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Let $T_3(A)$ count ordered solutions $a+c=2b$ in $A^3$, including the trivial ones. The hypothesis gives $T_3(A)\ge c{n}^2$. Its relation to additive energy is the promised easy Fourier step. Translate $A$ into $[0,M]$ and embed it in an odd prime cyclic group of order $P>4M$. No relevant integer equality wraps modulo $P$. With the unnormalized Fourier transform $\hat{f}(r)={\sum }_x{1}_A(x)e(-rx/P)$, character orthogonality givesSince multiplication by $2$ permutes the characters, Cauchy-Schwarz and Parseval identity on a finite group implyThus $E(A)\ge c^2{n}^3$. Equivalently, one can apply Cauchy-Schwarz directly to $T_3(A)={\sum }_{b\in A}{r}_{A+A}(2b)$.

By part (i), some $B\subseteq A$ has $|B|\ge c_1(c)n$ and $|B+B|\le K(c)|B|$. Part 3(ii) then bounds $|2B-2B|\le K(c{)}^4|B|$. Apply part (ii) with $k=2$: there is $B^{\prime }\subseteq B$, $|B^{\prime }|\ge |B|/3$, Freiman 2-isomorphic to a subset $C\subseteq {\mathbb{Z}}_N$ with $N\le 4K(c{)}^4|B|+1$. Consequently $C$ has a positive subset density depending only on $c$, and $N$ tends to infinity with $n$ because $N\ge |B^{\prime }|$.

View the representatives of $C$ as a subset of $\{0,\dots ,N-1\}$. The assumed four-term case of Szemerédi's theorem gives four distinct integer representatives in arithmetic progression once $n$ is sufficiently large in terms of $c$. If their preimages are $b_0,b_1,b_2,b_3$, the two model equalitieslift through the Freiman 2-isomorphism to the same equalities in $\mathbb{Z}$. Hence their consecutive differences agree; injectivity makes the common difference nonzero. Therefore $A$ contains a nonconstant four-term arithmetic progression. This proves dense three-term progressions force four-term progressions without an assumption that $A$ initially lies in a short interval.