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Choose $0<R<R_1$, $0<S<S_1$ and $0<T<T_1$. On this smaller closed polydisc, the holomorphic function $A$ and its $w$-derivative are bounded; write $|A|\le M$ and $|{\partial }_{w}A|\le L$. Choose $0<r<R$ with $rM<S/2$ and $rL<1$, replacing a zero bound by any positive upper bound if necessary, and choose $0<\epsilon <T$.

Consider functions $u(z,\alpha )$ continuous on $\overline{D(0,r)}\times \overline{D(0,\epsilon )}$, jointly holomorphic in the interior, with $u(0,\alpha )=0$ and $\Vert u{\Vert }_{\infty }\le S/2$. This is a closed subset of a Banach space in the supremum norm. The integral operatoris well defined on the straight segment and is jointly holomorphic. Indeed, it equals $z{\int }_0^{1}A(tz,u(tz,\alpha ),\alpha )dt$, an integral of jointly holomorphic functions locally uniformly bounded in both variables. Convexity of the $w$-disc and the derivative bound giveThus the Banach fixed-point theorem gives a fixed point $f(z,\alpha )$. More explicitly, the successive integral iterates starting from zero converge uniformly, and locally uniform convergence of holomorphic functions makes their limit jointly holomorphic. Differentiating the integral equation gives the required complex ordinary differential equation and initial value. This proves local existence, uniqueness and joint holomorphic dependence on $(z,\alpha )$. Uniqueness among arbitrary local solutions follows first on a small disc where their values stay within $|w|<S$, by the same contraction mapping estimate, and then throughout their common connected domain by the identity theorem.

For holomorphic dependence of ordinary differential equations on parameters in the initial value, fix an admissible initial value $w_{\ast }$ and put $u=f-w_0$. The equation becomesIts right-hand side is jointly holomorphic in $(z,u,w_0)$ near $(0,0,w_{\ast })$. The preceding construction supplies a common disc and a jointly holomorphic solution for $w_0$ near $w_{\ast }$. Consequently $f=u+w_0$ is jointly holomorphic in $z$ and its initial value $w_0$.

The Stone theorem for the circle group says that a strongly continuous unitary representation $U$ of $\mathbb{T}=\{z:|z|=1\}$ has an orthogonal decompositionEquivalently, $U_z={\sum }_n{z}^n{P}_n$ strongly, with mutually orthogonal projections $P_n$ summing strongly to the identity. Define these projections using normalized Haar measure:Haar invariance gives $U_w{P}_n=w^n{P}_n$. Changing $z$ to $z^{-1}$ in the adjoint integral proves $P_n^{\ast }=P_n$. Applying the defining integral to $P_m\xi$ gives $P_n{P}_m={\delta }_{nm}{P}_m$ by orthogonality of the characters $z^k$. Thus $P_n$ are orthogonal projections, and their ranges are exactly the indicated character subspaces.

It remains to prove completeness of these subspaces. The Fejér kernel has the finite expansionIt is nonnegative, has normalized integral one, and its integral outside any fixed neighborhood of the identity tends to zero, since the denominator in the usual geometric-sum formula is bounded away from zero there. Strong continuity therefore givessplit the integral into that neighborhood, where $\Vert U_z\xi -\xi \Vert$ is small, and its complement, where it is bounded by $2\Vert \xi \Vert$. The integral on the left is ${\sum }_{|n|<N}(1-|n|/N)P_n\xi$. Hence the closed span of all $H_n$ is $H$. Orthogonality then gives the unweighted strong sum and the stated representation formula.

Conversely, any orthogonal decomposition indexed by the integers gives this representation; continuity follows first on finite sums and then on all vectors by the uniform unitary norm bound. With $z=e^{it}$ its generator isThis diagonal operator is self-adjoint and $U_{e^{it}}=e^{itA}$. Thus the circle theorem is the integer-spectrum version of the one-parameter theorem.

The Stone theorem for one-parameter unitary groups states that every strongly continuous unitary representation $U_t$ of $\mathbb{R}$ is uniquely $U_t=e^{itA}$ for a possibly unbounded self-adjoint operator $A$. Its domain isConversely every self-adjoint $A$ gives such a group by the Borel functional calculus for a normal operator.

Prove the forward direction by putting $G=iA$, initially defined as the norm derivative at zero. Its domain is dense: for $f\in C_c^{\infty }(\mathbb{R})$, the vector ${\xi }_f=\int f(t)U_t\xi dt$ has derivative $G{\xi }_f=-\int f^{\prime }(t)U_t\xi dt$. Smooth approximate identities make ${\xi }_f\to \xi$. The group law gives $U_{t}D(G)=D(G)$ and $G{U}_t=U_{t}G$ on that domain. Differentiation of preserved inner products shows that $G$ is skew-symmetric, so $A=-iG$ is symmetric.

The derivative operator is closed. Indeed, if ${\xi }_n\to \xi$ and $G{\xi }_n\to \eta$, integrate the orbit derivative to obtainTaking limits gives the same identity with $\xi ,\eta$; dividing by $t$ at zero proves $\xi \in D(G)$ and $G\xi =\eta$.

Now define bounded Laplace integralsShifting the integration variable in $U_h{R}_{\pm }\xi$ and taking the derivative at zero shows that $R_{\pm }\xi \in D(G)$ andHence $I-G$ and $I+G$ are both surjective. Since $G=iA$, this means $A+iI$ and $A-iI$ are surjective.

Here is why these range conditions establish self-adjointness rather than just symmetry. If $\eta \in D(A^{\ast })$, solve $(A-iI)\xi =(A^{\ast }-iI)\eta$. Then $\eta -\xi \in \ker (A^{\ast }-iI)=\mathrm{Ran}(A+iI{)}^{\perp }=\{0\}$. Thus $D(A^{\ast })\subseteq D(A)$, and symmetry gives equality with $A^{\ast }=A$.

The permitted functional calculus now gives $V_t=e^{itA}$. It is unitary, strongly continuous by dominated convergence against spectral measures, and differentiable with derivative $iA{V}_t\xi$ on $D(A)$. Differentiate $V_{-t}{U}_t\xi$ for $\xi \in D(A)$; the two derivative terms cancel because both groups preserve the domain and commute with their generator. Its value is consequently the constant $\xi$. Density of the domain implies $U_t=V_t$ on all of $H$.

For completeness, the converse and the exact domain follow from the same calculus. If $\xi \in D(A)$, dominated convergence applied to $|(e^{it\lambda }-1)/t|\le |\lambda |$ gives the derivative $iA\xi$. If that norm derivative exists, Fatou's lemma applied to its squared norm gives $\int {\lambda }^{2}d⟨E_A(\lambda )\xi ,\xi ⟩<\infty$, so $\xi \in D(A)$. The derivative characterization also proves uniqueness. Thus

Choose a uniformly random permutation of $[n]$ and take its initial segments. This produces a Uniformly random maximal chain in a Boolean lattice. A fixed $r$-set belongs to this chain with probability $r!(n-r)!/n!={\left(\genfrac{}{}{0ex}{}{n}{r}\right)}^{-1}$. An antichain meets the chain at most once, so taking the expected value of the number of meetings proves the LYM inequality:In particular the Sperner theorem bound follows by multiplying each summand by at most the largest binomial coefficient.

For the vector request, an ordering by ordinary subset need not work: distinct vectors may cancel. Instead, write $s_A={\sum }_{i\in A}{x}_i$ and construct a separated block decomposition for vector subset sums. A block is separated if distinct labeled members have subset sums at distance at least one. We prove that the labeled subsets of $[n]$ can be partitioned into separated blocks, withblocks of size $n-2r+1$, for $0\le r\le \lfloor n/2\rfloor$, with $\left(\genfrac{}{}{0ex}{}{n}{-1}\right)=0$.

For $n=0$ the one subset forms one block. Suppose a block of sums $z_1,\dots ,z_{\ell }$ has already been constructed for $n-1$ vectors, and let $x=x_n$. Choose $z_{\ast }$ minimizing $z_j\cdot x$ in the block. Replace the two copies of the block, with and without the new coordinate, by the following two blocks:The old labels are retained, so these partition the two copies even if equal sums occur in different blocks. Translation preserves separation in the first translated portion, and the second block is a subset of the old separated block. For the new cross pairs, Cauchy-Schwarz inequality givesThus both new blocks are separated. Their sizes are $\ell +1$ and $\ell -1$, omitting an empty block. The resulting size-profile recurrence is $b_{n,r}=b_{n-1,r}+b_{n-1,r-1}$, and Pascal's identity verifies the displayed binomial-difference formula, including the even middle boundary. The total number of blocks telescopes to $\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$.

A signed sum associated with $A=\{i:{ϵ}_i=1\}$ is $2s_A-{\sum }_i{x}_i$. If two signed sums lie in the open ball of radius one, their difference has norm less than two, so their subset sums have distance less than one. At most one can therefore lie in any separated block. Hence the Littlewood-Offord inequality isThe counting is by sign choices, not by distinct numerical values of the sums. The open-ball condition handles the equality-distance boundary correctly.

For the last request take real coefficients. Absorb their signs into the choices ${ϵ}_i$, so all coefficients are positive and at least one. A signed-sum real interval of length four corresponds to a subset-sum real interval of length two. A strict three-member inclusion chain $A⊊B⊊C$ would have $s_C-s_A\ge 2$, which cannot occur in that open real interval. Thus its set family is a k-Sperner family with $k=2$. Every maximal chain in a Boolean lattice meets it at most twice, yielding ${\sum }_r|{\mathcal{F}}_r|/(\genfrac{}{}{0ex}{}{n}{r})\le 2$. Write $f_r=|{\mathcal{F}}_r|/(\genfrac{}{}{0ex}{}{n}{r})$, so $0\le f_r\le 1$ and ${\sum }_r{f}_r\le 2$. If $B_1\ge B_2$ are the two largest binomial coefficients and $f_{\ast }$ is the fraction at a largest level, then $|\mathcal{F}|\le B_1{f}_{\ast }+B_2(2-f_{\ast })\le B_1+B_2$. This proves the two-level Littlewood-Offord bound:It is the greatest possible number uniformly over choices of coefficients and center. For even $n=2m$, take every coefficient equal to one and center $a=-1$: the real interval $(-3,1)$ contains the sum levels $-2,0$, of multiplicities $\left(\genfrac{}{}{0ex}{}{n}{m-1}\right),\left(\genfrac{}{}{0ex}{}{n}{m}\right)$. For odd $n=2m+1$, take center zero, giving levels $-1,1$, each of multiplicity $\left(\genfrac{}{}{0ex}{}{n}{m}\right)$. These attain the bound. For $n=0$ the only sign choice gives the separate maximum one. Attainment here is at the exhibited centers; it need not hold at every prescribed center.

All graphs here are finite and simple. For the strong graph product $G⊠H$, two distinct graph vertex pairs are adjacent if in each coordinate the entries are equal or adjacent. Use $G^{⊠m}$ for its strong graph power; the exam's $G^m$ means this product power, not the distance-based graph power. The Shannon capacity of a graph iswhere $\alpha$ is the independence number. Products of independent sets are independent, so $a_{r+s}\ge a_r{a}_s$ for $a_m=\alpha (G^{⊠m})$. To see why the limit exists, fix $r$ and write $m=qr+s$, $0\le s<r$. Then $a_m\ge a_r^q$, using $a_s\ge 1$ and $a_0=1$, so ${\mathrm{liminf}}_m(\log a_m)/m\ge (\log a_r)/r$. Take the supremum over $r$; the reverse upper bound holds termwise. This proves the required form of Fekete lemma. For the empty graph on no graph vertices one sets graph capacity zero separately.

Associativity of the strong product givesAn orthonormal representation of a graph assigns a real unit vector $u_v$ to each graph vertex, with $u_v\perp u_w$ whenever distinct graph vertices $v,w$ are nonadjacent. Adjacent graph vertices need not receive orthogonal vectors. A handle is a unit vector $h$. The handle formulation of the Lovász theta function iswith a zero denominator giving infinity. Representations may use any finite-dimensional Euclidean space. This is the usual Lovász number in its orthonormal-representation formulation.

For an independent set $I$, the vectors $u_v$, $v\in I$, are mutually orthonormal. Directly,If $T={\max }_v|h\cdot u_v{|}^{-2}$ is finite, every summand is at least $1/T$, giving $|I|\le T$. For a strong power, use the vectors and handleTheir norms are one. Nonadjacency of two distinct words means a coordinate has distinct nonadjacent entries, so the corresponding tensor inner product has a zero factor. The handle inner products satisfyApplying the preceding calculation gives $\alpha (G^{⊠m})\le T^m$. Taking roots, then the infimum over representations and handles, provesThis is the tensor-product capacity bound from an orthonormal representation, and the displayed projection calculation supplies Bessel inequality explicitly.

For $C_5$, label graph vertices by $\mathbb{Z}/5\mathbb{Z}$, with edges at differences $\pm 1$. The five wordsare independent in its strong square: a difference $\pm 1$ in the first coordinate produces a nonedge difference $\pm 2$ in the second, while a first-coordinate difference $\pm 2$ is already a nonedge. Hence $c(C_5)\ge \sqrt{5}$.

For the matching upper bound set $\rho =1/\sqrt{5}$, $h=(0,0,1)$, andEach vector is unit. For a nonedge, the angle difference is $\pm 4\pi /5$, andbecause $\cos (4\pi /5)=-(1+\sqrt{5})/4$ and $\rho =1/\sqrt{5}$. Thus these vectors form an orthonormal representation. The handle has $|h\cdot u_j{|}^2=\rho$, giving $\vartheta (C_5)\le 1/\rho =\sqrt{5}$. Together the two bounds prove the pentagon capacity code result

For a discrete group, one standard definition of amenability is the fixed-point property: every continuous affine action on a nonempty compact convex set in a locally convex space has a fixed point. The equivalent state formulation, which may be stated without proof here, is that there is a left-invariant invariant mean on ${\ell }^{\infty }(\Gamma )$. Equivalently, every unital left-translation-invariant star-subalgebra of ${\ell }^{\infty }(\Gamma )$ admits a left-invariant state, with continuity for the supremum norm; one may equally pass to its norm closure. A state means a positive linear functional of norm one with value one on the constant function one. Restriction proves the subalgebra formulation from the full-algebra version, and taking the full algebra proves the converse. Such a mean is finitely additive on indicator functions, not necessarily countably additive.

Write $(\lambda (g)\mu )(h)=\mu (g^{-1}h)$ on ${\ell }^1(\Gamma )$ and $(L_{g}f)(h)=f(g^{-1}h)$ on bounded functions. We prove equivalence of this mean condition, the Reiter condition, and existence of a Følner sequence.

Assume first that an invariant mean $m$ exists. Fix a finite set $K\subseteq \Gamma$. In the real Banach space ${⨁}_{g\in K}{\ell }^1(\Gamma )$ with sum norm, consider the convex setIf zero were outside its norm closure, the Hahn-Banach separation theorem would give bounded real functions $f_g$ and $\delta >0$ such thatfor every such $\mu$. Testing point masses at $h$ showsApplying the positive invariant mean makes the left side zero and the right side at least $\delta$, a contradiction. Therefore zero is in the closure: for any $\epsilon >0$ a finitely supported probability measure makes the sum of the translation defects over $K$ less than $\epsilon$. Enumerate the countable group, take the first $n$ elements for $K$ and $\epsilon =1/n$, and obtain a sequence satisfying the Reiter condition. This proves (1) implies (2).

Next assume (2). An ${\ell }^1$ probability measure can be approximated in norm by finitely supported probability measures, by restricting to a large finite set and renormalizing. Replacing $\mu$ by $\nu$ changes each defect by at most $2\Vert \mu -\nu {\Vert }_1$. Thus simultaneous small defects for finitely many translations can be achieved with finite support.

For such $\nu$, set $F_t=\{h:\nu (h)>t\}$. The elementary identity $|r-s|={\int }_0^{\infty }|1_{r>t}-1_{s>t}|dt$ for $r,s\ge 0$ givesIf the sum of defects over $K$ is less than $\epsilon$, some $t$ with $F_t\ne \varnothing$ must satisfy ${\sum }_{g\in K}|F_t△g{F}_t|<\epsilon |F_t|$; otherwise integration would contradict that strict inequality. Applying this layer-cake extraction of Følner sets to successive finite sets of translations gives nonempty finite $F_n$ with each relative boundary tending to zero. Hence (2) implies (3).

Conversely, normalized counting measures ${\mu }_n=1_{F_n}/|F_n|$ satisfyso (3) implies (2). Finally, from (2) define states $m_n(f)={\sum }_h{\mu }_n(h)f(h)$. The dual unit ball of ${\ell }^{\infty }(\Gamma )$ is weak-star compact by the Banach-Alaoglu theorem; take a convergent subnet, since this compactness need not be metrizable. Its limit $m$ is positive with $m(1)=1$, andThus $m$ is invariant, proving (2) implies (1). Together these implications establish all three equivalences. Countability is used to obtain sequences from finite tests; it is not a justification for replacing the compactness subnet by a subsequence.

The topology needed here is the weak-star topology $\sigma (E^{\ast },E)$, meaning pointwise convergence on $E$. With the literal weak topology $\sigma (E^{\ast },E^{\ast \ast })$, the claimed compactness is false. For example, $E=c_0$ is separable and $E^{\ast }={\ell }^1$. Its unit vectors $e_n$ have no weak cluster point: every coordinate functional forces a cluster point to have all coordinates zero, while the bounded functional $x\mapsto {\sum }_j{x}_j$ has value one throughout. Thus the dual unit ball is not weakly compact. We prove the intended weak-star assertion in full.

Choose a countable dense set $(x_k)$ in the unit ball of $E$. On $E_1^{\ast }$ defineEach term is at most $2\cdot 2^{-k}$, so the series converges. The triangle inequality is immediate, and $d=0$ forces equality on the dense set and hence on all of $E$. It is therefore a metric. Convergence in this metric is equivalent to convergence on every $x_k$: one direction follows from any individual term, and the other by bounding the series tail uniformly. Because $\Vert \varphi \Vert ,\Vert \psi \Vert \le 1$, approximation of any unit vector by $x_k$ then gives convergence on all of $E$. The same finite-tail and approximation argument for neighborhoods proves equality with the weak-star topology, not just agreement of its convergent sequences.

Given a sequence $({\varphi }_n)$ in this ball, diagonal selection gives a subsequence converging at every $x_k$, since each coordinate lies in a compact disk. Uniform norm bounds then make it pointwise Cauchy at every $x\in E$. Define $\varphi (x)$ as the limit. Linearity passes to the limit, and $|\varphi (x)|\le \Vert x\Vert$, so $\varphi \in E_1^{\ast }$. The subsequence converges to it in the metric. Thus this metric space is sequentially compact, hence compact, proving

Put $F=\ker (I-U)$. For a unitary operator, $\ker (I-U^{\ast })=F$, soThis follows from the general identity $\mathrm{Ran}(A{)}^{\perp }=\ker A^{\ast }$. The averaging operators have $\Vert T_n\Vert \le 1$ and equal the identity on $F$. On a vector $(I-U)\eta$, telescoping givesThe uniform norm bound extends this convergence to the closure of the range. Thus $T_n$ tends to identity on $F$ and to zero on its orthogonal complement. Consequentlywhich is exactly convergence in the strong operator topology. This proves the Von Neumann mean ergodic theorem in the requested unitary case.

Let $\eta =|A|/N$ and $g=1_A-\eta$. In the norm convention, the quadratic uniformity of a set meansIf uniformity is parametrized instead by $\Vert g{\Vert }_{U^3}^8\le \alpha$, replace the parameter below by ${\alpha }^{1/8}$. We prove the genuine-progression conclusion by placing nonnegative weights on a region where wrapping and repeated entries are impossible.

First suppose $N\ge 16$. Put $M=\lfloor N/16\rfloor$, $J=\{0,\dots ,M-1\}$, $t=M/N$, and let $w=1_J\ast 1_J$ be the normalized convolution on a finite group. Define the weighted countBoth weights are nonnegative. Their supports require $0\le x\le 2M-2$ and $M\le d\le 3M-2$ in the ordinary representatives. In particular $d\ge 1$ and $x+3d\le 11M-8<N$. Every contributing arithmetic progression is therefore a set of four distinct integers in $\{0,\dots ,N-1\}$, with no modular wrap.

The averages of both weights are $t^2$. The convolution theorem on a finite group and the Parseval identity on a finite group giveTranslation multiplies coefficients by phases, so $w(d-M)$ has the same Fourier coefficient sum. Telescoping the product of the four indicators around ${\eta }^4$ produces four error terms, each with one factor $g$ and all other factors bounded by one.

To bound a weighted error, expand both weights in their finite Fourier series. For frequencies $r,s$, the extra character is $e^{2\pi i(rx+sd)/N}$. Write it asand absorb these two characters into the functions at slopes zero and one. Even if the balanced factor is one of those functions, its $U^3$ norm remains at most $\alpha$ by the character invariance proved in part (i). The cyclic bound of part (ii), with $C_0=6^{1/8}$, therefore bounds each unweighted Fourier term by $C_0\alpha$. Summing absolute coefficients costs at most $t^2$. ConsequentlySince $t\ge 1/32$ and $\eta \ge \delta$, this is positive whenever $\alpha <{\delta }^4/(4096C_0)$. A positive count supplies the required genuine arithmetic progression. This is the argument that triangular weights count genuine four-term progressions; it avoids losing powers of $\log N$ by cutting off a sharp interval.

For completeness, treat $4\le N<16$. If $A$ is a proper nonempty subset, then $\eta (1-\eta )\ge \delta /N$. The norm bound from part (i) givesChoose $\alpha$ below this bound as well. Then quadratic uniformity forces $A={\mathbb{Z}}_N$, which contains $\{0,1,2,3\}$. Thus, for every $N\ge 4$, a sufficient threshold isThe restriction $N\ge 4$ is necessary: for $N=1,2,3$, the full group has balanced subset indicator zero and is quadratically uniform for every positive parameter, but no subset has four elements. The printed conclusion needs this elementary size qualification.

Index the functions as $g_0,g_1,g_2,g_3$ and write $\Lambda ={\mathbb{E}}_{x,d}{\prod }_{j=0}^3{g}_j(x+jd)$. We first control $g_0$. Change variables to $y=x+3d$, average over $d$, and apply Cauchy-Schwarz in $y$, discarding the bounded factor $g_3(y)$. Expand the squared inner average with the second copy of $d$ written $d+h$, then put $x=y-3d$. This yieldsThe right side is real and nonnegative because it arose as an average of squared absolute values. Next use $y=x+2d$ and apply Cauchy-Schwarz in $(h,y)$, discarding the bounded $g_2$ derivative. Expanding with a displacement $k$ in $d$ givesFinally use $y=x+d$ and Cauchy-Schwarz in $(h,k,y)$, discarding the bounded $g_1$ double derivative. Expansion with displacement $r$ givesAll these changes of variables are bijections without any division in $G$.

Put $t_j=|G[j]|$, where $G[j]=\{z:jz=0\}$ is the torsion subgroup killed by an integer $j$. The variables $2k$ and $3h$ are uniform on $2G$ and $3G$, whose indices are $t_2,t_3$. Since the last squared integrand is nonnegative, its restricted average is at most $t_2{t}_3$ times the unrestricted average over the two increments. By part (i), that unrestricted average is $\Vert g_0{\Vert }_{U^3}^8$. Hence $|\Lambda |\le (t_2{t}_3{)}^{1/8}\Vert g_0{\Vert }_{U^3}$.

To control $g_1$, apply the same three eliminations in the order of slopes $3,0,2$. The surviving $g_1$ has increments $-2h,k,-r$. The last increment is uniform, and only the first is restricted to $2G$, so the resulting bound is $|\Lambda |\le t_2^{1/8}\Vert g_1{\Vert }_{U^3}$. Reversing the arithmetic progression by $(x,d)\mapsto (x+3d,-d)$ treats $g_3$ and $g_2$ respectively. This proves the four-term progression bound with torsion factors:For a fixed finite group, this positive constant tends to zero with $\alpha$. In ${\mathbb{Z}}_N$, $t_j=\gcd (j,N)$, so $\boxed{\beta =6^{1/8}\alpha \text{ suffices uniformly in }N}$. If multiplication by two and three are bijective, the sharper $\beta =\alpha$ holds.

A dependence on torsion is necessary if a bound uniform over all finite abelian groups is intended. In $G={\mathbb{F}}_2^m$, choose independent random signs $f(x)$. A cube with eight distinct vertices has sign product of expectation zero. Each of the twenty-eight possible collisions has probability $1/|G|$, so the expectation of $\Vert f{\Vert }_{U^3}^8$ is at most $28/|G|$. Some sign function therefore has norm at most $(28/|G|{)}^{1/8}$. But with factors $(f,1,f,1)$, every arithmetic progression is $(x,x+d,x,x+d)$ and its product is $f(x{)}^2=1$. Thus unbounded torsion obstructs uniform control of four-term progressions. The stated fixed-group bound, and the uniform cyclic-group bound needed in part (iii), are unaffected.

The expectations use normalized counting measure on a finite abelian group $G$. This is the finite-group convention implicit in the averaging notation; an arbitrary infinite group has no normalized counting probability measure on all its points. For $h\in G$, define ${\Delta }_{h}f(x)=f(x)\overline{f(x+h)}$, and let $\mathcal{C}$ denote complex conjugation. The Gowers U3 norm isAveraging first over $h_3$ gives the Derivative identity for the Gowers U3 normIndeed, if $u(x)={\Delta }_{h_1}{\Delta }_{h_2}f(x)$, the average over $(x,h_3)$ is ${\mathbb{E}}_{x,h_3}u(x)\overline{u(x+h_3)}=|{\mathbb{E}}_{x}u(x){|}^2$. The eighth root is consequently well-defined. Homogeneity follows because the cube has four unconjugated and four conjugated factors, so $\Vert af{\Vert }_{U^3}=|a|\Vert f{\Vert }_{U^3}$. If the norm is zero, every squared term in the finite average vanishes. Taking $h_1=h_2=0$ gives ${\mathbb{E}}_x|f(x){|}^4=0$, so $f=0$ pointwise. This proves positive definiteness.

For the triangle inequality, we prove the needed Gowers-Cauchy-Schwarz inequality. For eight functions $F_{\omega }$ on a product of finite sets $X_1\times X_2\times X_3$, form the box norm multilinear averageFix a coordinate $j$. After averaging its two variables separately, this is $\mathbb{E}A\overline{B}$, where $A$ uses all vertices with ${\omega }_j=0$ and $B$ those with ${\omega }_j=1$, with the remaining parity conjugations. The Cauchy-Schwarz inequality giveswhere $S_j^b$ replaces the function at each vertex by the original function with its $j$th index set to $b$. Each right-hand average is a squared absolute value averaged over the other coordinates, hence nonnegative. Apply this operation successively for $j=1,2,3$, to every resulting factor. The eight final families each use one original $F_{\omega }$ at all vertices. Taking roots gives the box Cauchy-Schwarz inequalityNow take $X_j=G$ and $F_{\omega }(a,b,c)=f_{\omega }(a+b+c)$. With $x=a_0+b_0+c_0$ and $h_j=x_j^1-x_j^0$, these independent variables induce uniform $(x,h_1,h_2,h_3)$: each such tuple has exactly $|G{|}^2$ preimages. Therefore each box norm is the corresponding $U^3$ norm, and the mixed cube obeysExpand the eight factors in the cube expression for $f+g$. Its $2^8$ mixed terms are bounded by this inequality, and their bounds sum to $(\Vert f{\Vert }_{U^3}+\Vert g{\Vert }_{U^3}{)}^8$. Thus $\boxed{\Vert f+g{\Vert }_{U^3}\le \Vert f{\Vert }_{U^3}+\Vert g{\Vert }_{U^3}}$, completing the norm proof.

Two consequences will be useful below. Multiplication by a character of a finite abelian group leaves the norm unchanged, because the alternating character product around a three-dimensional cube equals one. Alsowhere the last inequality uses the $h=0$ term and $\Vert u{\Vert }_{U^2}^4={\sum }_{\chi }|\hat{u}(\chi ){|}^4\ge |\mathbb{E}u{|}^4$. Hence $\Vert f{\Vert }_{U^3}\ge |G{|}^{-1/8}\Vert f{\Vert }_2$.