# Sumsets and dilates: does $|A+\lambda A|<|A+A|$ ever hold?

The following problem is somehow hidden in this recently asked question, but I believe that it deserves to be asked explicitly.

Is it true that for any finite set $A$ of real numbers, and any real $\lambda\notin\{0,-1\}$, one has $$|A+\lambda A| \ge |A+A|$$ (where $\lambda A=\{\lambda a\colon a\in A\}$ is the dilate of $A$ by the factor $\lambda$, and $A+B=\{a+b\colon a\in A,\ b\in B\}$ is the sumset of $A$ and $B$)?

The energy version seems equally interesting to me. Let $T_A(\lambda)$ denote the number of representations of $\lambda$ in the form $\frac{a_1-a_2}{a_3-a_4}$ with $a_1,a_2,a_3,a_4\in A$. It is easily seen that $T_A(-\lambda)=T_A(\lambda)$ and $T_A(\lambda)<T_A(0)=|A|^2(|A|-1)$ for any real $\lambda\ne 0$.

Is it true that for any finite set $A$ of real numbers, and any real $\lambda\ne 0$, one has $$T_A(\lambda) \le T_A(1) ?$$

I have not done any computations, so maybe it is possible to find a counterexample just by a computer search.

Both questions have now received nice and exhaustive answers thanks to Boris Bukh, Kevin Costello, and Terry Tao (who has actually answered even before the question got asked). Unfortunately, I cannot accept more than one answer; so, I am accepting only that which, for some reason, got less votes.

It seems the energy version is true if make the additional assumption that $$\lambda=c/d$$ is rational, meaning that $$T_A(\lambda)$$ counts the number of solutions in $$A$$ to $$d(a_1-a_2)-c(a_3-a_4)=0 \, \, \, \, \, \, (*).$$ Using a Freiman isomorphism together with translation invariance, we can assume that $$A$$ is a subset of integers lying in $$[0,n]$$.

Let $$p$$ be a prime larger than $$2(c+d)n$$. We now think of $$A$$ as a subset of $$Z_p$$, which does not change the solution count. Let $$f$$ be the characteristic function of $$A$$, and $$\hat{f}$$ its Fourier transform. The number of solutions to $$(*)$$ is then given by $$\sum_{\xi \in Z_p} \hat{f}(d \xi) \hat {f} (-d \xi) \hat{f}(c \xi) \hat{f}(-c \xi)=\sum_{\xi \in Z_p} |\hat{f} (d \xi)|^2 |\hat{f} (c \xi)|^2$$

Since $$c$$ and $$d$$ are nonzero, the two sequences $$|\hat{f} (d \xi)|^2$$ and $$|\hat{f} (c \xi)|^2$$ are permutations of each other. So it follows from the rearrangement inequality that the sum is maximized when $$c=d$$, which corresponds to $$\lambda=1$$.

It seems intuitive that moving from the rational to the reals can't increase the number of solutions, but I don't see an obvious way of extending this.

It is false. Let $\lambda$ be any integer other than $0$ and $\pm 1$. First, I will construct an example mod $p$ for every large prime $p$, which will then be upgraded to $\mathbb{Z}$ by taking Cartesian products.

For each pair $\{a,a'\}$ satisfying $a+\lambda a'=0$ toss a coin, which decides which element is "eliminated". (Note that there is no rule against double jeopardy!). Let $A$ be the set of surviving elements. It is clear that $0\not\in A+\lambda A$. On the other hand, $A+A=\mathbb{Z}/p\mathbb{Z}$ with high probability. Indeed, fix $x$ and consider the event $x\in A+A$. We can easily find linearly many pairs $\{a,x-a\}$ such that the sets of coin tosses on which they depend are disjoint. So, $\Pr[x\not\in A+A]=O(c^{-p})$ and the union bound completes the argument modulo $p$.

Let $p_1,p_2,\dotsc$ be distinct primes and let $A_{p_1},A_{p_2},\dotsc$ be the sets modulo these primes such that $0\not \in A+\lambda A$, but $A+A=\mathbb{Z}/p\mathbb{Z}$. Let $P=\prod p_i$ and $P'=\prod (p_i-1)$. Let $A=A_{p_1}\times A_{p_2}\times \dotsb A_{p_k}\subset \mathbb{Z}/P\mathbb{Z}$ (I am using the Chinese Remainder theorem). Then $|A+A|=P$ and $|A+\lambda A|\leq P'$. Since the sum of reciprocals of the primes diverges, we can make the ratio $P/P'$ as large as we wish, and in particular larger than $|\lambda|+1$. If we then treat $A$ as a subset of $\mathbb{Z}$, then the size of $A+\lambda A$ can be at most $(|\lambda|+1)P'$ which is still smaller than $P$.

A similar argument ought to give a counterexample for algebraic values of $\lambda$, but I have not worked out the details.