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Every nonprincipal Dirichlet character modulo the prime $q$ is primitive, so its finite Fourier coefficients have modulus one off zero; the coefficient at zero is zero by character orthogonality. Fourier inversion theorem givesThe finite geometric series givesThe printed hint omits $n$ from the exponential; its literal constant summand would not obey the bound for arbitrary $N$. The geometric-series calculation proves the needed estimate independently. Pairing $a$ with $q-a$ givesThe last sum is a harmonic number. This proves the Pólya–Vinogradov inequality uniformly in $M$ and $N$; complete blocks of length $q$ also vanish by Orthogonality of Dirichlet characters.

By the Weierstrass M-test, absolute summability of the Fourier coefficients makes ${\sum }_n\hat{f}(n)e^{int}$ uniformly convergent to a continuous periodic function $H$. Termwise integration is justified by that uniform convergence, and orthogonality givesPart (ii) then implies $H=f$. Thus the Fourier series converges uniformly to the original function, not merely to some continuous limit. In particular,

Apply part (i) to the continuous difference $h=f-g$. Every Fourier coefficient of $h$ vanishes, so every Fourier partial sum is zero. The orthogonal projection convergence from part (i) therefore gives $\Vert h{\Vert }_2=0$.

If $h(t_0)\ne 0$, continuity gives an interval on which $|h|$ is bounded below by a positive number. That interval would contribute positively to $\Vert h{\Vert }_2^2$, a contradiction. HenceThe continuity hypothesis upgrades equality almost everywhere to pointwise equality.

Use normalized Fourier coefficients and the inner productThe functions $e_n(t)=e^{int}$ are orthonormal. Hence the Fourier partial sum $S_{N}f={\sum }_{|n|\le N}\hat{f}(n)e_n$ is the orthogonal projection onto the trigonometric polynomials of degree at most $N$. For any such polynomial $P$, orthogonality givesso $\Vert f-S_{N}f{\Vert }_2\le \Vert f-P{\Vert }_2$.

Given $\epsilon >0$, the permitted density result supplies a trigonometric polynomial $P$ with $\Vert f-P{\Vert }_{\infty }<\epsilon$. Once $N$ includes its degree,Therefore the Fourier partial sums converge to $f$ in the normalized $L^2$ norm, giving exactly the stated mean-square limit.

Use the transform convention $F(\lambda )={\int }_{\mathbb{R}}f(t)e^{-i\lambda t}dt$. The given Fourier inversion theorem givesHere $F$ is continuous, vanishes at both endpoints and belongs to $L^2[-\pi ,\pi ]$. It therefore defines a continuous periodic function. Its Fourier coefficient at index $-n$ is $f(n)$. By part (i),Pair this convergence with $e^{it\lambda }$. Since that function has normalized $L^2$ norm one, the Cauchy-Schwarz inequality yields, uniformly in $t$,The elementary integral is the sinc function,Thus the sampling expansion by periodic Fourier projection isThere is no pointwise interchange with an unproved Fourier series: the calculation first uses finite sums and then an $L^2$ limit.

The convergence can also be made absolute. Parseval's identity gives ${\sum }_n|f(n){|}^2=\Vert F{\Vert }_2^2$, and Bessel's inequality applied to $e^{it\lambda }$ gives ${\sum }_n|D(t-n){|}^2\le 1$. Henceuniformly in $t$. At an integer argument, $D$ is one at zero and zero at the other integers, so the expansion interpolates the samples exactly.

For the Hurwitz proof of the planar isoperimetric inequality, take a positively oriented regular simple closed curve of length $L$ and enclosed area $A$. Write its complex position as $z(t)=x(t)+iy(t)$, with $0\le t\le 2\pi$ proportional to arc length. Then $|z^{\prime }(t)|=L/(2\pi )$. Translate the curve to make its mean position zero, and write its Fourier coefficients as $c_n$, with $c_0=0$.

Green's theorem gives the signed area, and Parseval's identity computes it:Periodic integration by parts and Parseval's identity applied to $z^{\prime }$ giveSince $n\le n^2$ for every integer $n$,The sums converge absolutely: $\sum |n||c_n{|}^2\le (\sum |c_n{|}^2{)}^{1/2}(\sum n^2|c_n{|}^2{)}^{1/2}$. Equality forces $c_n=0$ unless $n=0$ or $1$; after the mean translation, $z(t)=c_1{e}^{it}$ is a circle. Conversely a circle attains equality. Thus circles uniquely attain equality, up to translation and orientation.

The same proof applies to a rectifiable simple closed curve using its Lipschitz arc length parametrization. Its derivative exists almost everywhere and belongs to $L^2$; periodic mollification converges to the curve in the function and derivative $L^2$ norms. This justifies the derivative coefficient identity, Parseval's identity and area integral by approximation. Reversing orientation, if necessary, makes the enclosed area positive. The printed name “Hurewitz” is read as Hurwitz.

Use the uniform bound for harmonic sine polynomialsFor completeness, reduce to $0<t\le \pi$ and split at $K=\min (m,\lfloor 1/t\rfloor )$. The first part is bounded by $Kt\le 1$. Geometric-series summation bounds every interval sum of $\sin kt$ by $1/\sin (t/2)\le \pi /t$. Summation by parts bounds the remaining harmonic-weighted tail by $\pi /[t(K+1)]\le \pi$. Negative $t$ follows by oddness and $t=0$ is immediate.

Let $H_m={\sum }_{k=1}^{m}1/k$. Choose $m_j$ so large that $H_{m_j}\ge C_0{2}^{j+2}$, and positive integers $L_j>m_j$ such that the intervals $[L_j-m_j,L_j+m_j]$ are strictly separated and increase. DefineThe bound $\Vert Q_j{\Vert }_{\infty }\le 2^{-j-2}$ makes this a continuous function with $g(0)=0$.

Each Fourier coefficient of $P_m$ has modulus $1/(2k)$, so the sum of the absolute coefficients is $H_m$. Therefore every prefix of a normalized block $Q_j$ has norm at most one. At any Fourier cutoff, all earlier blocks are complete, at most one block is partial and all later blocks are absent. HenceAt zero, a completed block contributes zero, but its prefix through frequency $L_j$ consists of the negative-frequency sine coefficients and equals $i/2$. ThusBoth index sequences tend to infinity. The uniformly bounded partial sums fail to converge at the origin, even though the function is continuous and zero there.

Write the finite abelian group additively. A character of a finite abelian group is a group homomorphism $\chi :G\to \{z:|z|=1\}$. We first prove that there are exactly $|G|$ such characters of a finite abelian group, without assuming a structure theorem.

Use extension of a character across a cyclic quotient. Given a subgroup $H$ and $a\notin H$, let $k$ be the least positive integer with $ka\in H$. The subgroup $K=H+⟨a⟩$ has $k$ cosets of $H$. If $\varphi$ is a character of a finite abelian group on $H$, choose any of the $k$ roots ${\lambda }^k=\varphi (ka)$ and defineThis is well-defined: two representations differ by an integer multiple of $ka$, and the root equation exactly cancels that difference. It is a character of a finite abelian group, and every extension arises from one of the $k$ choices of $\lambda$. Build a chain from the trivial subgroup to $G$ by adjoining elements. The character of a finite abelian group count multiplies by the same factor as the subgroup order at every step, so $|\hat{G}|=|G|$.

For a nontrivial character of a finite abelian group $\eta$, choose $a$ with $\eta (a)\ne 1$. Translating the group sum showsso the sum is zero. Applied to $\chi \overline{\psi }$, this provesThe $|G|$ orthogonal nonzero characters of a finite abelian group therefore form a basis of all complex functions on $G$, a vector space of dimension $|G|$.

With the unnormalized Fourier coefficients $\hat{f}(\chi )={\sum }_{x\in G}f(x)\overline{\chi (x)}$, expansion in that basis gives the Fourier inversion on a finite groupUnder a normalized forward-transform convention the prefactor would instead be one. Thus the unspecified constant is determined by the convention, and the inversion itself follows directly from character of a finite abelian group counting and orthogonality.

The endpoint agreement $f(1)=f(0)=0$ makes the boundary terms vanish when integrating the Fourier coefficients of $f^{\prime }$ by parts. If $a_n^{\prime },b_n^{\prime }$ denote these coefficients, thenThese are exactly the coefficients obtained by differentiating the preceding Fourier series:On every compact subinterval of $(0,1)$ the differentiated series converges uniformly by the Dirichlet test, since partial sums of the sine factors are uniformly bounded there. The Termwise differentiation of a Fourier series is therefore valid away from integers and gives $1-2x$ on $(0,1)$. At an integer the periodic original has a corner, so the differentiated series is not asserting the existence of an ordinary derivative there.

The constant and first Fourier coefficients of ${\varphi }^8$ are $1$ and $-16$. In the weight-four Eisenstein basis at level two, these forceFor $n>0$, its $q^n$ coefficient is $-16{\sigma }_3(n)+256{\sigma }_3(n/2)$, where the second divisor sum is zero if $n$ is odd. The even divisors have cube sum $8{\sigma }_3(n/2)$, soOn the other hand, expanding the eighth power of the theta series of integer squares counts ordered integer eight-tuples of square sum $n$. Their sign is $(-1{)}^{{\sum }_j{m}_j}=(-1{)}^n$, since $m_j^2\equiv m_j(\mathrm{mod}2)$. If $r_8(n)$ denotes that count, thenSeparately $r_8(0)=1$, from the all-zero tuple. The positive-divisor formula is not a formula at zero. This is the eight-square representation formula.

Use the product printed in the original PDF, with factors $(1-q^n{)}^2$ and $(1-q^{11n}{)}^2$. The Fricke involution normalizes ${\Gamma }_0(11)$ and preserves its modular cusp space. Therefore $g=-11^{-1}{\tau }^{-2}f(-1/(11\tau ))$ lies in the given one-dimensional space, so $g=cf$ for a constant $c$.

At the Fricke fixed point ${\tau }_{\ast }=i/\sqrt{11}$, the prefactor $-11^{-1}{\tau }_{\ast }^{-2}$ is one. Thus $g({\tau }_{\ast })=f({\tau }_{\ast })$. The product has $0<q_{\ast }<1$, every factor is positive, and its limit is nonzero since ${\sum }_n{q}_{\ast }^n+{\sum }_n{q}_{\ast }^{11n}<\infty$. Hence $f({\tau }_{\ast })>0$, forcing $c=1$. This Fricke sign from a nonvanishing fixed-point value provesPart (b) now gives $\Lambda (f,s)=\Lambda (f,2-s)$. Its Taylor series at one contains only even powers, so its order of vanishing is even. The function is not identically zero, since its first Fourier coefficient is one. In this example the product is positive on the entire positive imaginary axis, and its Mellin integral at $s=1$ is positive. Thus the stronger conclusion isThe TeX aid duplicates and corrupts the product in this part; neither corrupted expression is used.

Use the orthonormal Fourier series basis $e_n(x)=(2\pi {)}^{-1/2}{e}^{inx}$, indexed by $n\in \mathbb{Z}$, and the Hilbert space inner product $⟨f,g⟩={\int }_0^{2\pi }\overline{f(x)}g(x)dx$. The integral operator is a periodic convolution operator. Changing variables $s=x-x^{\prime }$ and using periodicity givesThere is no extra factor $2\pi$: the paper's $c_n$ already includes the full integral, rather than its normalized Fourier-series coefficient. The adjoint kernel is $\overline{K(x^{\prime }-x)}$, so $A^{\ast }{e}_n=\overline{c_n}{e}_n$.

A singular system of a periodic convolution operator is thereforeIndeed $A{v}_n={\sigma }_n{u}_n$ and $A^{\ast }{u}_n=(c_n/|c_n|)\overline{c_n}{e}_n={\sigma }_n{v}_n$. The unit factor determined by the complex argument of $c_n$ in $u_n$ is necessary when the complex Fourier coefficients are not positive real numbers. Both families are orthonormal and complete, since every $c_n\ne 0$. One may enumerate $\mathbb{Z}$ by $\mathbb{N}$, or order the positive singular values by decreasing magnitude.

The continuous kernel on a finite square makes $A$ a Hilbert-Schmidt operator, hence compact. Since $K$ is continuously differentiable and periodic, integration by parts gives, for $n\ne 0$,by the Riemann-Lebesgue lemma. In particular the singular values tend to zero. Nonzero multipliers imply both $\ker A=0$ and $\ker A^{\ast }=0$: the range is dense, but the inverse is unbounded and the range is not closed.

The Fourier coefficient of the velocity-integrated density is its mixed transform at zero velocity frequency. ThusThe constant ${\rho }_{\infty }$ contributes only to the zero Fourier mode, where it equals ${\hat{f}}_0(0,0)$. ThereforeIt is important to remove the zero mode, which is conserved rather than mixed away.

Take the positive integer $n\ge 1$ and put $m=n+1\ge 2$. From the preceding estimate and the absent zero Fourier mode, for $t\ne 0$,Here $\zeta$ is the Riemann zeta function, and its displayed series is finite because $m>1$. One may replace the last derivative norm by the given full mixed Sobolev norm to obtain the requested constant depending only on $n$ and $f_0$.

An absolutely summable sequence of Fourier coefficients gives a uniformly and absolutely convergent Fourier series, with supremum bounded by the sum of their absolute values. Its sum agrees almost everywhere with $r_t$ by uniqueness of Fourier coefficients for integrable periodic functions. HenceThis is the uniform phase-mixing bound from velocity derivatives: uniform convergence for the continuous representative of the density, with rate $O(|t{|}^{-n-1})$. No uniform decay of the full phase-space distribution is asserted. The free-transport phase mixing acts by shifting nonzero spatial modes to large velocity frequency. If a convention allows $0\in \mathbb{N}$, that endpoint needs separate assumptions or an argument: the harmonic series in this proof would diverge.

Use probability measure-preserving systems: $T:X\to X$ is measurable and $\mu (T^{-1}A)=\mu (A)$. Its invariant sigma-algebra isErgodicity means that every invariant measurable set has measure zero or one:Equivalently, every invariant measurable function is constant almost everywhere. Indeed all rational sublevel sets of a real invariant function belong to $\mathcal{I}$, so the zero-or-one alternative forces its distribution to concentrate at one value. Apply this to real and imaginary parts for complex functions. Conversely, an invariant set has an invariant indicator, so the function criterion implies the set criterion. This is the invariant-function characterization of ergodicity.

For an irrational rotation of the circle, $T(x)=x+\alpha (\mathrm{mod}1)$ with $\alpha \notin \mathbb{Q}$, an invariant $g\in L^2(m)$ has Fourier coefficients satisfyingFor $k\ne 0$ the multiplier is nonzero. Completeness of the Fourier series basis therefore makes $g$ constant. Apply this to an invariant indicator to prove the irrational rotation is ergodic. A second example is the cyclic permutation of $q$ points with equal masses: a nonempty invariant set contains the whole cycle and hence has measure one.

The identity on $[0,1]$ with Lebesgue measure is not ergodic, since $[0,1/2]$ is invariant and has measure $1/2$. Nor is rotation by $1/5$ on the circle: the setis invariant and has measure $1/2$. These examples also show why preservation of measure does not imply ergodicity.

Let $U=U_T$ be the Koopman operator, $Uf=f\circ T$, and setThe Von Neumann mean ergodic theorem says that, for $f\in L^2(\mu )$, these Cesaro averages converge in $L^2$ to the orthogonal projection onto the invariant functions:For an invertible system, $U$ is a unitary operator. The Hilbert-space theorem states more generally that the Cesaro averages of any unitary operator converge strongly to its fixed-space projection.

Here is the requested invertible-case proof. Since $U^{\ast }=U^{-1}$,Thus the orthogonal decomposition for unitary ergodic averages isOn the first summand, $A_{N}f=f$. On a vector $(I-U)g$, telescoping givesBecause $\Vert A_N\Vert \le 1$, approximation extends this convergence to the closure of the range. The orthogonal decomposition now proves the theorem for every $f$. An invariant $L^2$ function is $\mathcal{I}$-measurable, and an $\mathcal{I}$-measurable function is invariant by approximation with indicators; hence the projection is the stated conditional expectation.

In an ergodic system, invariant functions are constants by the criterion already proved. Their orthogonal projection is obtained by taking the inner product with $1$, whose norm is one. Therefore the ergodic limit is the space average:

The Birkhoff ergodic theorem states that, for $f\in L^1(\mu )$,The limit is invariant, integrable, and has the same integral as $f$. On a probability space convergence also holds in $L^1$. In an ergodic system the limit is $\int fd\mu$ almost everywhere.

For the topological sharpening, take a continuous map $T$ on a compact metric space. Unique ergodicity means there is exactly one invariant Borel probability measure $\mu$. The uniform ergodic convergence for uniquely ergodic systems says that every continuous real or complex $f$ satisfiesThus convergence holds at every point and uniformly in that point. The continuity, compactness and continuous-observable hypotheses belong to this theorem; it is not a claim of uniform convergence for all measurable $L^1$ functions.