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The common amplitude $a>0$ is uniquely determined by any entry's absolute value. After choosing it, each entry independently has sign $+1$ or $-1$. Thus the correspondenceis a bijection from $(0,\infty )\times \{-1,1{\}}^{\mathbb{N}}$ onto the set. The positive reals have cardinal $\mathfrak{c}$, and binary sign sequences have cardinal $2^{{\aleph }_0}=\mathfrak{c}$. HenceHere “independently” refers to unconstrained deterministic choices of coordinates, not probabilistic independence; no probability distribution is being specified.

The Weierstrass approximation theorem asserts that every real continuous function on a compact interval is a uniform limit of polynomials. An affine change of variable reduces the interval to $[0,1]$. There are two complementary routes: construct positive approximation operators, or construct an algebra dense enough to approximate absolute values and hence pointwise maxima and minima.

For the first route, the Korovkin theorem reduces convergence of positive linear operators on continuous functions $L_n$ to the three tests $1,t,t^2$. Here is its proof. Given $f\in C[0,1]$ and $\epsilon >0$, uniform continuity supplies a $\delta >0$ such that $|f(t)-f(x)|<\epsilon$ for $|t-x|<\delta$. For all $t,x$ one then hasApply positivity to the two real inequalities obtained from this bound. At $x$ this givesThe expression in brackets is nonnegative and tends uniformly to zero by the three test assumptions. Also $L_{n}1\to 1$ uniformly. Taking the upper limit of the supremum norm and then letting $\epsilon \downarrow 0$ proves convergence for every $f$. This is the quadratic barrier estimate for positive approximation operators.

The Bernstein polynomial supplies the constructive corollaryThe binomial theorem makes its nonnegative weights sum to one. Differentiating the binomial theorem, or using the first two binomial distribution moments, givesThe Korovkin theorem proves $B_{n}f\to f$ uniformly, and each $B_{n}f$ is a polynomial, proving the Weierstrass approximation theorem. The same criterion gives the Schoenberg spline operator convergence in Solution 3. For periodic functions its counterpart uses $1,\cos t,\sin t$: the nonnegative barrier $1-\cos (t-x)$ is bounded away from zero outside any fixed small arc. This periodic Korovkin test set gives convergence of the normalized Fejér sums and density of trigonometric polynomials.

The Lebesgue convergence criterion for approximation operators addresses stability. Suppose bounded linear operators $L_n:C[a,b]\to C[a,b]$ converge to the identity on every polynomial. Then they converge on every continuous function if and only if ${\sup }_n\Vert L_n\Vert <\infty$. Sufficiency is the extension of approximation operators from a dense subspace: for a polynomial $p$ with $\Vert f-p\Vert <\eta$ and a common bound $M$,First let $n\to \infty$, then $\eta \downarrow 0$. Necessity follows from the Uniform boundedness principle, since a convergent $L_{n}f$ is bounded for every $f$. Applied to approximation projectors, the Lebesgue inequality for approximation projectors isIndeed $P_{n}v=v$ makes $f-P_{n}f=(f-v)-P_n(f-v)$; take the infimum. Thus density of the ranges and a uniform bound on the Lebesgue constants of interpolation imply convergence. Density alone does not control a sequence of interpolation projectors.

There is also a kernel version of this stability criterion, useful for trigonometric polynomial approximation. If integrable periodic kernels $K_n$ have uniformly bounded $L^1$ norms, their integrals tend to one, and ${\int }_{|t|\ge \delta }|K_n(t)|dt\to 0$ for every $\delta >0$, then convolution with $K_n$ converges uniformly to every continuous periodic function. Subtract $f(x)\int K_n$ and split the error into a small arc, bounded by $\omega (f,\delta ){\sup }_n\Vert K_n{\Vert }_1$, and its complement, bounded by $2\Vert f{\Vert }_{\infty }{\int }_{|t|\ge \delta }|K_n|$. The remaining mass error tends to zero. This approximate identity argument shows directly why nonnegative unit-mass Fejér kernels work even though ordinary Fourier partial sums need not converge uniformly for every continuous target.

Lebesgue's constructive approach to the Weierstrass approximation theorem starts instead with the absolute value. On $[-1,1]$ define the polynomial recurrence for the absolute valueInduction gives $0\le p_r(x)\le |x|$, monotonicity, andThe pointwise limit satisfies $p(x{)}^2=x^2$ and is nonnegative, hence equals $|x|$. Dini's theorem makes the convergence uniform. More explicitly, the continuous decreasing errors have, for any positive tolerance, open sublevel sets covering the compact interval; a finite subcover supplies one index valid everywhere. Every continuous piecewise linear function has the formThe $c_j$ are its changes in slope. Scaling and translating the absolute-value approximation shows that $L$ is uniformly approximable by polynomials. Solution 4, or polygonal interpolation and uniform continuity, makes these piecewise linear functions dense in $C[a,b]$. Approximating $f$ first by $L$ and then by a polynomial is another complete proof of the Weierstrass approximation theorem.

The Stone-Weierstrass theorem generalizes from intervals to a compact Hausdorff space $K$: a real subalgebra $A\subseteq C(K,\mathbb{R})$ containing the constants and separating points is uniformly dense. Its proof uses the same absolute-value mechanism. The uniform closure of a function algebra is an algebra. If $a$ belongs to this closure, approximation of $|t|$ by polynomials on a bounded interval containing its range proves that $|a|$ also belongs to the closure. Therefore the closure containsPoint separation and constants allow exact interpolation of any two real values: choose $h\in A$ with $h(s)\ne h(t)$ and solve for $\alpha ,\beta$ in $\alpha h+\beta$. The lattice approximation from two-point approximation proved in Solution 4 now makes the closure all of $C(K,\mathbb{R})$.

The complex version requires a unital, point-separating subalgebra closed under complex conjugation. Its real-valued members separate points using real or imaginary parts, so the real result approximates both parts of every complex target. Conjugation closure is essential: polynomials in $z$ alone on the closed unit disk cannot approximate $\overline{z}$ uniformly, because ${\int }_{|z|=1}p(z)dz=0$ whereas ${\int }_{|z|=1}\overline{z}dz=2\pi i$.

Among the corollaries are density of multivariate polynomials on compact subsets of real Euclidean space, since coordinates separate points, and density of trigonometric polynomials on the circle, since sine and cosine separate points there. On a compact interval, polynomials with rational coefficients are a countable dense family, giving separability of $C[a,b]$. The real and complex approximation results above concern uniform approximation of continuous targets; neither asserts convergence of arbitrary interpolation schemes or arbitrary Fourier series.

The mean value theorem states that, if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, with $a<b$, there is $c\in (a,b)$ such thatTo deduce it from Rolle's theorem, subtract the secant line: $k(t)=f(t)-f(a)-\frac{f(b)-f(a)}{b-a}(t-a)$. This function has the same continuity and differentiability properties and satisfies $k(a)=k(b)=0$. Rolle's theorem gives an interior point with $k^{\prime }(c)=0$, which is precisely the required equality.

If $h^{\prime }=0$ on $\mathbb{R}$, apply the mean value theorem on any $[x,y]$ with $x<y$. It gives $h(y)-h(x)=(y-x)h^{\prime }(c)=0$. Hence $h$ is constant.

For the differential equation, the product rule givesThe preceding result makes this product constant. Thus all solutions are $f(x)=C{e}^{ax}$, and differentiation verifies every such solution.

For the integral assertion, continuity ensures that $f$ is Riemann integrable on every finite closed interval. Use the usual orientation for an integral with negative upper endpoint. For $h\ne 0$, additivity of the integral givesThe absolute value of this expression is at most the supremum of $|f(t)-f(x)|$ on the segment joining $x$ and $x+h$. Continuity at $x$ makes this bound tend to zero for either sign of $h$. Consequently $F^{\prime }(x)=f(x)$, proving this form of the fundamental theorem of calculus.

Finally, a differentiable solution of the integral equation is continuous, so the fundamental theorem of calculus yields $g^{\prime }=g$; setting $x=0$ gives $g(0)=A$. The differential-equation calculation above now forcesConversely, $A+{\int }_0^{x}A{e}^{t}dt=A+A(e^x-1)=A{e}^x$, so this function satisfies the original integral equation. It is unique because every differentiable solution has already been forced to this form and this initial condition.

Use normalized Fourier-Walsh transform coefficients $\hat{g}(\xi )=2^{-n}{\sum }_{x}g(x)(-1{)}^{x\cdot \xi }$. The Hamming weight $|\xi |$ counts the nonzero coordinates. Norms on the input cube use uniform probability, and a frequency integral here is counting over $\xi \in {\mathbb{F}}_2^n$. A common alternative normalization multiplies both sides of the requested comparison by the same factor and does not change it.

Beckner's inequality states that for $1<p\le q<\infty$ and $0\le \rho \le \sqrt{(p-1)/(q-1)}$, the noise operator on the Boolean hypercube, with coefficients $\widehat{T_{\rho }g}(\xi )={\rho }^{|\xi |}\hat{g}(\xi )$, satisfies $\Vert T_{\rho }g{\Vert }_q\le \Vert g{\Vert }_p$. In particular choose $q=2$, $p=4/3$, $\rho =1/\sqrt{3}$.

Let $g=1_A$ and $\alpha =|A|/2^n$. Then $\Vert g{\Vert }_{4/3}^2={\alpha }^{3/2}$. On frequencies of weight at most two, ${\rho }^{2|\xi |}\ge 1/9$, so Beckner's inequality and Parseval's identity give the low-degree Fourier mass of a sparse Boolean set estimateThe indicator function is real and the cube characters are real, so the displayed squares equal the squared absolute values. Total Fourier weight is $\sum \hat{g}(\xi {)}^2=\Vert g{\Vert }_2^2=\alpha$. If $2^n>2003$, then $0<\alpha \le 1/2003$ and $9\sqrt{\alpha }\le 9/\sqrt{2003}<1/2$, since $2003>18^2$. ThereforeThis establishes the strict comparison for every power $N=2^n>2003$, not merely for an unspecified very large threshold. The original PDF supplies the density condition and Beckner request, both missing from the converted TeX.

Let $K$ be a number field of degree $d$, and normalize its absolute values so that the product formula is ${\prod }_v|a{|}_v^{n_v}=1$ for $a\ne 0$, where $n_v=[K_v:{\mathbb{Q}}_v]$; at a complex place use the ordinary modulus with weight two. For $P=[a_0:\cdots :a_N]$, defineOnly finitely many factors differ from one. Rescaling all coordinates multiplies the local maxima by $|a{|}_v$, whose weighted product is one, so the projective height is well-defined. It is at least one by comparison with any nonzero coordinate. On enlarging $K$ to $L$, the identity ${\sum }_{w\mid v}[L_w:{\mathbb{Q}}_v]=[L:K][K_v:{\mathbb{Q}}_v]$ makes the normalized product unchanged. Thus this is an absolute height on algebraic projective points. For coprime rational integer coordinates it is simply their maximum absolute value.

For a projective variety $V/K$ and an embedding $i:V\hookrightarrow {\mathbb{P}}^N$, take $h_i=h\circ i$. Changing the basis of the same projective linear system changes the height by a bounded amount: the coordinate transformation and its inverse have fixed local coefficient bounds, equal to one outside finitely many places. If $L$ is ample and $L^m$ is very ample, set $h_L=m^{-1}{h}_i$ for an embedding from $L^m$, up to a bounded function. A general line bundle can be written $(L\otimes A)\otimes A^{-1}$ with both $A$ and $L\otimes A$ very ample, defining its height as the difference of their embedding heights.

This is the Weil height machine, withThe first relation follows from the Segre embedding, whose local maximum is the product of the two maxima; tensor powers similarly use the Veronese embedding. The second follows by pulling back the defining sections. These constructions make the choices equivalent modulo bounded functions. For a fixed homogeneous morphism with no base point, upper local height bounds come from its coefficients, and lower bounds from the Projective Nullstellensatz expressing coordinate powers in the ideal of its defining forms. The constants are one outside a finite set of places. This gives the uniform bounded errors and shows why heights belong to line-bundle classes rather than being independent of every projective embedding.

The Northcott lemma states that for fixed $B$ and $D$, the algebraic points of ${\mathbb{P}}^N$ with $H(P)\le B$ and $[\mathbb{Q}(P):\mathbb{Q}]\le D$ form a finite set. In particular the $K$-rational points of bounded projective height are finite. Here is a proof, including the degree bound.

First let $\alpha$ have degree $e\le D$ and $H(\alpha )=H([1:\alpha ])\le B$. Write its primitive minimal polynomial as $a_e{\prod }_{j=1}^e(T-{\alpha }_j)$, with positive integer $a_e$. The height-Mahler measure formula isIts finite-place part follows from multiplicativity of the nonarchimedean polynomial coefficient norm: a primitive integer polynomial has norm one at each prime, so factoring it bounds the product of the roots' local maxima by the inverse leading-coefficient norm. The archimedean part is the displayed product over conjugates. Combining them with the product formula gives the identity.

Expanding the polynomial now bounds each integer coefficient byThere are finitely many degrees and finitely many such integer coefficient lists, each with finitely many roots. Hence there are only finitely many possible $\alpha$. This proves the minimal-polynomial coefficient proof of Northcott finiteness for algebraic numbers.

On the projective chart $a_j\ne 0$, each ratio ${\alpha }_i=a_i/a_j$ has degree at most the point's field degree andThe equality uses the product formula. Each coordinate ratio therefore belongs to a finite set, and there are finitely many charts. This proves the projective Northcott theorem. Restricting to $V$ gives the result for a chosen embedding. For an ample line-bundle height, the comparison $m{h}_L=h_i+O(1)$ reduces its bounded-height sets to this case.

Both qualifications matter: a degree bound cannot be dropped, since infinitely many roots of unity have height one; and the finiteness assertion need not hold for a nonample line bundle, such as the trivial bundle's bounded height on a variety with infinitely many rational points.

Write the hyperbolic toral automorphism as $T_A([v])=[Av]$ on ${\mathbb{T}}^2={\mathbb{R}}^2/{\mathbb{Z}}^2$, where $A\in \mathrm{GL}(2,\mathbb{Z})$. Its determinant is $\pm 1$. The eigenvalues cannot be a nonreal conjugate pair: their product would have absolute value one, forcing each to have absolute value one, contrary to hyperbolicity. The same product condition and hyperbolicity imply two distinct real eigenvalues, labelled so thatWe will proveThis is the entropy of a hyperbolic toral automorphism; the absolute value is important when the expanding eigenvalue is negative.

Choose real eigenvectors $v_s,v_u$ and the adapted norm $\Vert a{v}_s+b{v}_u{\Vert }_{\ast }=\max (|a|,|b|)$. Its quotient metric on the torus isIt is a compatible metric, so part (a) allows its use. Let $\rho ={\min }_{0\ne \ell \in {\mathbb{Z}}^2}\Vert \ell {\Vert }_{\ast }>0$ and $D=\Vert A{\Vert }_{\ast }=|{\lambda }_u|$. Fix $ϵ>0$ small enough that $2ϵ<\rho$ and $(D+1)ϵ<\rho$.

If $[v]$ lies in the Bowen ball $B_n(0,ϵ)$, each $T_A^k[v]$ has a unique lift $z_k$ with $\Vert z_k{\Vert }_{\ast }<ϵ$. Uniqueness follows because two such lifts differ by a lattice vector of norm less than $2ϵ<\rho$. Moreover, $z_{k+1}-A{z}_k$ is a lattice vector, andIt must be zero. Thus $z_k=A^k{z}_0$: there is no hidden lattice jump between successive lifted iterates.

Write $z_0=a{v}_s+b{v}_u$. The simultaneous bounds $\Vert A^k{z}_0{\Vert }_{\ast }<ϵ$ for $0\le k<n$ are exactlyThe stable coordinate is largest at time zero and the unstable coordinate at time $n-1$. Conversely, every vector in this rectangle satisfies all the Bowen ball bounds. The quotient is injective on this rectangle by $2ϵ<\rho$. If $J=|\det (v_s,v_u)|$, its normalized Haar measure is thereforeThe quotient metric is translation invariant and $T_A$ is linear, so $d_n(x,y)=d_n(0,y-x)$; every Bowen ball has this same measure, independently of its centre.

Take an $ϵ$-separated set of largest cardinality $s_n(d,ϵ)$. It is maximal under inclusion, so its radius-$ϵ$ Bowen balls cover the torus: a point outside all these balls could be added. The covering bound givesIts radius-$ϵ/2$ Bowen balls are disjoint, by the triangle inequality for $d_n$. Their measures sum to at most one, givingTaking logarithms, dividing by $n$, and letting $n\to \infty$ yields $h_d(T_A,ϵ)=\log |{\lambda }_u|$ for every sufficiently small $ϵ$. The limit as $ϵ\downarrow 0$ proves the answer. Contraction in the stable direction contributes no exponential growth; only expansion in the unstable direction does.

For the compact set $K\subseteq B\subseteq U$ chosen in the preceding step, the Urysohn lemma gives a continuous function $0\le \phi \le 1$ equal to one on $K$ and zero on $X\setminus U$. In a metric space this cutoff can also be written explicitly, when $K$ and $X\setminus U$ are both nonempty:The two sets are disjoint and closed, so the denominator never vanishes, and their distance functions are continuous. If $K$ is empty, take $\phi =0$; if $U=X$ and $K$ is nonempty, take $\phi =1$. These choices satisfy the required boundary conditions in the exceptional cases.

The difference $\phi -1_B$ vanishes on $K$ and outside $U$, and its absolute value is at most one everywhere. ThereforeThis is the precise continuous approximation required by both proofs of ergodicity, and also gives $|\int \phi d\mu -\mu (B)|<ϵ$.

For $r>0$, define the forward and backward one-sided interval averaging operatorsand set $A_0^{\pm }f=f$. Each positive-radius average is bounded in absolute value by $m_u(f)(x)$. Although $x$ is an endpoint of its averaging interval, enlarge that interval slightly so that $x$ is interior, apply the maximal bound and let the enlargement shrink to zero.

To dominate the limiting identity too without assuming the differentiation result we are proving, use the sublinear operatorThe maximal estimate and Chebyshev inequality implyThus $S$ has weak type (1,1) and dominates every $A_r^{\pm }$, including $r=0$.

Smooth compactly supported functions are dense in $L^1$, obtained by truncating and then smoothing with a compactly supported approximate identity. For each such continuous function both averages tend to its value at every point. Applying part (ii) to the two families, and intersecting their full-measure convergence sets, gives $A_r^{\pm }f(x)\to f(x)$ almost everywhere.

The forward difference quotient of the primitive is $A_r^{+}f(x)$, and the backward quotient is $A_r^{-}f(x)$:They have the same limit, proving differentiability almost everywhere, with the stronger conclusionThis proves differentiation of an indefinite Lebesgue integral directly from the maximal estimate and density, without assuming $m_u(f)\ge |f|$ as an unproved prerequisite.

A norm estimate controls the worst possible refined differences for every choice of the initial controls. Submultiplicativity turns a length-$p$ contraction into a bound for arbitrary long products, so it proves a certain number of continuous derivatives. Its absolute values can lose cancellations; failure of the estimate is inconclusive. It therefore gives a lower bound on smoothness.

Conversely, an observable eigenmode of a repeated local product exhibits a particular sequence of shrinking neighborhoods with inadequate decay. General-position controls have a nonzero coefficient in that mode, so greater derivative continuity is impossible. Special symmetric or polynomial data may suppress the bad mode; this is why the general-position qualification matters. The obstruction gives an upper bound on smoothness, while failure to find an obstruction proves nothing. A lower bound on spectral growth becomes an upper bound on smoothness because large surviving differences correspond to rough behavior.

For example, the binary Cardinal cubic B-spline subdivision mask has symbol $M(z)=(1+z{)}^4/8$, up to an irrelevant index shift. At derivative order $r=2$, the first-difference subdivision mask is $B_2=(1+z)/2$, whose maximum parity sum is $1/2$: the limit is at least $C^2$. At $r=3$, $B_3=1$, and the surviving difference mode does not decay. For generic controls the third derivative is piecewise constant with nonzero jumps, so the subdivision curve is not $C^3$. Here the lower and upper tests determine the exact integer continuity, but they need not coincide for an arbitrary subdivision mask. Stability, sum rules and borderline modes must be handled explicitly before interpreting any numerical estimate as a continuity level.

For a bounded function on an interval $I$, write ${\omega }_I(f)={\sup }_{I}f-{\inf }_{I}f$ for its oscillation of a function on an interval. For a partition $P$, the difference between its upper Darboux sum and lower Darboux sum is ${\sum }_{I\in P}|I|{\omega }_I(f)$. The Riemann integrability criterion says that this can be made arbitrarily small.

Since ${\omega }_I(f_1+f_2)\le {\omega }_I(f_1)+{\omega }_I(f_2)$, choose good partitions for the two functions and take their common refinement. Darboux sum refinement monotonicity makes both oscillation sums small on this refinement, so their sum is Riemann integrable. The same argument handles differences and scalar multiples.

Both $t\mapsto \max (t,0)$ and $t\mapsto |t|$ are Lipschitz continuous with constant one. Thus ${\omega }_I(f^{+})\le {\omega }_I(f)$ and ${\omega }_I(|f|)\le {\omega }_I(f)$. This calculation shows that Lipschitz composition preserves Riemann integrability, so the positive part and the absolute value are integrable.

The converse for the absolute value is false. Define $f=1$ on rational numbers and $f=-1$ on irrational numbers. Then $|f|=1$ is integrable, but every interval has supremum one and infimum minus one, so every Darboux difference for $f$ is $2(b-a)$. Hence $f$ is not Riemann integrable.

Finally,is integrable by the closure properties just proved. This establishes that Riemann integrability is closed under addition and maximum.

A countable supremum can fail, even for a uniformly bounded family. Enumerate the rational points of $[a,b]$ as $q_1,q_2,\dots$ and set $f_n=1_{\{q_n\}}$. Each singleton indicator function is Riemann integrable with integral zero: isolate its single nonzero point in intervals of arbitrarily small total length. Butis the nonintegrable Dirichlet function. This proves that a pointwise supremum need not preserve Riemann integrability.

Assume $V>0$ and that its derivatives do not change substantially over the field interval traversed in the estimate. With initially negligible kinetic energy, $H^2\simeq V/(3M_{\mathrm{Pl}}^2)$. The force $-V_{,\varphi }$ changes the field velocity on the Hubble friction time. Keeping $H$ and $V_{,\varphi }$ approximately constant for that local estimate, and starting at rest, solves the field equation explicitly:After one Hubble time,Writing $A=M_{\mathrm{Pl}}^2{V}_{,\varphi }^2/V^2$, the condition $A<1$ keeps this kinetic-to-potential ratio below about $0.31$, comfortably inside the inflationary bound. Integrating once more gives the fractional potential change estimateover the same interval. Thus the characteristic time for an order-one potential change is $H^{-1}/A$, up to constants of order one. The requested local Hubble-timescale criterion isUsing a strict inequality with one as the order-one threshold is conventional; its coefficient is not an exact universal end-of-inflation bound. Much smaller $A$ gives the controlled slow-roll regime and many Hubble times of cosmic inflation if the bounds persist.

After the transient in relaxation to the slow-roll attractor, $3H\dot{\varphi }\simeq -V_{,\varphi }$. Differentiating this approximate relation givesTherefore the standard sufficient conditions for negligible acceleration throughout slow roll are ${ϵ}_V\ll 1$ andThe absolute value matters: large negative curvature also drives rapid evolution. The more precise local acceleration criterion is $|{\eta }_V-{ϵ}_V|\ll 1$, with the usual separate smallness conditions avoiding a tuned cancellation. If a field starts exactly at rest on a nonzero slope, its initial acceleration is not negligible; the displayed exponential transient describes its approach to the attractor.

These are local timescale and slow-roll conditions, rather than a global duration theorem from derivatives at a single point. A smooth potential can have a narrow flat shoulder followed by a steep drop, ending cosmic inflation soon despite small starting derivatives. The conditions must hold over the traversed interval. This is the additional smooth-evolution assumption implicit in the estimate.

We prove the Roth theorem on three-term arithmetic progressions by a density increment on an ordinary integer arithmetic progression. The required increment will be small, but depend only on the current subset density.

First establish the increment step. Suppose $A\subseteq [1,L]$ has subset density $\rho =|A|/L>0$ and no nonconstant three-term arithmetic progression. Put $M=4L+1$, regard $[1,L]$ as a subset $I$ of the cyclic group ${\mathbb{Z}}_M$, and set $f=1_A-\rho 1_I$. Then $\mathbb{E}f=0$ and $|f|\le 1$. Use the normalized Fourier coefficient on a finite abelian groupThe character orthogonality geometric-series identity ${\mathbb{E}}_x{e}^{2\pi irx/M}=1$ for $r=0$ and zero otherwise gives Fourier inversion on a finite group, Parseval identity on a finite group, andSince $M$ is odd, multiplication by two permutes its frequencies. If one argument is $f$ and the other two have absolute value at most one, Cauchy-Schwarz and the Parseval identity on a finite group give $|T|\le {\max }_r|\hat{f}(r)|$.

A modular arithmetic progression whose three entries lie in $I$ satisfies the corresponding integer equation: the absolute value of $x+z-2y$ is less than $2L<M$, so congruence to zero forces equality. Thus $T(1_A,1_A,1_A)=|A|/M^2\le 1/(16L)$, since only constant arithmetic progressions remain. Meanwhile, the choices $1\le x\le \lfloor L/2\rfloor$ and $0\le h\le \lfloor L/4\rfloor$ give at least $L^2/16$ arithmetic progressions in $I$ for $L\ge 8$. As $M\le 5L$, this gives $T(1_I,1_I,1_I)\ge 1/400$. Expand $1_A=\rho 1_I+f$. The seven terms other than the main term are each bounded by $\max |\hat{f}|$. For $L\ge 50{\rho }^{-3}$,Consequently a nonzero frequency $r$ obeys $|\hat{f}(r)|\ge \gamma$, where $\gamma ={\rho }^3/6000$. It is nonzero because $\hat{f}(0)=0$.

We turn this coefficient into a local increase. Let $Q=\lfloor \sqrt{L}\rfloor$. Partition the unit circle into $Q$ equal intervals and consider the $Q+1$ points $jr/M$, $0\le j\le Q$, modulo one. Two occupy the same interval, yielding an integer $1\le q\le Q$ with $\Vert qr/M{\Vert }_{\mathbb{R}/\mathbb{Z}}\le 1/Q$. Set $\ell =\lfloor \gamma Q/200\rfloor$. If $L$ exceeds a sufficiently large absolute multiple of ${\rho }^{-6}$, then $\ell \ge 1$ and each residue-class arithmetic progression of step $q$ in $[1,L]$ has at least $\ell$ terms. Partition each into pieces of lengths between $\ell$ and $2\ell$, distributing the final remainder among the pieces.

On each such arithmetic progression $P$, the character of a finite abelian group $\chi (x)=e^{-2\pi irx/M}$ differs from its value at the first point by at most $4\pi \ell /Q<\gamma /4$. Write $s_P={\sum }_{x\in P}f(x)$. Thereforeso ${\sum }_P|s_P|\ge 3\gamma L/4$. But ${\sum }_P{s}_P={\sum }_{I}f=0$, and hence the sum of the positive $s_P$ is at least $3\gamma L/8$. Some piece must consequently satisfy $s_P/|P|\ge \gamma /4$. On that piece,Also, once $\gamma Q\ge 400$, $|P|\ge \ell \ge \gamma \sqrt{L}/800$. This proves the one-frequency density increment on an integer interval: for absolute constants $c,C>0$, if $L\ge C{\rho }^{-6}$, there is a arithmetic progression $P\subseteq [1,L]$ of length at least $c{\rho }^3\sqrt{L}$ on which the subset density rises by at least $c{\rho }^3$. For example one may decrease $c$ to $10^{-7}$ and choose $C$ large enough for the displayed elementary inequalities.

Now fix $0<\delta \le 1$ and suppose the required conclusion fails. On each new arithmetic progression, identify its points in order with $[1,L^{\prime }]$. This affine identification preserves all integer arithmetic progressions, so its copy of $A\cap P$ is still progression-free. All densities remain at least $\delta$. Put $\epsilon =c{\delta }^3$, $H=C{\delta }^{-6}$, and choose an integer $t>1/\epsilon$. Every increment raises the subset density by at least $\epsilon$ and changes the length by $L_{j+1}\ge \epsilon \sqrt{L_j}$. Iterating the latter inequality givesChoose $N>(H/{\epsilon }^2{)}^{2^t}$. Then every length up to step $t$ exceeds $H$, so every increment is available. After $t$ increments the subset density exceeds one, a contradiction. ThusFor $\delta >1$ there are no sets satisfying the subset density assumption. The argument above proves the substantive case without invoking the theorem being asked for.

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 $\Vert f{\Vert }_{U^3}\ge |G{|}^{-1/8}\Vert f{\Vert }_2$.

Since $f(0)=0$, its difference quotient isThe absolute value of cosine is at most $1$, so this quotient lies between $0$ and $|h|$. The squeeze theorem proves