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

Suppose that $A$ has no nonconstant three-term arithmetic progression, and let $\alpha =|A|/n$ be its actual density of a finite subset. Thus $0<\delta \le \alpha \le 1$. Put $I=[n]$, $N=2n+1$, and embed $I$ in the cyclic group $G=\mathbb{Z}/N\mathbb{Z}$. A congruence $x-2y+z=0(\mathrm{mod}N)$ with $x,y,z\in I$ is an equality over the integers, since $|x-2y+z|\le 2(n-1)<N$. Thus this embedding creates no extra three-term arithmetic progressions. Since $N$ is odd, multiplication by two permutes $G$.

Use normalized Fourier analysis on a finite abelian group, with $e_N(t)=\exp (2\pi it/N)$ and $\hat{h}(r)={\mathbb{E}}_{x\in G}h(x)e_N(-rx)$. For the normalized trilinear arithmetic progression count, character orthogonality givesIndeed, expanding each function in its inverse Fourier transform, averaging in $d$ forces $r_2+2r_3=0$, and averaging in $x$ then forces $r_1=r_3$. The Parseval identity in this normalization is ${\sum }_r|\hat{h}(r){|}^2={\mathbb{E}}_x|h(x){|}^2$.

Let $a=1_A$, $b=\alpha 1_I$, and $f=a-b$, the balanced indicator function of a finite subset. Then $\hat{f}(0)=0$. Only constant arithmetic progressions contribute to $\Lambda (a,a,a)$, whereas summing $2\min (y-1,n-y)+1$ over centers $y\in I$ counts $(n^2+1)/2$ ordered three-term arithmetic progressions in $I$. HenceSet $M={\max }_r|\hat{f}(r)|$. Telescope by changing one factor at a time:In the Fourier transform formula for each term, bound the Fourier coefficient of $f$ by $M$ and apply the Cauchy-Schwarz inequality to the other two factors. The permutation $r\mapsto -2r$ preserves the required squared sums. Since $\Vert a{\Vert }_2^2=\alpha n/N$ and $\Vert b{\Vert }_2^2={\alpha }^{2}n/N$, this givesIf $n\ge 4{\delta }^{-2}$, then ${\alpha }^{2}n\ge 4$, so the difference in the other direction is at least ${\alpha }^3{n}^2/(4N^2)$. Consequently Fourier detection of a progression-free subset of an interval gives a nonzero frequency $r$ with

To turn this into a density increment, take $Q=\lfloor \sqrt{n}\rfloor$. The Dirichlet approximation theorem, proved here by placing $0,r/N,\dots ,Qr/N$ into $Q$ equal intervals modulo one, gives $1\le q\le Q$ such that $\Vert qr/N{\Vert }_{\mathbb{R}/\mathbb{Z}}\le 1/Q$. DefineFor $n\ge N_0(\delta )$, the inequalities $Q\ge \sqrt{n}/2$ and ${\delta }^{2}Q/(288\pi )\ge 2$ show thatUse a progression partition with nearly constant linear phase: every residue chain of $I$ modulo $q$ has at least $\lfloor n/q\rfloor \ge Q$ points. Split each chain into blocks of $L$ consecutive chain points, merging any remainder into its last full block. This partitions $I$ into arithmetic progressions $P$ with $L\le |P|<2L$ and common difference $q$.

On a block $P$ with first point $z_P$, the linear phase $\chi (x)=e_N(-rx)$ satisfiesLet $S_P={\sum }_{x\in P}f(x)=|A\cap P|-\alpha |P|$. Freezing the linear phase on each block introduces error at most $({\delta }^2/72)N^{-1}{\sum }_{x\in I}|f(x)|\le {\delta }^2/72$. ThereforeThe block sums total zero. Their positive part consequently totals at least $N{\delta }^2/144$. Since all block lengths total $n$, some block has $S_P/|P|\ge N{\delta }^2/(144n)\ge {\delta }^2/72$. Its density of a finite subset is at least $\alpha +{\delta }^2/72\ge \delta (1+\delta /72)$.

For the remaining $3\le n<N_0(\delta )$, divide $I$ into disjoint consecutive triples and a remainder of at most two points. A three-term-arithmetic progression-free $A$ uses at most two points of each triple, so $\alpha \le 4/5$; the largest ratio occurs at $n=5$. Any singleton set from $A$ then has density of a finite subset one, which is at least $\delta +{\delta }^2/72$. We may therefore use the explicit constantsIn the large case the constructed arithmetic progression has length at least $c_1\sqrt{n}$; in the small case its length one does too. Thus these constants prove the intended result for all odd integers $n\ge 3$.

Write $d(U,V)=e(U,V)/(|U||V|)$ for the edge density of a bipartite graph between nonempty disjoint vertex sets. A regular pair of vertex sets $(U,V)$ is $\epsilon$-regular if $|d(U^{\prime },V^{\prime })-d(U,V)|\le \epsilon$ whenever $U^{\prime }\subseteq U$, $V^{\prime }\subseteq V$, $|U^{\prime }|\ge \epsilon |U|$, and $|V^{\prime }|\ge \epsilon |V|$.

The Szemerédi regularity lemma says that for every $\epsilon >0$ and positive integer $m_0$ there is $M=M(\epsilon ,m_0)$ such that every finite simple graph on $N\ge m_0$ vertices has a set partitionwhere $V_1,\dots ,V_k$ have equal positive size and at most $\epsilon k^2$ unordered pairs $(V_i,V_j)$, $i<j$, fail to be regular pairs of vertex sets. The version requiring only sufficiently large $N$ is equivalent, since the finitely many smaller admissible sizes can use singleton sets as classes. It suffices to prove this for $0<\epsilon \le 1/2$: reducing a larger parameter to $1/2$ gives stronger conclusions.

Here is a complete energy proof, including restoration of equal class sizes. For a set partition with exceptional set $V_0$, use the following version of equitable regularity energy, omitting pairs incident with $V_0$:For $i=j$, define $d(V_i,V_i)$ using the ordered adjacency indicator function, with diagonal entries zero. Thus the same formula works on every rectangle. The refinement variance identity for regularity energy iswhere $A,B$ run over the refined cells. Expanding the square proves the identity because their weighted mean edge density of a bipartite graph is $d(U,V)$. In particular, refinement never decreases the energy. If a union of refined cells $U^{\prime }\times V^{\prime }$ has discrepancy greater than $\epsilon$, the Cauchy-Schwarz inequality bounds its contribution below by $|U^{\prime }||V^{\prime }|{\epsilon }^2/N^2$. For an irregular equal-class pair of size $L$, witnesses have sizes at least $\epsilon L$, so the increase exceeds ${\epsilon }^4{L}^2/N^2$ in each orientation.

Choose $k_0\ge \max \{m_0,2\}$ so large that $2^{-k_0}\le {\epsilon }^6/100$. Initially partition into $k_0$ equal classes and fewer than $k_0$ exceptional vertices. For sufficiently large $N$ this exceptional set has size at most $\epsilon N/2$. Suppose there are more than $\epsilon k^2$ irregular unordered pairs at a subsequent step, with equal class size $L$. Choose one witnessing pair of subsets for every such pair. Split each class according to membership in all its witness subsets; each class produces at most $2^k$ atoms. The preceding refinement variance identity for regularity energy shows that the energy increase exceedsas long as the exceptional set has size at most $\epsilon N\le N/2$.

Apply equalization with a controlled exceptional set. Set $\ell =\lfloor L/4^k\rfloor$, cut every atom into blocks of size $\ell$, and send its leftover points to $V_0$. Provided $L\ge 2\cdot 4^k$, the new number $k^{\prime }$ of classes satisfiesThe lower bound holds because every old class retains at least one full block: its leftovers total less than $2^k\ell <L$. The discarded fraction is at mostCutting blocks is a further refinement before discarding. Discarding changes only ordered pairs with at least one discarded vertex, so it loses at most $2r$ in the energy. The net increase is therefore at least ${\epsilon }^5/2-2{\epsilon }^6/100\ge {\epsilon }^5/4$.

There can be fewer than $T=\lceil 4{\epsilon }^{-5}\rceil$ such strict energy increases, because the energy stays in $[0,1]$. Through $T$ steps the exceptional fraction is at mostThis also verifies the exceptional-set bound used at every preceding step. Iterating the finite recursion $k\mapsto 2k{4}^k$ at most $T$ times bounds all class counts by some $K=K(\epsilon ,m_0)$. Choosing $N$ sufficiently large in terms of $K$, for instance with $N\ge 4K{4}^K$ and $N\ge 2k_0/\epsilon$, guarantees $L=(N-|V_0|)/k\ge N/(2K)\ge 2\cdot 4^k$ throughout. Thus the process terminates with the desired regular pairs of vertex sets. Choose a finite integer threshold $N_{\ast }$ for these requirements and set $M\ge \max \{K,N_{\ast }\}$. For $m_0\le N<N_{\ast }$, use $N$ singleton sets; every pair is regular because its only admissible nonempty subsets are the whole pair. This completes the Szemerédi regularity lemma for all admissible $N$.

For the half graph, a class of size $s$ has exactly $\lceil s/10\rceil$ elements with fewer than $s/10$ predecessors, and the same number with fewer than $s/10$ successors. Thus its boundary contains at most $2\lceil s/10\rceil$ elements. Let $k_X,k_Y$ be the numbers of classes. The boundary rank obstruction in a half graph gives the union boundbecause $k_X,k_Y\le n/100$. There is therefore an index $m$ away from both boundaries. In its classes $X_i,Y_j$, all four subsets $X^{-}=\{x<m\}$, $X^{+}=\{x>m\}$, $Y^{-}=\{y<m\}$, $Y^{+}=\{y>m\}$ have at least one tenth of their respective class sizes. These subsets are taken within $X_i,Y_j$, which need not be intervals.

In the half graph, $d(X^{-},Y^{+})=1$ and $d(X^{+},Y^{-})=0$. If $(X_i,Y_j)$ were a $1/10$-regular pair of vertex sets, both values would be within $1/10$ of $d(X_i,Y_j)$, implying $1\le 2/10$, a contradiction. At least one class pair is not $1/10$-regular.

True. For each $y\in x$, the singleton set $\{y\}$ is a finite subset of $x$, so the hypothesis givesTaking the supremum over all $y\in x$ yields

Irreducible degrees divide the order of a finite $2$-group and satisfyThe number of linear characters is $|G/G^{\prime }|$. For a nontrivial finite $2$-group this is at least $4$ unless the group is abelian, and every nonlinear degree is at least $2$. A degree $4$ constituent is therefore impossible in the nonabelian case. If there are $\ell$ linear and $m$ degree-two characters, thenThus the only degree collections arewith respectively $r=16,10,7$.

For $1^{16}$ take the cyclic group $G=C_{16}=⟨g⟩$. Every conjugacy class is a singleton, and its character table iswhere ${\zeta }_{16}=e^{2\pi i/16}$.

For $1^8,2^2$, take $G=D_8\times C_2$, where
$D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$ and
$C_2=⟨z⟩$. On the five classesthe four linear characters of $D_8$, indexed by
$\epsilon ,\delta \in \{\pm 1\}$, and its degree-two character areThe ten classes of $G$ are $(C,1)$ and $(C,z)$. Its full table consists ofwhere $\chi$ runs over the above five rows and
$\eta \in \{1,-1\}$ runs over the two characters of $C_2$. This explicitly gives eight linear and two degree-two rows.

For $1^4,2^3$, take the dihedral groupOrder its seven classes asThe four linear rows areThe three degree-two rows, for $k=1,2,3$, areThe row norms and mutual inner products, weighted by the displayed class sizes, verify irreducibility and completeness.

First form the singleton set $\{b\}=F_1(b,b)$ and then use $F_2$ to take a set union:The unused second argument of $F_2$ may be any term. Since $x\cap c=x\setminus (x\setminus c)$, a term using only the prescribed operation symbols isAfter substituting the displayed term for both occurrences of $U$, this is literally a term in $F_1,F_2,F_3$, and its value is $(a\cup \{b\})\cap c$.