Proof sketch via Fourier analysis

In 1953, Roth used Fourier analysis to prove an upper bound of ${\displaystyle r_{3}([N])=O\left({\frac {N}{\log \log N}}\right)}$. Below is a sketch of this proof.

Define the Fourier transform of a function ${\displaystyle f:\mathbb {Z} \rightarrow \mathbb {C} }$ to be the function ${\displaystyle {\widehat {f}}}$ satisfying

${\displaystyle {\widehat {f}}(\theta )=\sum _{x\in \mathbb {Z} }f(x)e(-x\theta )}$,

where ${\displaystyle e(t)=e^{2\pi it}}$.

Let ${\displaystyle A}$ be a 3-AP-free subset of ${\displaystyle \{1,\dots ,N\}}$. The proof proceeds in 3 steps.

1. Show that a ${\displaystyle A}$ admits a large Fourier coefficient.
2. Deduce that there exists a sub-progression of ${\displaystyle \{1,\dots ,N\}}$ such that ${\displaystyle A}$ has a density increment when restricted to this subprogression.
3. Iterate Step 2 to obtain an upper bound on ${\displaystyle |A|}$.

Proof sketch via graph regularity

Below is an outline of a proof using the Szemerédi regularity lemma.

Let ${\displaystyle G}$ be a graph and ${\displaystyle X,Y\subseteq V(G)}$. We call ${\displaystyle (X,Y)}$ an ${\displaystyle \epsilon }$-regular pair if for all ${\displaystyle A\subset X,B\subset Y}$ with ${\displaystyle |A|\geq \epsilon |X|,|B|\geq \epsilon |Y|}$, one has ${\displaystyle |d(A,B)-d(X,Y)|\leq \epsilon }$.

A partition ${\displaystyle {\mathcal {P}}=\{V_{1},\ldots ,V_{k}\}}$ of ${\displaystyle V(G)}$ is an ${\displaystyle \epsilon }$-regular partition if

${\displaystyle \sum _{(i,j)\in [k]^{2},(V_{i},V_{j}){\text{ not }}\epsilon {\text{-regular}}}|V_{i}||V_{j}|\leq \epsilon |V(G)|^{2}}$.

Then the Szemerédi regularity lemma says that for every ${\displaystyle \epsilon >0}$, there exists a constant ${\displaystyle M}$ such that every graph has an ${\displaystyle \epsilon }$-regular partition into at most ${\displaystyle M}$ parts.

We can also prove that triangles between ${\displaystyle \epsilon }$-regular sets of vertices must come along with many other triangles. This is known as the triangle counting lemma.

Triangle Counting Lemma: Let ${\displaystyle G}$ be a graph and ${\displaystyle X,Y,Z}$ be subsets of the vertices of ${\displaystyle G}$ such that ${\displaystyle (X,Y),(Y,Z),(Z,X)}$ are all ${\displaystyle \epsilon }$-regular pairs for some ${\displaystyle \epsilon >0}$. Let ${\displaystyle d_{XY},d_{XZ},d_{YZ}}$ denote the edge densities ${\displaystyle d(X,Y),d(X,Z),d(Y,Z)}$ respectively. If ${\displaystyle d_{XY},d_{XZ},d_{YZ}\geq 2\epsilon }$, then the number of triples ${\displaystyle (x,y,z)\in X\times Y\times Z}$ such that ${\displaystyle x,y,z}$ form a triangle in ${\displaystyle G}$ is at least

${\displaystyle (1-2\epsilon )(d_{XY}-\epsilon )(d_{XZ}-\epsilon )(d_{YZ}-\epsilon )\cdot |X||Y||Z|}$.

Using the triangle counting lemma and the Szemerédi regularity lemma, we can prove the triangle removal lemma, a special case of the graph removal lemma.^[6]

Triangle Removal Lemma: For all ${\displaystyle \epsilon >0}$, there exists ${\displaystyle \delta >0}$ such that any graph on ${\displaystyle n}$ vertices with less than or equal to ${\displaystyle \delta n^{3}}$ triangles can be made triangle-free by removing at most ${\displaystyle \epsilon n^{2}}$ edges.

This has an interesting corollary pertaining to graphs ${\displaystyle G}$ on ${\displaystyle N}$ vertices where every edge of ${\displaystyle G}$ lies in a unique triangle. In specific, all of these graphs must have ${\displaystyle o(N^{2})}$ edges.

Take a set ${\displaystyle A}$ with no 3-term arithmetic progressions. Now, construct a tripartite graph ${\displaystyle G}$ whose parts ${\displaystyle X,Y,Z}$ are all copies of ${\displaystyle \mathbb {Z} /(2N+1)\mathbb {Z} }$. Connect a vertex ${\displaystyle x\in X}$ to a vertex ${\displaystyle y\in Y}$ if ${\displaystyle y-x\in A}$. Similarly, connect ${\displaystyle z\in Z}$ with ${\displaystyle y\in Y}$ if ${\displaystyle z-y\in A}$. Finally, connect ${\displaystyle x\in X}$ with ${\displaystyle z\in Z}$ if ${\displaystyle (z-x)/2\in A}$.

This construction is set up so that if ${\displaystyle x,y,z}$ form a triangle, then we get elements ${\displaystyle y-x,{\frac {z-x}{2}},z-y}$ that all belong to ${\displaystyle A}$. These numbers form an arithmetic progression in the listed order. The assumption on ${\displaystyle A}$ then tells us this progression must be trivial: the elements listed above are all equal. But this condition is equivalent to the assertion that ${\displaystyle x,y,z}$ is an arithmetic progression in ${\displaystyle \mathbb {Z} /(2N+1)\mathbb {Z} }$. Consequently, every edge of ${\displaystyle G}$ lies in exactly one triangle. The desired conclusion follows. ${\displaystyle \blacksquare }$

Improving bounds

There has also been work done on improving the bound in Roth's theorem. The bound from the original proof of Roth's theorem showed that

${\displaystyle r_{3}([N])\leq c\cdot {\frac {N}{\log \log N}}}$

for some constant ${\displaystyle c}$. Over the years this bound has been continually lowered by Szemerédi,^[12] Heath-Brown,^[13] Bourgain,^[14][15] and Sanders.^[16][17] The current (July 2020) best bound is due to Bloom and Sisask^[11] who have showed the existence of an absolute constant c>0 such that

${\displaystyle r_{3}([N])\leq {\frac {N}{(\log N)^{1+c}}}.}$

In February 2023 a preprint by Kelley and Meka^[18][19] gave a new bound of

${\displaystyle r_{3}([N])\leq 2^{-O((\log N)^{c})}\cdot N}$.

and four days later Bloom and Sisask simplified the result and with a little improvement to ${\displaystyle r_{3}([N])\leq \exp(-c(\log N)^{1/11})N}$.^[20] Several months later, Bloom and Sisask obtained a further improvement to ${\displaystyle r_{3}([N])\leq \exp(-c(\log N)^{1/9})N}$, and stated (without proof) that their techniques can be used to show ${\displaystyle r_{3}([N])\leq \exp(-c(\log N)^{5/41})N}$.^[21]

There has also been work done on the other end, constructing the largest set with no three-term arithmetic progressions. The best construction has barely been improved since 1946 when Behrend^[22] improved on the initial construction by Salem and Spencer and proved

${\displaystyle r_{3}([N])\geq N\exp(-c{\sqrt {\log N}})}$.

Due to no improvements in over 70 years, it is conjectured that Behrend's set is asymptotically very close in size to the largest possible set with no three-term progressions.^[11] If correct, the Kelley-Meka bound will prove this conjecture.

Roth's theorem in finite fields

As a variation, we can consider the analogous problem over finite fields. Consider the finite field ${\displaystyle \mathbb {F} _{3}^{n}}$, and let ${\displaystyle r_{3}(\mathbb {F} _{3}^{n})}$ be the size of the largest subset of ${\displaystyle \mathbb {F} _{3}^{n}}$ which contains no 3-term arithmetic progression. This problem is actually equivalent to the cap set problem, which asks for the largest subset of ${\displaystyle \mathbb {F} _{3}^{n}}$ such that no 3 points lie on a line. The cap set problem can be seen as a generalization of the card game Set.

In 1982, Brown and Buhler^[23] were the first to show that ${\displaystyle r_{3}(\mathbb {F} _{3}^{n})=o(3^{n}).}$ In 1995, Roy Mesuhlam^[24] used a similar technique to the Fourier-analytic proof of Roth's theorem to show that ${\displaystyle r_{3}(\mathbb {F} _{3}^{n})=O\left({\frac {3^{n}}{n}}\right).}$ This bound was improved to ${\displaystyle O(3^{n}/n^{1+\epsilon })}$ in 2012 by Bateman and Katz.^[25]

In 2016, Ernie Croot, Vsevolod Lev, Péter Pál Pach, Jordan Ellenberg and Dion Gijswijt developed a new technique based on the polynomial method to prove that ${\displaystyle r_{3}(\mathbb {F} _{3}^{n})=O(2.756^{n})}$.^[26][27][28]

The best known lower bound is approximately ${\displaystyle 2.218^{n}}$, given in 2022 by Tyrrell.^[29]

Roth's theorem with popular differences

Another generalization of Roth's theorem shows that for positive density subsets, there not only exists a 3-term arithmetic progression, but that there exist many 3-APs all with the same common difference.

Roth's theorem with popular differences: For all ${\displaystyle \epsilon >0}$, there exists some ${\displaystyle n_{0}=n_{0}(\epsilon )}$ such that for every ${\displaystyle n>n_{0}}$ and ${\displaystyle A\subset \mathbb {F} _{3}^{n}}$ with ${\displaystyle |A|=\alpha 3^{n},}$ there exists some ${\displaystyle y\neq 0}$ such that ${\displaystyle |\{x:x,x+y,x+2y\in A\}|\geq (\alpha ^{3}-\epsilon )3^{n}.}$

If ${\displaystyle A}$ is chosen randomly from ${\displaystyle \mathbb {F} _{3}^{n},}$ then we would expect there to be ${\displaystyle \alpha ^{3}3^{n}}$ progressions for each value of ${\displaystyle y}$. The popular differences theorem thus states that for each ${\displaystyle |A|}$ with positive density, there is some ${\displaystyle y}$ such that the number of 3-APs with common difference ${\displaystyle y}$ is close to what we would expect.

This theorem was first proven by Green in 2005,^[30] who gave a bound of ${\displaystyle n_{0}={\text{tow}}((1/\epsilon )^{O(1)}),}$ where ${\displaystyle {\text{tow}}}$ is the tower function. In 2019, Fox and Pham recently improved the bound to ${\displaystyle n_{0}={\text{tow}}(O(\log {\frac {1}{\epsilon }})).}$^[31]

A corresponding statement is also true in ${\displaystyle \mathbb {Z} }$ for both 3-APs and 4-APs.^[32] However, the claim has been shown to be false for 5-APs.^[33]