partial }_j^2$. In these finite-dimensional inner product spaces, the orthogonal complement of $r^2{\mathcal{P}}_{n-2}$ is exactly ${\mathcal{H}}_n$. Hence the Fischer decomposition isMultiplication by $r^2$ is injective, while restriction of a homogeneous polynomial to the sphere is injective: homogeneity recovers its value at every nonzero point. Counting monomials now gives the eigenvalue multiplicitywhere the second term is zero for $n<2$.

It remains to establish that these are all the eigenvalues and their full multiplicities. Iterating the Fischer decomposition expresses every restricted polynomial as a sum of the restricted harmonic polynomials. Restricted polynomials separate points and contain the constants, so the Stone-Weierstrass theorem makes their span dense in continuous functions on the sphere, hence dense in its L2 space. The Laplace-Beltrami operator is self-adjoint, and the distinct numbers ${\lambda }_n$ strictly increase for $d\ge 1$. Its different eigenspaces are therefore orthogonal. Any additional eigenfunction, or any additional vector in one of these eigenspaces orthogonal to its known spherical harmonics, would be orthogonal to a dense subspace and hence zero. The standard compact elliptic spectral theorem then shows that the spectrum of the round sphere is precisely ${\lambda }_n=n(n+d-1)$ with multiplicity $m_n$. For $S^1$ this gives multiplicity one at zero and two at $n^2$ for $n\ge 1$. If the degenerate case $d=0$ is included, $S^0$ has two points and its Laplacian is zero on a two-dimensional function space.

The discrete Bochner–Herglotz theorem says that a sequence $r:\mathbb{Z}\to \mathbb{C}$ is a positive-definite function, meaningif and only if there is a unique finite positive Borel measure $\sigma$ on $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ such thatThe mass is $\sigma (\mathbb{T})=r(0)$. The necessity of positive definiteness follows by integrating $|{\sum }_j{c}_j{e}^{2\pi ijt}{|}^2$.

For sufficiency, positive definiteness makes the relevant Gram matrices Hermitian positive semidefinite, so $r(-n)=\overline{r(n)}$ and $|r(n)|\le r(0)$. DefineTaking $c_j=e^{-2\pi ijt}$ in the positive-definiteness condition gives $p_N(t)\ge 0$, while ${\int }_{\mathbb{T}}{p}_N(t)dt=r(0)$. Moreover,Thus the integrals against every trigonometric polynomial converge. These polynomials are uniformly dense in continuous functions on the circle: smooth periodic approximation followed by Fourier inversion for smooth functions gives this density. The uniform mass bound $r(0)$ then makes the integrals against every continuous $\phi$ converge as well. Their limit is a positive linear functional $L$ with $|L(\phi )|\le r(0)\Vert \phi {\Vert }_{\infty }$. The standard Riesz-Markov-Kakutani representation theorem gives a finite positive Borel measure $\sigma$ with $L(\phi )=\int \phi d\sigma$. The preceding coefficient computation proves its required Fourier coefficients. Finally, two such measures agree on all trigonometric polynomials, hence on all continuous functions by uniform density, and therefore are identical. This proves the Bochner–Herglotz theorem, including existence and uniqueness; if $r(0)=0$, the estimate $|r(n)|\le r(0)$ gives the zero measure directly.

Now let $U$ be the Koopman operator $Uf=f\circ T$ on the complex L2 space of the probability measure-preserving system. We use $⟨f,g⟩=\int f\overline{g}d\mu$, linear in the first entry. Since $U$ is an isometry, the sequence $r_f(n)=⟨U^{n}f,f⟩$ for $n\ge 0$, extended by $r_f(-n)=\overline{r_f(n)}$, is positive-definite: its matrix is the Gram matrix of $f,Uf,\dots$. The Bochner–Herglotz theorem gives its spectral measure of a Koopman observable ${\sigma }_f$.

We first prove exactly how an atom of a measure detects an eigenfunction. Fix $\zeta =e^{2\pi i{t}_0}$ and setThe mean ergodic theorem, applied to the isometry $V={\zeta }^{-1}U$, says these averages converge in L2 space to the orthogonal projection $P_{\zeta }f$ onto $\ker (U-\zeta I)$. Its usual Hilbert-space proof works without invertibility: the orthogonal complement of $\overline{\mathrm{ran}(I-V)}$ is $\ker (I-V^{\ast })=\ker (I-V)$, and the averages of $(I-V)h$ telescope to $(h-V^{N}h)/N\to 0$. The equality of the two kernels follows from $V$ being an isometry, by expanding $\Vert Vh-h{\Vert }^2$ when $V^{\ast }h=h$.

Expanding the averaged norm through the spectral measure of a Koopman observable givesThe integrand is at most one and converges to $1_{\{t_0\}}$. By the dominated convergence theorem,Thus a nonzero atom corresponds to a nonzero Koopman operator eigenfunction, with eigenvalue $\zeta$, in the cyclic spectral data of $f$.

We also need the Wiener Fourier coefficient lemma, which here has a short proof. For any finite positive circle measure $\sigma$ with coefficients $r(n)$,The bracket is bounded in absolute value by one and tends to the indicator of the diagonal $t=s$. Another application of the dominated convergence theorem givesIn particular the limit is zero for an atomless measure.

Suppose there are no nonconstant Koopman operator eigenfunctions. For every $f$ in the mean-zero L2 space, all its projections $P_{\zeta }f$ vanish: a nonzero projection would be a nonconstant eigenfunction, since it has mean zero. Hence ${\sigma }_f$ is atomless, and the Wiener Fourier coefficient lemma givesThe Cauchy-Schwarz inequality also gives vanishing averaged absolute autocorrelations. For two mean-zero observables $f,g$, the sesquilinear polarization identitythen gives vanishing averaged absolute cross-correlations. Taking $f=1_A-\mu (A)$ and $g=1_B-\mu (B)$ proveswhich is weak mixing.