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Use the product topology on $\mathcal{C}=[k{]}^{\mathbb{Z}}$, with $[k]$ discrete, and the left shift $(\mathcal{L}c)(j)=c(j+1)$. A basic cylinder set specifies finitely many coordinates. This is a compact metric space, and $\mathcal{L}$ is a homeomorphism. Write $Y=\overline{\{{\mathcal{L}}^{n}c:n\ge 0\}}$ for the forward orbit closure. A minimal point means that $Y$ is a minimal dynamical system, equivalently that every point of $Y$ has a dense forward orbit in $Y$; it need not be a fixed point.

There is a convention needed in the source: the bounded-gaps property concerns every finite integer interval $I$, hence every finite word over an alphabet. Literally allowing $I=\mathbb{Z}$ would make the displayed condition impossible for finite $U$, although a constant coloring is a minimal point. Under the standard finite-word interpretation, the property is precisely uniform recurrence.

Suppose first that $Y$ is a minimal dynamical system. Its nonempty compact forward-invariant subset $\mathcal{L}(Y)$ must equal $Y$. As the ambient left shift is invertible, all integer translates of $c$ belong to $Y$. Let $I=[a,b]\cap \mathbb{Z}$ have length $\ell =b-a+1$, and let $V=\{z\in Y:z{|}_I=c{|}_I\}$. This is a nonempty clopen set. Each forward orbit in $Y$ meets $V$, so $\{{\mathcal{L}}^{-n}V:n\ge 0\}$ covers $Y$. By compactness, finitely many suffice; let $D$ be the largest index in this finite cover. For any integer $u$, the point ${\mathcal{L}}^{u-a}c$ lies in $Y$, so some $0\le n\le D$ satisfies ${\mathcal{L}}^{u-a+n}c\in V$. The prescribed word therefore occurs at positions $[u+n,u+n+\ell -1]$, entirely inside $[u,u+D+\ell -1]$. Taking $M=D+\ell$ proves the bounded-gaps property for every interval of length at least $M$.

Conversely suppose $c$ is uniformly recurrent. Every finite word from any integer translate of $c$ occurs arbitrarily far to the right, so every such translate belongs to $Y$. Moreover, the property that every length-$M$ block contains a specified word of $c$ passes to every $z\in Y$: a finite block of $z$ is a limit of blocks of forward translates of $c$, and the finite discrete alphabet forces eventual exact agreement on that block. Given $I=[a,b]\cap \mathbb{Z}$, look for its word inside $z([a,a+M-1])$. An occurrence starts at $a+n$ for some $n\ge 0$, so ${\mathcal{L}}^{n}z$ agrees with $c$ on $I$. Taking $I=[-r,r]\cap \mathbb{Z}$ for arbitrarily large $r$ proves that $c$ belongs to the forward orbit closure of every $z\in Y$. That closure is a closed forward-invariant subset of $Y$, so it also contains every forward translate of $c$ and hence all of $Y$. Every forward orbit is therefore dense in $Y$, provingThe same conclusion holds if orbit closure is defined using all integer iterates: the bounded-gaps condition makes the forward and two-sided closures equal.

Put $L={\mathcal{O}}_X(D)$ and let $s_D$ be the canonical global section cutting out the effective Cartier divisor $D$. The divisor restriction exact sequence givesThe hypothesis says $L{|}_D$ is an ample line bundle. By Serre vanishing, $H^1(D,L^m{|}_D)=0$ for all sufficiently large $m$, so the long exact sequence in sheaf cohomology makessurjective for all such $m$. These are finite-dimensional vector spaces. Their dimensions form a nonincreasing sequence of nonnegative integers, so the maps are isomorphisms from some point on. Exactness then implies that the restrictionis surjective for all sufficiently large $m$.

Choose such an $m$ for which $L^m{|}_D$ is also a globally generated line bundle. Lift a generating collection of its global sections to $X$. At each point of $D$, one lift has nonzero image in the one-dimensional residue-field fibre, hence generates the stalk of $L^m$ by Nakayama lemma. Outside $D$, $s_D^m$ is nowhere zero and generates $L^m$. Together these sections generate it everywhere. ThereforeThis proof works on an arbitrary projective scheme because the defining section of an effective Cartier divisor is a non-zero-divisor. If $D$ is empty, $L\cong {\mathcal{O}}_X$ and the conclusion is immediate. Semiampleness is the conclusion: the pullback of a line avoiding the centre of a blowup of a smooth algebraic surface of ${\mathbb{P}}^2$ satisfies the hypothesis on its support but has zero intersection with the exceptional curve, and therefore is not ample.

Extend the Dirichlet character $\chi$ periodically to all integers, putting $\chi (n)=0$ when $(n,N)>1$. Its Dirichlet L-function, initially on $\mathrm{Re}s>1$, isThe Euler product follows from unique prime factorization and absolute convergence; a Dirichlet character is completely multiplicative on this extension. Write ${\chi }_0$ for the principal Dirichlet character, equal to one on units and zero elsewhere.

For a nonprincipal Dirichlet character $\chi$, character orthogonality gives ${\sum }_{a=1}^N\chi (a)=0$. Explicitly, multiplication by a unit $u$ with $\chi (u)\ne 1$ permutes the unit residues and multiplies this sum by $\chi (u)$, forcing it to vanish. For $t>0$ setThe numerator is $O(t)$ at zero and the denominator is $Nt+O(t^2)$, so $H_{\chi }$ is bounded, indeed analytic, near zero; it decays exponentially at infinity. Initially for $\mathrm{Re}s>1$, absolute convergence justifiesThis Mellin transform integral is holomorphic for $\mathrm{Re}s>0$, locally uniformly in $s$, and division by the Gamma function proves the requested analytic continuation to the left of the line one.

In fact, the same argument proves Mellin continuation of a nonprincipal Dirichlet L-function to the entire plane. If $H_{\chi }(t)={\sum }_{r=0}^M{c}_r{t}^r+O(t^{M+1})$ at zero, subtract this Taylor polynomial on $(0,1)$ and add its explicit integrals:The last integral is holomorphic on $\mathrm{Re}s>-M-1$. Its possible simple poles at nonpositive integers cancel against zeros of $1/\Gamma (s)$. Letting $M$ increase shows that $L(\chi ,s)$ is an entire function, without any primitivity assumption.

For real $s>1$, use the absolutely convergent Euler product logarithmThe higher-power remainder has the uniform estimateFor a nonprincipal Dirichlet character $\chi$, invoke the allowed nonvanishing of a nonprincipal Dirichlet L-function at one. Its holomorphy and nonvanishing give a holomorphic logarithm on a small disk about one. On the connected real interval $1<s<1+\eta$, $B_{\chi }(s)$ differs from this logarithm by a fixed element of $2\pi i\mathbb{Z}$: the difference is continuous with exponential one. Thus the prime-character sum near one is bounded. This branch argument is needed for complex-valued Dirichlet characters.

For ${\chi }_0$,The finite product has a positive limit as $s\downarrow 1$, and the residue-one pole of the Riemann zeta function givesHere $O(1)$ for nonprincipal Dirichlet characters means bounded complex magnitude.

Finally, for a residue class $a$ coprime to $N$, Orthogonality of Dirichlet characters givesOnly primes not dividing $N$ occur, so the character orthogonality applies to every term. This sum diverges as $s\downarrow 1$. A finite collection of primes would give a bounded sum, a contradiction. Every reduced residue class contains infinitely many primes. This is the Dirichlet theorem on primes in arithmetic progressions; the coprimality hypothesis is essential.

Use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-ix\xi }dx$. An orthonormal multiresolution analysis is a family of closed vector subspaces $V_j\subset L^2(\mathbb{R})$, indexed by integers, with $V_j\subset V_{j+1}$, $\overline{{\bigcup }_j{V}_j}=L^2(\mathbb{R})$, and ${\bigcap }_j{V}_j=\{0\}$. Its dilation condition is $f\in V_j$ if and only if $f(2\cdot )\in V_{j+1}$. The space $V_0$ is invariant under integer function translations and admits a scaling function $\phi$ whose integer function translations form an orthonormal basis. Consequently ${\phi }_{j,k}=2^{j/2}\phi (2^j\cdot -k)$ form an orthonormal basis of $V_j$.

Since $\phi \in V_1$, expansion in that orthonormal basis gives the scaling refinement equationThe series converges in the L2 norm. The associated low-pass filter of a multiresolution analysis is the periodic Fourier series symbolInitially the symbol is defined almost everywhere. Orthonormal integer function translations give the quadrature mirror filter identity $|m(\xi ){|}^2+|m(\xi +\pi ){|}^2=1$ almost everywhere. For a Lebesgue integrable scaling function, its Fourier transform is continuous and $|\hat{\phi }(0)|=1$; choosing a constant phase makes $\hat{\phi }(0)=1$ and the continuous representative near zero has $m(0)=1$. Different conventions absorb $\sqrt{2}$ into the refinement coefficients; the displayed convention fixes that ambiguity.

Let $W_j=V_{j+1}\ominus V_j$ denote the orthogonal complement of $V_j$ inside $V_{j+1}$. The Meyer-Mallat theorem constructs a wavelet whose integer function translations form an orthonormal basis of $W_0$. Dilation then gives the orthogonal decomposition into orthogonal complements $L^2(\mathbb{R})={⨁}_{j\in \mathbb{Z}}{W}_j$, so its translates and dilates form an orthonormal wavelet basis.

Using the MRA low-pass filter, choose the high-pass symbol $n(\xi )=e^{-i\xi }\overline{m(\xi +\pi )}$. The quadrature mirror filter identity makes the two analysis channels orthonormal. The resulting Fourier transform formula isEquivalently, with the refinement coefficients from the scaling refinement equation, put $g_k=(-1{)}^{k-1}\overline{h_{1-k}}$ and $\psi (x)=\sqrt{2}{\sum }_k{g}_k\phi (2x-k)$. This index choice gives precisely the displayed high-pass symbol. Changing all $g_k$ by one constant unit phase gives the same orthonormal wavelet construction. In particular the often-used $(-1{)}^k\overline{h_{1-k}}$ convention differs only by a global minus sign. The basis is ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$ for $j,k\in \mathbb{Z}$.

A wavelet with $p$ vanishing moments satisfies $x^k\psi \in L^1$ and $\int x^k\psi (x)dx=0$ for $0\le k<p$. The moment differentiation of the Fourier transform theorem states that these weighted integrability hypotheses make $\hat{\psi }\in C^{p-1}$, with ${\hat{\psi }}^{(k)}(\xi )=\int (-ix{)}^k\psi (x)e^{-ix\xi }dx$. Hence ${\hat{\psi }}^{(k)}(0)=0$. The MRA projection Fourier identity, together with density of the approximation spaces and continuity of $\hat{\phi }$ at zero, gives $|\hat{\phi }(0)|=1$. These are the two analytic theorems needed in addition to the quadrature mirror filter construction.

For the ordinary derivative conclusion, assume also that the MRA low-pass filter is $C^{p-1}$ near $\pi$ and $\hat{\phi }$ is $C^{p-1}$ near zero. These hypotheses hold, for example, for a compactly supported scaling function with a finite refinement filter. The wavelet identity givesThe denominator is nonzero near zero. Its reciprocal and the exponential are $C^{p-1}$, so the product rule and the zero Taylor polynomial of $\hat{\psi }$ give the smooth-mask vanishing-moment criterion

The printed integrability hypothesis alone does not guarantee ordinary higher derivatives. It gives only a continuous $\hat{\phi }$. Without added smoothness, Taylor theorem for $\hat{\psi }$ still gives $m(\pi +t)=o(|t{|}^{p-1})$ for the representative defined by the quotient: a Peano zero. This establishes the value at $\pi$, and for $p\ge 2$ the first derivative there, but it does not automatically establish iterated derivatives.

Here is a lacunary scaling-phase regularity counterexample for $p\ge 3$. Start with a compactly supported Daubechies wavelet having $p$ vanishing moments, with scaling function ${\phi }_0$ and finite MRA low-pass filter $m_0$. Define the periodic phaseThe Fourier coefficients of $\theta$ are absolutely summable. The Wiener algebra is closed under multiplication and the exponential series, so $a$ and $a^{-1}$ have absolutely summable Fourier coefficients. Thus $\phi$ is an absolutely summable combination of integer function translations of ${\phi }_0$, belongs to $L^1\cap L^2$, and generates the same $V_0$. Unimodularity preserves orthogonality and unit norm of its integer function translations; the inverse phase preserves their complete span. The same approximation spaces therefore give a multiresolution analysis.

The identity $\theta (2t)=\theta (t)+\theta (t+\pi )$ gives $a(2t)=a(t)a(t+\pi )$, so the periodic phase change of a scaling function produces $m(t)=a(t+\pi )m_0(t)$. In the high-pass formula the phases cancel:The associated wavelet is therefore exactly the original one, with unchanged vanishing moments.

At any dyadic point $t_0=2\pi k/2^j$, the difference quotient of the terms with $j\le n\le {\log }_2(1/|h|)$ is $1+O((2^{n}h{)}^2)$ per term. The sum of these errors and the remaining tail divided by $h$ are bounded; the finitely many earlier terms have finite limits. Hence $(\theta (t_0+h)-\theta (t_0))/h={\log }_2(1/|h|)+O(1)$, which has no finite limit. The exponential has the same failure of differentiability. Dyadic points are dense, and $m_0(\pi +t)$ is nonzero for sufficiently small $t\ne 0$. Thus $m$ is not differentiable on any neighborhood of $\pi$, so an ordinary second derivative at $\pi$, understood as the derivative of a locally defined first derivative, need not exist. The Peano zero remains valid. This separates the intended smooth-mask theorem from what the literal assumptions establish.

Under the smoothness hypotheses in the smooth-mask vanishing-moment criterion, write any nonzero integer $j$ as $j=2^r\ell$, where $r\ge 0$ and $\ell$ is odd, possibly negative. Iterating the scaling refinement equation exactly $r+1$ times givesThe last MRA low-pass filter factor is $m(\pi \ell +t/2^{r+1})$. Its derivatives of all orders less than $p$ vanish at $t=0$, because $\ell$ is odd and the MRA low-pass filter is $2\pi$-periodic. If the remaining factors are $C^{p-1}$ at the displayed arguments, the product rule forces every derivative of order less than $p$ of the product to vanish. In particular compact support and a finite MRA low-pass filter supply all these hypotheses. Therefore the integer-frequency zeros of a scaling function are

Under only the printed assumptions, the exact same finite refinement argument gives the Peano zero $\hat{\phi }(2\pi j+t)=o(|t{|}^{p-1})$. Indeed, the last MRA low-pass filter factor has that estimate, the other MRA low-pass filter factors are bounded by one almost everywhere by the quadrature mirror filter identity, and the last Fourier transform factor is bounded because $\phi \in L^1$. The estimate initially holds almost everywhere and extends to every $t$ by continuity of $\hat{\phi }$. It does not imply arbitrary ordinary higher derivatives. In the lacunary scaling-phase regularity counterexample, $\hat{\phi }=a{\hat{\phi }}_0$ is nondifferentiable at dense dyadic points next to the isolated integer-frequency zeros of the smooth base transform. Thus the same regularity qualification is necessary here.

For $0<\alpha \le 1$, uniform Lipschitz-$\alpha$ means that one constant $M$ satisfies $|f(x)-f(y)|\le M|x-y{|}^{\alpha }$ for every $x,y\in [0,1]$. The resulting Hölder space has normFor the subsequent estimates at general $\alpha >0$, use the higher-order Hölder class convention: write $\alpha =r+\beta$ with $r\ge 0$ an integer and $0<\beta \le 1$, and require $f\in C^r([0,1])$ with $f^{(r)}$ uniformly Lipschitz-$\beta$. ThusAt integer $\alpha =k$, this uses $C^{k-1,1}$; the alternative classical $C^k$ convention is stronger on this compact closed interval and also suffices for the estimates.

For exponents above one, a first-difference inequality alone forces the function to be constant. Partition $[x,y]$ into $n$ equal pieces and use the triangle inequality: $|f(y)-f(x)|\le M|y-x{|}^{\alpha }{n}^{1-\alpha }\to 0$. Consequently a nontrivial higher-order definition is needed for the full range of exponents in the later wavelet estimates. Equivalently, higher-order Hölder class gives a local Taylor polynomial with remainder bounded by $C\Vert f{\Vert }_{C^{\alpha }}|x-x_0{|}^{\alpha }$.

Absolute values are missing in the printed formula. For signed real blocks, the intended block mixed norm must beWithout them, at $q=1$ a block $(-1,1)$ has value zero although it is nonzero, and a negative scalar has negative value. For nonintegral $q$, a negative component may not even have a real power. The literal expression therefore is not a norm and does not define the claimed general convex problem. Even integer exponents do not fix all other $q$ in the stated range.

With the corrected norm, let $q^{\prime }$ be the Holder conjugate exponent: $q^{\prime }=q/(q-1)$ for $q>1$ and $q^{\prime }=\infty$ for $q=1$. Blockwise Holder inequality, with equality from a norming block, givesThe duality of block mixed norms identifies the closed compact convex set and the projection residual from the Moreau decompositionFor $q=2$, the dual blocks are Euclidean balls; for $q=1$, they are cubes given by componentwise bounds $|p_{i,j}^r|\le \lambda$. No injectivity or surjectivity of $A$ is required: compactness of the block product makes its linear image compact, and the strictly convex fidelity still gives a unique primal minimizer. For general $q^{\prime }$, the Euclidean projection onto a dual block ball is not generally obtained by radial scaling, so the special clipping formula from the isotropic case should not be reused without further analysis.

Let $g$ have order $n$ in a finite group. In a complex group representation, its matrix is diagonalizable with eigenvalues ${\zeta }_j$ satisfying ${\zeta }_j^n=1$. If each $g^i$ with $\gcd (i,n)=1$ is conjugate to $g$, the character of a representation is constant on these powers. Henceusing the stated integrality of each Ramanujan sum. Therefore $\chi (g)$ is rational. It is also an algebraic integer, being a sum of roots of unity; a rational algebraic integer is an integer. Thus $\boxed{\chi (g)\in \mathbb{Z}}$.

In a symmetric group, raising a permutation to a power coprime to its order preserves each cycle length and therefore its entire cycle type. Since cycle type determines the conjugacy class, the hypothesis holds, and all character values of symmetric groups are integers.

Suppose $\chi$ is an irreducible character of a group of odd order and $\chi =\overline{\chi }$. Write $d=\chi (1)$. Every nonidentity element pairs with its distinct inverse, since there is no element of order $2$. Also $\chi (g^{-1})=\overline{\chi (g)}=\chi (g)$. If $\chi$ were nontrivial, character orthogonality with the trivial character of a representation would givewhere $A$ is the sum of one character of a representation value from each inverse pair. This $A$ is an algebraic integer, so $A=-d/2\in \mathbb{Q}$ is an integer. Hence $d$ is even.

But an irreducible character degree divides the group order, so $d$ must be odd. One quick justification of that divisibility is that the central class sum of a conjugacy class $C$ acts by the algebraic integer ${\lambda }_C=|C|\chi (g_C)/d$. Its integrality follows from its integer matrix on the regular group representation. Character orthogonality then givesa rational algebraic integer, hence an integer. The parity contradiction proves $\boxed{\chi =1_G}$: the trivial character is the only self-conjugate irreducible character of an odd-order group.