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We take $\alpha ,\beta >0$, as is necessary for the assertion. The nonvacuous cases have $0<\alpha \le 1/2$ and $0<\beta \le 1$. Put $n=|A|$ andWe prove a lower bound for $K$ by transferring small sumset growth through algebraic operations on the scalars. This gives a self-contained polynomial skew-sumset expansion over a prime field proof using the permitted Ruzsa triangle inequality, Ruzsa covering lemma and Plünnecke-Ruzsa inequality.

Since $|B|>1$, choose $b_0\in B\setminus \{0\}$. The bound $|b_{0}A+A|\le K|b_{0}A|$ and the mixed-set Plünnecke-Ruzsa inequality giveIn particular $|A+A|\le K^{2}n$ and $|3A-2A|\le K^{5}n$. Here repeated-set sums such as $3A$ allow independent choices from each copy.

For a scalar $t$, write $D(t)=|A+tA|/n$, and put $Q=A-A$ for the difference set. The Ruzsa covering lemma gives $tA\subseteq T_t+Q$ with $|T_t|\le D(t)$. Indeed choose a maximal family of disjoint translates $u+A$, with $u\in tA$; counting their union bounds the family size, and maximality forces every remaining translate to meet one of them.

For any $r,s$, the containments $A+(r+s)A\subseteq A+rA+sA$ and $A+rsA\subseteq r{T}_s+A+rQ$ give, after covering each copy of $rA,sA$,For the second bound, specifically $rQ=rA-rA\subseteq T_r-T_r+2Q$, and $A+2Q=3A-2A$. The Ruzsa triangle inequality also gives $|A-rA|\le |A+A||A+rA|/n$, so $D(-r)\le K^{2}D(r)$. These are the closure bounds for small skew-sumset scalars.

We next prove a growth lemma for the scalar sets. For $S\subseteq {\mathbb{F}}_p$ with $m=|S|\ge 2$, letIf $R\ne {\mathbb{F}}_p$, then $R+1$ is not contained in $R$: otherwise equal cardinality would make $R$ invariant under addition of one, which generates the entire prime field. Choose $r=1+(a-b)/(c-d)\notin R$. The map $(x,y)\mapsto x+ry$ on $S^2$ is injective. Indeed a collision with different second coordinates would express $r$ as a ratio of two differences from $S$, while equal second coordinates force equal first coordinates. Clearing the nonzero denominator puts its image insideso the latter set has at least $m^2$ elements. Here $SS$ is the product set, and $3SS-3SS$ denotes three independent product terms minus three others, not multiplication by the scalar three.

If $R={\mathbb{F}}_p$, let $E_t$ count the collisions $x_1+t{y}_1=x_2+t{y}_2$ with all four variables in $S$. Summing over $t$ givesThe first term comes from equal second coordinates, which force equal first coordinates. For unequal second coordinates, each quadruple determines exactly one scalar. Some $t$ consequently satisfies $E_t\le m^2+m^4/p$. The Cauchy-Schwarz inequality on the representation counts givesRepresenting $t$ as a ratio from $R$ and clearing its denominator embeds a dilate of this sumset in $2SS-2SS\subseteq 3SS-3SS$, padding with the same product in the positive and negative sums. We have proved the quadratic growth of a sixfold product difference set estimate

Starting from $S_0=B$, define $S_{j+1}=3S_j{S}_j-3S_j{S}_j$. ChooseFor $p\ge \max (4,2^{2/\beta })$, the growth lemma implies $|S_k|\ge p/2$. To see the constants explicitly, put $v_j=|S_j|/2$. Then $v_0\ge p^{\beta /2}$ and $v_{j+1}\ge \min (p/4,v_j^2)$. Induction gives $v_j\ge \min (p/4,p^{\beta 2^{j-1}})$, and the chosen $k$ makes the exponent at least one.

Every $t\in S_0$ has $D(t)\le K$. If $D(t)\le K^{L_j}$ on $S_j$, the closure bounds show that each product has exponent at most $3L_j+5$, its negative at most $3L_j+7$, and a sum of six such terms has exponent at most $18L_j+67$. Thus, withwe have $|A+tA|\le K^{L_k}n$ for every $t\in S_k$. The integer $L_k$ depends only on $\beta$.

Now apply the collision count to $A$ instead of $S$. Summing over all scalars gives at most $p{n}^2+n^4$. Since $|S_k|\ge p/2$, some $t\in S_k$ has collision count at most $2n^2+2n^4/p$. The Cauchy-Schwarz inequality therefore givesIt follows thatFor $p\ge 4^{2/\alpha }$ this gives $K\ge p^{\alpha /(2L_k)}\ge n^{\alpha /(2L_k)}$, proving the desired positive exponent for all sufficiently large primes.

For completeness, the same conclusion with a possibly smaller exponent holds for the remaining primes. Choose an integerIf $p<P_0$, choose any nonzero $b\in B$. We have $1<n<p$. If $|A+bA|=n$, all its translates $A+u$ for $u\in bA$ would coincide. Two distinct such $u$ would make $A$ invariant under a nonzero translation, forcing $A={\mathbb{F}}_p$, a contradiction. Thus $|A+bA|\ge n+1\ge n^{1+c_{\alpha ,\beta }}$ by the definition of the second exponent. Together with the large-prime case this provesPositivity of the parameters cannot be dropped: at $\alpha =0$, taking $A={\mathbb{F}}_p$ prevents any expansion; at $\beta =0$, the allowed set $B=\{0\}$ prevents expansion of any proper $A$ with more than one element. No external sum-product theorem was invoked: the scalar growth and transfer arguments were proved above.

For each element $x$ of the product set $A\cdot B$, let $s_x$ count its representations as $ab$ with $(a,b)\in A\times B$. The ratio equality $a/b=c/d$ is equivalent to $ad=cb$. Swapping the two $B$ coordinates is a bijection between the ratio-equality quadruples and the equal-product quadruples. Therefore the multiplicative energy satisfiesThe Cauchy-Schwarz inequality gives the energy lower bound:For the upper bound, place the Cartesian product $P=A\times B$ in the strictly positive quadrant. For every occupied Euclidean ray from the origin let $P_{\lambda }$ be its points and $r_{\lambda }=|P_{\lambda }|$. The slope is $\lambda =b/a$, soTake $\log$ to be the natural logarithm and put $L=\lceil \log |B|\rceil \ge 1$, using the non-triviality assumption $|B|\ge 2$. Partition the possible occupancies into $L$ classes: $[e^j,e^{j+1})$ for $0\le j<L-1$, and $[e^{L-1},|B|]$ for the last class. This last closed endpoint ensures that the partition works even when the top endpoint is attained. Within each class the ratio of any two occupancies is at most $e$.

Fix one class and order its occupied Euclidean rays by increasing slope, with occupancies $r_1,\dots ,r_t$. Write $S=|A+A||B+B|$, the cardinality of $(A+A)\times (B+B)$. If $t=1$, then $r_1^2\le |A||B|\le S$, since adding any fixed element gives an injection of $A$ into its sumset $A+A$, and similarly for $B$.

If $t\ge 2$, the sums $P_i+P_{i+1}$ contain exactly $r_i{r}_{i+1}$ different points, by injectivity of sums on two distinct rays. Indeed, the two direction vectors are linearly independent, so the coefficients of a sum uniquely recover its two summands. Positivity puts every sum strictly inside the open planar sector between those two Euclidean rays. The sectors between successive selected Euclidean rays are disjoint, even if there are additional unselected rays between them. All these sums belong to $(A+A)\times (B+B)$, and henceFor neighbouring occupancies their ratio lies between $e^{-1}$ and $e$, soSumming covers every $r_i^2$ at least once, and gives ${\sum }_i{r}_i^2\le 4S$. The same bound holds for empty or singleton classes by the preceding observations. Adding over the $L$ classes proves the multiplicative energy sumset bound $E(A,B)\le 4LS$. Combining both bounds yields the sum-product conclusion:If instead $\log$ is interpreted as base two, use the same $L=\lceil {\log }_2|B|\rceil$ occupancy classes with $2$ in place of $e$; $2+2^{-1}<4$ proves that convention as well. Positivity is essential to the sector argument.

For $S\subseteq {\mathbb{F}}_p$ with $|S|\ge 2$, put $R=(S-S)/(S-S)$ using nonzero denominators. If $R$ is proper, some $r\in R+1$ lies outside $R$, since invariance under addition of one would give all of the prime field. The map $(x,y)\mapsto x+ry$ on $S^2$ is injective. Clearing the denominator in a representation of $r-1$ puts a dilate of its image inside $3SS-3SS$. If $R$ is the full field, averaging the collision count of $S+rS$ gives some $r$ with $|S+rS|\ge \frac{1}{2}\min (p,|S{|}^2)$. Clearing its denominator puts a dilate inside $2SS-2SS\subseteq 3SS-3SS$. Here $SS$ is a product set, and the prefactors denote repeated sumsets, not scalar multiplication.