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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.