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Realizing a fixed pattern. Choose an integer $\Omega\ge1$ bounding $|x|$ and $|y|$ on $S$. Then every $d=y-x$ satisfies $|d|\le2\Omega$. Fix $3\le\alpha<4$ and choose an integer $M\ge2\Omega$ such that

$$(4-\alpha)M\ge16\Omega+1.$$

By Green and Tao's Theorem 1.1, there are integers $a\ge2$, $h\ge1$ such that $a+jh$ is prime for every $0\le j\le M+2\Omega$. Put $q=a+Mh$. For every cell of $S$,

$$q+h(y-x)=a+(M+y-x)h$$

is one of these primes. The indices lie between $0$ and $M+2\Omega$, and distinct cells have distinct indices by the injectivity of $y-x$. On the dilated cell set $hS$, take $\omega=h\Omega$. We have

$$(4-\alpha)q\ge (4-\alpha)Mh \ge(16\Omega+1)h\ge16\omega+1.$$

Thus L198 applies to $hS$ and $q$. Multiplication by the nonzero integer $h$ preserves the cardinalities of all three projections. L198 therefore gives precisely the required count in a block of $\lfloor\alpha P\rfloor$ consecutive positive integers, where $P$ is the largest chosen prime. This proves the first assertion.

The progression minimum. Define $F'_3(u)$ exactly as in the claim, using distinct positive common differences. The minimum exists because the attainable cardinalities form a nonempty subset of the positive integers.

For the lower bound, take any $u$ primes with largest member $P$ and any block $J$ of $K=\lfloor\alpha P\rfloor$ consecutive integers. Since $K\ge3P$, for every chosen prime $p$ there are three consecutive multiples $m,m+p,m+2p$ in $J$. The set of counted integers therefore contains progressions with $u$ distinct positive differences, so its size is at least $F'_3(u)$.

For the upper bound, let $A\subset\mathbb Z$ attain $F'_3(u)$. Select exactly $u$ of its distinct positive differences $d_i$, and choose $a_i,a_i+d_i,a_i+2d_i\in A$ for each. The cells $S=\{(a_i,a_i+d_i):1\le i\le u\}$ have injective differences. Their three projections are subsets of $A$, so their cardinalities sum to at most $3|A|$. The first assertion realizes this cell set as primes and a block with that sum as its exact count. Consequently

$$F'_3(u)\le G_\alpha(u)\le3F'_3(u).$$

Finally any block of $\lfloor\alpha P\rfloor\ge3P$ integers contains a subblock of $3P$ integers. Its count is therefore at least $G_3(u)$. Combining this with $F'_3(u)\le G_3(u)$ gives $G_3(u)\le G_\alpha(u)\le3G_3(u)$. No limit in $u$ or uniform bound on the sizes of the realizing primes is needed.