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Use the unnormalized Fourier transform in the question and the Plancherel theorem, with inverse factor $1/(2\pi )$. For a general $L^2$ scaling function, the transform is interpreted in the $L^2$ sense; the printed integral need not be absolutely convergent. All frequency identities are almost-everywhere statements.

Changing variables in the Fourier transform givesThus taking the transform of the scaling refinement equation gives $f(\xi )=m(\xi /2)f(\xi /2)$, orConversely the same calculation and injectivity of the Fourier transform recover refinement, with convergence understood in $L^2$.

To justify both the orthogonality assertion and this convergence precisely, put $P(t)={\sum }_{k\in \mathbb{Z}}|f(t+2\pi k){|}^2$. It belongs to $L^1[-\pi ,\pi ]$ by monotone integration. The Plancherel theorem givesHence orthonormality of all integer translates is equivalent, by uniqueness of Fourier coefficients in L1, to the periodized-energy conditionWhen this holds, the squared norm of ${\sum }_n{a}_n\varphi (2\cdot -n)$ is $\frac{1}{2}{\sum }_n|a_n{|}^2$, so square-summable coefficient series converge in $L^2$. In the converse direction the refinement identity and $P=1$ give, by splitting the periodization of $|f(2t){|}^2$ into even and odd translates,Thus $m$ is bounded and its Fourier coefficients $a_n/2$ are square summable. If $m_N$ are its Fourier partial sum, thenThis verifies the transformed refinement series converges to $f$, completing the equivalence of the two pairs of conditions without imposing an unnecessary $L^1$ assumption on $\varphi$.

The half-open intervals $[-\pi +2\pi k,\pi +2\pi k)$ tile the line, so exactly one term contributes to ${\sum }_k|f(t+2\pi k){|}^2$. It equals one almost everywhere.

Choose the Shannon scaling mask, a $2\pi$-periodic low-pass filter of a multiresolution analysis which equals one on $[-\pi /2,\pi /2)$ and zero on the rest of $[-\pi ,\pi )$. On the support of $f$ its product with $f(t)$ equals $f(2t)$; outside that support both sides vanish. Its Fourier coefficients giveso it also has the required symbol representation. The inverse Fourier transform gives the Shannon scaling functionThis sinc function has $L^2$ norm one. It illustrates why the Fourier transform convention in part B must allow $L^2$ transforms: the Shannon scaling function is not absolutely integrable on the line.

Use the Fourier coefficientsThey are continuous for $0\le y\le 1$ and bounded by $2\Vert u{\Vert }_{\infty }$. On compact subintervals of $0<y<1$, the equation is uniformly elliptic with smooth coefficients. Local elliptic regularity at the flat vertical sides with zero boundary values permits differentiation and integration by parts there. ThusFor completeness, this identity also follows in the distributional sense without assuming boundary derivatives initially. Integrate on $(\epsilon ,\pi -\epsilon )$ against $\sin (nx)$ and a test function of $y$ with support away from zero and one. The interior gradient estimate near a zero Dirichlet side gives $|u_x|\le C{\epsilon }^{-1}\sup |u|$ on a nearby ball of radius comparable to $\epsilon$; continuity and the zero side value make that supremum $o(1)$ uniformly over the support in $y$. Therefore the boundary products $u_x\sin (nx)$ and $u\cos (nx)$ tend to zero. Move the $y$ derivatives to the test function before taking $\epsilon \downarrow 0$. This proves the ordinary differential equation, which makes each $b_n$ smooth in $0<y<1$ and yields the same classical identity.

The Cauchy-Euler differential equation has indicial rootsFor $n\ge 1$ the negative root is strictly negative. Boundedness as $y\downarrow 0$ forces $B_n=0$. Continuity at $y=1$ identifiesBecause $p_n^{+}>n$, the series ${\sum }_{n\ge 1}{A}_n{y}^{p_n^{+}}\sin (nx)$ converges absolutely and uniformly for $0\le y\le \rho <1$. On compact subsets with $y>0$, differentiated series converge as well. At each fixed $0<y<1$, its Fourier coefficients equal those of $u(\cdot ,y)$; completeness of the Fourier sine basis in $L^2(0,\pi )$ and continuity in $x$ make the two functions equal. ThereforeThis is also the requested sum over integers: take all coefficients with $n\le 0$ to be zero. Negative indices give redundant Fourier modes, and the $n=0$ sine vanishes. No pointwise convergence of the ordinary Fourier series on the top boundary is needed.

An important consequence is a forced zero trace at a quadratically degenerate boundary. Indeed the absolute sum is bounded by $2\Vert u{\Vert }_{\infty }{\sum }_{n\ge 1}{y}^n=2\Vert u{\Vert }_{\infty }y/(1-y)$ for $y<1$, which tends to zero as $y\downarrow 0$, uniformly in $x$. Thus $u(x,0)=0$ on the entire lower side. It follows also from $b_n(0)=0$ for every sine coefficient and completeness.

The linear elastic wave equation is $\rho {\mathbf{u}}_{tt}=(\lambda +\mu )\nabla (\nabla \cdot \mathbf{u})+\mu \Delta \mathbf{u}$. For a shear-horizontal wave, take $\mathbf{u}=(0,v(x,z,t),0)$ with no $y$ dependence. Its divergence vanishes, andRigid fixed boundaries require $v=0$ at $z=0,h$. The modes are $v=A\sin (q_{n}z)\cos (kx-{\omega }_{n}t)$ with $q_n=n\pi /h$, $n=1,2,\dots$, andThere is no $n=0$ displacement mode with these clamped boundary conditions. For $k>0$, the phase velocity and group velocity areThe phase velocity decreases from infinity to $c_s$, while the group velocity increases from zero to $c_s$; their product is $c_s^2$.

For a mode of frequency $\omega$, the appropriate time and cross-sectional average is $⟨A⟩=\omega /(2\pi h){\int }_0^{2\pi /\omega }{\int }_0^{h}Adzdt$, per unit transverse width. Its average kinetic energy density is $\rho A^2{\omega }^2/8$. The only nonzero strains are $e_{xy}=e_{yx}=v_x/2$ and $e_{yz}=e_{zy}=v_z/2$, while $e_{kk}=0$. Thus the elastic energy density is $\mu (v_x^2+v_z^2)/2$, whose average is $\mu A^2(k^2+q_n^2)/8$. The dispersion relation proves equality of average kinetic and elastic energies.

For the ray asymptotics of a clamped elastic waveguide mode, expand the localized initial displacement in transverse sine modes and Fourier transform in $x$. For a displacement released from rest, each mode evolves with $\cos ({\omega }_n(k)t)$. At $x=Vt$ the oscillatory phases are $t(kV\pm {\omega }_n(k))$. If $0<|V|<c_s$, one phase has a stationary point withThe stationary phase method gives a generic oscillatory amplitude proportional to $t^{-1/2}$ for each contributing mode. If the initial Fourier coefficient vanishes at the stationary point, that mode can decay faster; smooth localized data make the modal sum well behaved. If $|V|>c_s$, there is no stationary point. Repeated integration by parts gives rapid decay for smooth rapidly decaying data; for initially compactly supported displacement, finite propagation speed makes the displacement on that ray exactly zero after a sufficiently long time. These are the requested subsonic and super-shear-speed regimes.

A continuous map $F$ of a compact metric space is uniquely ergodic if it has exactly one invariant measure that is a Borel probability measure. The uniqueness requirement ranges over all invariant Borel probability measures.

For an irrational rotation of the circle, let $R_{\alpha }x=x+\alpha$ on $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, and let $\nu$ be any invariant Borel probability measure. For each integer $r$, set ${\chi }_r(x)=e^{2\pi irx}$. Invariance givesFor $r\ne 0$, irrationality forces $e^{2\pi ir\alpha }\ne 1$, so the corresponding Fourier coefficient is zero. For $r=0$ it is one. These are exactly the Fourier coefficients of normalized Lebesgue measure $m$. By the Stone-Weierstrass theorem, trigonometric polynomials are uniformly dense in the continuous functions on the circle group; hence $\nu$ and $m$ integrate every continuous function equally and are the same Borel probability measure. Since $m$ is invariant, the irrational rotation of the circle is uniquely ergodic, with unique measure $m$.

For the irrational skew shift on the two-dimensional torus, writeThe map is invertible, with $T^{-1}(x,y)=(x-\alpha ,y-x+\alpha )$ modulo one, and it preserves $m_2$ as allowed in the question. We first prove the ergodic transformation property by the invariant-function characterization of ergodicity.

Let $F\in L^2(m_2)$ satisfy $F\circ T=F$. Its expansion in the Fourier basis is $F={\sum }_{r,s}{c}_{r,s}{e}_{r,s}$ in $L^2$. Direct calculation of the Koopman operator givesUniqueness of the Fourier coefficients therefore impliesFor $s\ne 0$, the magnitudes of the Fourier coefficients along all distinct indices $(r+js,s)$, $j\in \mathbb{Z}$, are equal. By the Bessel inequality they are square summable, so every such coefficient must be zero. When $s=0$, the relation becomes $c_{r,0}=e^{2\pi ir\alpha }{c}_{r,0}$, which forces $c_{r,0}=0$ for $r\ne 0$. Only $c_{0,0}$ remains. Thus every invariant $L^2$ function is constant, and

To prove unique ergodicity, we will establish uniform averages for every continuous function directly. No theorem on unique ergodicity of skew products is needed. Induction on $n$, using the old first coordinate in the second coordinate of $T$, yieldsIn particular, for a Fourier basis element,If $s=0$ and $r\ne 0$, this is a geometric series in $n$, anduniformly in $(x,y)$.

For $s\ne 0$, we give the finite Van der Corput inequality for finite scalar sequences and its proof. For $|u_n|\le 1$, extend $u_n$ by zero outside $0\le n<N$. Fix an integer $1\le H\le N$ and set $C_h={\sum }_{n=0}^{N-h-1}{u}_{n+h}\overline{u_n}$. Every original summand appears $H$ times in the identityThe Cauchy-Schwarz inequality, followed by expansion of the squared window sums, givesConsequently the Van der Corput inequality for finite scalar sequences isThe finite prefactor is important; the order of limits will be $N\to \infty$ with $H$ fixed, followed by $H\to \infty$.

For the irrational skew shift character sequence above, differencing cancels the quadratic term:For every fixed $h\ge 1$, $s\alpha h$ is irrational, so another geometric series estimate givesuniformly in $(x,y)$. With fixed $H$, the Van der Corput inequality for finite scalar sequences therefore impliesLetting $H\to \infty$ proves that the averages of every nonconstant Fourier basis element converge uniformly to zero. The constant character has average one. This establishes uniform equidistribution of an irrational skew shift on all trigonometric polynomials.

The Stone-Weierstrass theorem makes these trigonometric polynomials uniformly dense in $C({\mathbb{T}}^2)$. If $p$ approximates a continuous $g$ with $\Vert g-p{\Vert }_{\infty }<\epsilon$, thenTaking $N\to \infty$ and then $\epsilon \downarrow 0$ proves uniform convergence to $\int gd{m}_2$ for every continuous $g$.

Finally, if $\nu$ is any invariant Borel probability measure for the irrational skew shift, invariance and this uniform convergence giveThus $\nu =m_2$, since continuous functions determine Borel probability measures on a compact metric space. We concludeThe uniform-average proof also shows that every starting point has the same limiting continuous-function averages, a stronger conclusion than the almost-everywhere assertion provided by the pointwise ergodic theorem.

A flat torus is the quotient ${\mathbb{R}}^d/\Lambda$, where $\Lambda$ is a full-rank Euclidean lattice, with the Riemannian metric induced from the Euclidean metric. Translation by a lattice vector is a Riemannian isometry, so this metric is well defined and has zero curvature. Its volume is the covolume of $\Lambda$.

The dual lattice is ${\Lambda }^{\vee }=\{w:⟨w,v⟩\in \mathbb{Z}\text{ for all }v\in \Lambda \}$. For $w\in {\Lambda }^{\vee }$, the function $e_w(x)=\exp (2\pi i⟨w,x⟩)$ descends to the flat torus, and direct differentiation givesConversely, write $\Lambda =B{\mathbb{Z}}^d$ and $x=By$. Ordinary Fourier series on ${\mathbb{R}}^d/{\mathbb{Z}}^d$ have frequencies $n\in {\mathbb{Z}}^d$, corresponding to $w=B^{-\mathsf{T}}n\in {\Lambda }^{\vee }$. After normalization they form a complete orthonormal basis in $L^2$, and the Fourier coefficients of a smooth function decay rapidly. Applying the Laplace-Beltrami operator term by term shows that a smooth eigenfunction with eigenvalue $\lambda$ has nonzero coefficients only where $4{\pi }^2|w{|}^2=\lambda$. Thus the spectrum of a flat torus isThe formula counts multiplicity over either the real or complex numbers: the two frequencies $w,-w$ give the real sine and cosine functions.

For the rigidity assertion in dimension two, it suffices to reconstruct a rank-two Euclidean lattice $L={\Lambda }^{\vee }$ from its vector-length multiset. The multiset first determines its covolume $A$. Indeed, a bounded fundamental parallelogram, and comparison of the cells meeting a disk with slightly larger and smaller disks, give the lattice-point asymptotics by fundamental cellsThe spectrum determines $N_L(R)$, including multiplicities, so it determines $A$.

Let $a$ be the smallest nonzero vector length, and choose $v\in L$ with $|v|=a$. This vector is primitive: $v=ku$ with $|k|>1$ would contradict minimality. From the length multiset subtract the two vectors $kv,-kv$ at every positive length $ka$, for $k=1,2,\dots$. The shortest remaining length $b$ is exactly the length of a shortest vector $w\notin \mathbb{Z}v$. This subtraction is legitimate even if several directions have length $a$: it subtracts just two copies, and then $b=a$. It also proves that $b$ is independent of our choice of shortest direction. Replace $w$ by $w-kv$ to arrangeThe replacement stays outside $\mathbb{Z}v$ and cannot be shorter than $b$, so a reduced choice still has length $b$.

The shortest-independent-vector basis lemma says that $v,w$ are a basis of $L$. To prove it, choose coordinates with $v=(a,0)$. Because $v$ is primitive, $L/\mathbb{Z}v\cong \mathbb{Z}$; let $h>0$ be the smallest positive vertical coordinate in $L$. Write $w=(x,mh)$, with $m\ge 1$ after changing its sign. If $m\ge 2$, a vector $u$ of vertical coordinate $h$, reduced horizontally modulo $v$, hasThis contradicts the definition of $b$, since $u\notin \mathbb{Z}v$. Hence $m=1$ and $v,w$ generate $L$.

The Gram matrix of this basis is $\left(\begin{array}{cc}{a}^2& c\\ c& b^2\end{array}\right)$, with $c=v\cdot w$, and its determinant is $A^2$. ThereforeChanging the sign of one basis vector allows $c\ge 0$. The numbers $a,b,A$, all audible from the spectrum, consequently determine the Gram matrix, and hence $L$, up to an orthogonal transformation. Taking dual lattices gives the same conclusion for $\Lambda$. Isospectral flat two-dimensional tori are isometric. The argument uses the special basis property of rank-two Euclidean lattices, so it does not claim higher-dimensional rigidity.

Hecke operators give a commuting family of arithmetic symmetries of modular forms. They preserve the weight and cusp forms, and diagonalizing them turns the Fourier coefficients of a modular form into multiplicative arithmetic data. Odd weight spaces are zero by the action of $-I$; in weight zero the operators act on constants by $T_{n}1={\sigma }_{-1}(n)$. For the substantive theory, fix an even positive weight $k$ and $\Gamma =S{L}_2(\mathbb{Z})$, and use the following normalization consistently.

Extend the slash operator for modular forms to $G{L}_2^{+}(\mathbb{Q})$ byIt is a right action and positive scalar matrices act trivially. Let ${\mathcal{D}}_n$ be the integral matrices of determinant $n>0$. The determinant-n matrix representatives for Hecke operators for $\Gamma \backslash {\mathcal{D}}_n$ areInteger row operations reduce the first column to $(a,0)$, after which a row shear reduces $b$ modulo $d$; uniqueness follows from these same conditions. The Hecke operator isRight multiplication by $\Gamma$ permutes the left cosets, so $T_{n}f$ is again invariant under the weight-$k$ slash operator for modular forms. Each summand is a holomorphic function on the complex upper half-plane, and the coefficient formula below proves holomorphy at a cusp. Thus $T_n$ acts on $M_k(\Gamma )$ and preserves $S_k(\Gamma )$. Geometrically, these operators sum over index-$n$ sublattices, or equivalently the associated finite-degree correspondences between complex tori; the factors above compensate for the weight's response to lattice scaling.

Write $f={\sum }_{r\ge 0}{a}_r{q}^r$. The sum over $b$ is zero unless $d\mid r$, in which case it is $d$. Substituting into the displayed definition proves the Fourier coefficients of a composite-index Hecke operator formulawith $\gcd (n,0)=n$. In particular, $a_0(T_{n}f)={\sigma }_{k-1}(n)a_0(f)$ and $a_1(T_{n}f)=a_n(f)$. At a prime this becomeswhere $a_{r/p}=0$ if $p\nmid r$.

The Hecke multiplication relations areThey can be checked by applying the coefficient formula twice and regrouping the divisors. More explicitly, coprime indices factor independently, so $T_m{T}_n=T_{mn}$ if $(m,n)=1$. At a prime, separating the terms in the divisor sum according to whether one more factor $p$ is available givesIteration gives $T_{p^a}{T}_{p^b}={\sum }_{j=0}^{\min (a,b)}{p}^{j(k-1)}{T}_{p^{a+b-2j}}$; combining the prime factors gives the full relation. Hence all Hecke operators commute, $T_1=1$, and the Hecke algebra of modular forms is generated by the $T_p$ together with scalars.

The Petersson inner product on the finite-dimensional cusp form space isThe integrand is invariant and the integral converges because cusp forms decay exponentially at the cusp. The Hecke operators are self-adjoint for the Petersson inner product at level one. To see the underlying adjoint calculation, unfold the finite correspondence for a determinant-$n$ matrix and change variable $\tau \mapsto \alpha \tau$. The invariant hyperbolic measure and the determinant-normalized slash factors transfer $\alpha$ from one side of the inner product to ${\alpha }^{-1}$. Scalar matrices act trivially, so one can replace ${\alpha }^{-1}$ by the integral adjugate $n{\alpha }^{-1}$, which again has determinant $n$. This reverses the correspondence and permutes the same collection of terms, with the same real prefactor $n^{k/2-1}$. Summing givesCommuting self-adjoint operators on a finite-dimensional inner-product space admit a common orthonormal basis of eigenvectors. Thus $S_k(\Gamma )$ has a basis of simultaneous Hecke eigenforms, with real eigenvalues.

A nonzero simultaneous cuspidal Hecke eigenform has $a_1\ne 0$: if $T_{n}f={\lambda }_{n}f$ and $a_1=0$, comparison of the first coefficients gives $a_n={\lambda }_n{a}_1=0$ for every $n\ge 1$, so $f=0$. Normalize $a_1=1$. ThenConsequently the full Fourier series is determined by the eigenvalues, and a simultaneous eigenvalue system has a one-dimensional eigenspace. The Hecke multiplication relations translate intoIn particular the coefficients are multiplicative at coprime indices and are determined by the prime coefficients. This is the elementary multiplicity-one statement for normalized level-one Hecke eigenforms.

The Hecke eigenvalues are algebraic integers. One way to explain the arithmetic input is to use the rational structure supplied by the graded ring $M_{\ast }(\Gamma ,\mathbb{Q})=\mathbb{Q}[E_4,E_6]$. The rational cusp form space has a basis with integral Fourier coefficients, for example from appropriate monomials in $E_4,E_6,\Delta$. Its subgroup of forms whose entire expansions are integral is a full lattice: a finite initial coefficient map is injective by the valence formula for the modular group, so it embeds this subgroup into a finite-rank integer module. The coefficient formula shows that every $T_n$ preserves this lattice, and therefore has a monic integral characteristic polynomial. Its eigenvalues are algebraic integers. Simultaneous diagonalization also makes their number field finite over $\mathbb{Q}$; conjugating the rational coefficient data gives another eigenform, whose eigenvalues are again real by self-adjointness. Thus the coefficient field is totally real.

For a normalized cuspidal Hecke eigenform, define its L-function of a cusp form by $L(f,s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$. The coefficient relations yield the Euler product of a Hecke eigenformIndeed the prime-power recurrence gives ${\sum }_{r\ge 0}{a}_{p^r}{X}^r=(1-a_{p}X+p^{k-1}{X}^2{)}^{-1}$, and multiplicativity assembles these factors. These identities hold analytically in an initial right half-plane, as well as formally. For instance boundedness of $y^{k/2}|f(\tau )|$ on the full complex upper half-plane gives $a_n=O(n^{k/2})$, so $\mathrm{Re}s>k/2+1$ suffices for absolute convergence.

The Mellin transform of a cusp-form L-function relates this arithmetic to the modular transformation:Splitting at one and using $f(i/y)=i^k{y}^{k}f(iy)$ givesExponential cusp decay makes the last integral entire in $s$. This proves the analytic continuation and functional equation of the L-function of a cusp form, complementing its Euler product. The root number is $i^k=(-1{)}^{k/2}$.

Two examples show the scope of the theory. For even $k\ge 4$, the Eisenstein series $E_k$ is an eigenform with $T_n{E}_k={\sigma }_{k-1}(n)E_k$. This follows from its divisor-sum expansion and the same multiplicativity relations; its associated positive-index Dirichlet series is proportional to $\zeta (s)\zeta (s-k+1)$. The constant coefficient gives $M_k(\Gamma )=\mathbb{C}{E}_k\oplus S_k(\Gamma )$, so an eigenbasis of the cusp space together with $E_k$ gives an eigenbasis of the whole modular-form space. The ordinary Petersson inner product above is only used on the cusp space, where it converges. In weight twelve the valence formula for the modular group gives $\dim S_{12}=1$: cancellation of the first coefficients of two independent cusp forms would give a nonzero form with order at infinity at least two, exceeding $12/12$. Thus the normalized modular discriminant is a Hecke eigenform; writing $\Delta ={\sum }_{n\ge 1}\tau (n)q^n$ gives the Ramanujan tau function relations $\tau (mn)=\tau (m)\tau (n)$ for coprime $m,n$ and $\tau (p^{r+1})=\tau (p)\tau (p^r)-p^{11}\tau (p^{r-1})$.

The constant Fourier coefficient is $A_0(y,s)={\int }_0^{1}G(x+iy,s)dx$. The $m=0$ terms give $2\zeta (2s)y^s$. For fixed $m\ne 0$, unfolding the sum over $n$ givesIndeed, the substitution $u=mx+n$ introduces a factor $1/|m|$, while the translated intervals of length $|m|$ cover the real line exactly $|m|$ times, cancelling that factor. Now scale $u=|m|yt$. The beta function integral yieldswhich can be checked by inserting $(1+t^2{)}^{-s}=\Gamma (s{)}^{-1}{\int }_0^{\infty }{v}^{s-1}{e}^{-v(1+t^2)}dv$ and evaluating the inner Gaussian integral. Thus the fixed-$m$ contribution is $|m{|}^{1-2s}{y}^{1-s}\sqrt{\pi }\Gamma (s-1/2)/\Gamma (s)$. Summing over positive and negative $m$ proves the constant term of a nonholomorphic Eisenstein seriesWith the printed normalization $\xi (u)={\pi }^{-u/2}\Gamma (u/2)\zeta (u)$ of the completed Riemann zeta function, this is exactlyBoth the unfolding and the original series calculation are justified for $\mathrm{Re}s>1$. Beyond this region the identities are understood meromorphically: $G(\tau ,s)$ is the Epstein zeta function of the unit-covolume lattice $(\mathbb{Z}\tau +\mathbb{Z})/\sqrt{y}$, so question 4 supplies its continuation. The symbol $\xi$ here has the completed-zeta normalization displayed above, which has poles at zero and one; it does not include the extra polynomial factor sometimes used to define an entire xi function.

Oddness gives zero cosine coefficients. Integration by parts gives the sine Fourier coefficientsThe Fourier series of the piecewise smooth periodic extension converges to its value at interior continuity points. Termwise integration from 0 to $x$ is valid: on the interval between these points, which stays inside $(-\pi ,\pi )$, the partial sums of $\sum (-1{)}^{n+1}\sin (nt)$ are uniformly bounded by the geometric-sum formula, since $|1+e^{it}|$ is bounded away from zero. The Dirichlet test therefore gives uniform convergence there. Integrating givesThe integrated series is absolutely and uniformly convergent. ConsequentlyIndependently, the constant Fourier coefficient is $a_0={\pi }^{-1}{\int }_{-\pi }^{\pi }{x}^{2}dx=2{\pi }^2/3$. Comparison yieldsThe original sine series is not uniformly convergent on the whole open interval; that stronger assertion is unnecessary for the justified integration.

A continuous transformation $T$ of a compact metric space $X$ is uniquely ergodic when there is exactly one $T$-invariant Borel probability measure on $X$. We use the usual compact-space convention for unique ergodicity; the compactness and continuity hypotheses matter in the assertion about all starting points.

Let $R_{\alpha }(x)=x+\alpha (\mathrm{mod}1)$ on the circle group, with $\alpha$ irrational. Normalized Lebesgue measure $m$ is invariant under this irrational rotation of the circle. If $\nu$ is any invariant Borel probability measure, define its Fourier coefficients by $c_k=\int e^{2\pi ikx}d\nu (x)$. Invariance givesFor $k\ne 0$, irrationality makes $e^{2\pi ik\alpha }\ne 1$, so $c_k=0$; also $c_0=1$. These are the Fourier coefficients of $m$. Hence $\nu$ and $m$ have the same integrals against every trigonometric polynomial. Such polynomials are uniformly dense in the continuous functions by the Stone-Weierstrass theorem, so the measures agree on every continuous test function and therefore agree as Borel measures. Thus

For the general uniquely ergodic system, fix $x\in X$ and form the empirical measuresOn a compact metric space, the Borel probability measures are compact for weak convergence of probability measures. Any subsequential limit $\nu$ is invariant: for every continuous $g$,Here $g$ is bounded and $g\circ T$ is continuous, so the identity passes to the limit. Uniqueness of the invariant Borel probability measure gives $\nu =\mu$. Every subsequential limit is therefore $\mu$, and compactness implies convergence of the entire sequence. Testing against $f$ provesIn fact this proves uniform ergodic convergence for uniquely ergodic systems: if convergence were not uniform in $x$, choose $N_j\to \infty$ and $x_j$ where the discrepancy stays above a fixed positive number. The same compactness and telescoping argument applied to ${\nu }_{N_j,x_j}$ forces a subsequence to converge to $\mu$, a contradiction. This also makes clear why an almost-everywhere Birkhoff ergodic theorem alone would not establish the requested everywhere assertion. Without the compact-space hypothesis the assertion need not hold: on the discrete space $\{p\}\bigsqcup {\mathbb{Z}}_{\ge 0}$, set $T(p)=p$ and $T(n)=n+1$. The only invariant probability is ${\delta }_p$, but the continuous bounded function which is zero at $p$ and one on the integer orbit has orbit average one there.

For the decimal application put $\alpha ={\log }_{10}2$. This is irrational: if $\alpha =p/q$ with positive integers $p,q$, then $2^q=10^p=2^p{5}^p$, contradicting unique prime factorization. Writing $n\alpha =m_n+t_n$ with $m_n=\lfloor n\alpha \rfloor$ and $t_n\in [0,1)$ gives $2^n=10^{m_n}10^{t_n}$. Its leading decimal digit is seven exactly whenThe half-open upper endpoint correctly excludes powers whose leading digit is eight, including $2^3=8$.

The indicator of $I$ is not continuous, so an extra step is needed. For any $\epsilon >0$, choose continuous functions $a_{\epsilon },b_{\epsilon }$ on the circle with $0\le a_{\epsilon }\le 1_I\le b_{\epsilon }\le 1$ and $\int (b_{\epsilon }-a_{\epsilon })dm<\epsilon$, by tapering in small neighbourhoods of the two endpoints. Applying the everywhere averaging result to these functions traps the lower and upper limits of the interval frequency between their integrals. Letting $\epsilon \downarrow 0$ proves everywhere interval frequency under an irrational rotation. ConsequentlyThis is the leading-seven case of Benford frequencies for powers of an integer.

Use the usual convention that a three-term arithmetic progression must be nonconstant. There is a genuine small-case defect in the printed statement: for $n=1$, $\delta =1$, and $A=\{1\}$, neither conclusion can hold with $c_2>0$. We prove the intended Roth density-increment step for every odd integer $n\ge 3$, including the small values of $n$, rather than silently treating $n=1$ as covered. Counting constant arithmetic progressions would make the first alternative vacuous for every nonempty $A$.