I am trying to solve exercise 2.4.3 from Tao-Vu book
Prove Corollary 2.24. What value of the implicit constant in the $O()$
notation do you get?Corollary 2.24 (Asymmetric sum set inequalities,
preliminary version) Let $A,B$ be additive sets with common ambient
group $Z$. Then we have estimates $$d(n_1A-n_2A+n_3B-n_4B,
n_5A-n_6A+n_7B-n_8B)=O((n_1+\dots+n_8)d(A,B))$$ for any $n_1,\dots,
n_8\in \mathbb{N}$.
An attemp of proof: By definition of Ruzsa distance we have $$d(n_1A-n_2A+n_3B-n_4B,
n_5A-n_6A+n_7B-n_8B)=$$ $$=\log \dfrac{|(n_1+n_6)A-(n_2+n_5)A+(n_3+n_8)B-(n_4+n_7)B|}{|n_1A-n_2A+n_3B-n_4B|^{1/2}|n_5A-n_6A+n_7B-n_8B|^{1/2}}\leq$$ $$\leq\log \dfrac{|(n_1+n_6)A-(n_2+n_5)A||(n_3+n_8)B-(n_4+n_7)B|}{|n_1A-n_2A+n_3B-n_4B|^{1/2}|n_5A-n_6A+n_7B-n_8B|^{1/2}}. \quad \quad(*)$$ Now we can use inequality $(2.17)$ on page 73 which says that $|mA-nA|\leq \delta[A]^{5(n+m-1)}|A|$ for all $m,n\geq 1$. Hence $$(*)\leq \log \dfrac{\delta[A]^{5(n_1+n_2+n_5+n_6-1)}\delta[B]^{5(n_3+n_4+n_7+n_8-1)}|A||B|}{|n_1A-n_2A+n_3B-n_4B|^{1/2}|n_5A-n_6A+n_7B-n_8B|^{1/2}}=$$ $$=5(n_1+n_2+n_5+n_6-1)d(A,A)+5(n_3+n_4+n_7+n_8-1)d(B,B)+$$ $$+\log \dfrac{|A||B|}{|n_1A-n_2A+n_3B-n_4B|^{1/2}|n_5A-n_6A+n_7B-n_8B|^{1/2}}.$$ Ruzsa's triangle inequality tell us that $d(A,A), d(B,B)\leq 2 d(A,B).$ Hence $$(*)\leq 10(n_1+\dots+n_8)d(A,B)+\log \dfrac{|A||B|}{|n_1A-n_2A+n_3B-n_4B|^{1/2}|n_5A-n_6A+n_7B-n_8B|^{1/2}}.$$ But the second term above can be estimated above roughly by $\log \sqrt{|A||B|}$.
I was trying to solve this probem all day but stucked. Would be grateful for any hints!
Best Answer
This is not an answer but this is what I've gotten so far.
By definition of Ruzsa distance we have: $$d(n_1A-n_2A+n_3B-n_4B, n_5A-n_6A+n_7B-n_8B)=$$ $$=\log \dfrac{|(n_1+n_6)A-(n_2+n_5)A+(n_3+n_8)B-(n_4+n_7)B|}{|n_1A-n_2A+n_3B-n_4B|^{1/2}|n_5A-n_6A+n_7B-n_8B|^{1/2}}\leq$$ $$\leq \log \dfrac{|\alpha A-\alpha A+\alpha B-\alpha B|}{|n_1A-n_2A+n_3B-n_4B|^{1/2}|n_5A-n_6A+n_7B-n_8B|^{1/2}}, \quad \quad (*)$$ where $\alpha=\sum \limits_{k=1}^8n_k.$ Hence $$(*)\leq\log \dfrac{|\alpha(A-B)-\alpha(A-B)|}{|A|^{1/2}|B|^{1/2}}.$$ Also we know that $|\alpha(A-B)-\alpha(A-B)|\leq \delta[A-B]^{10\alpha}|A-B|.$ So we obtain that $$\leq \log \dfrac{\delta[A-B]^{10\alpha}|A-B|}{|A|^{1/2}|B|^{1/2}}=10\alpha \cdot d(A-B,A-B)+d(A,B).$$ That means that we need to obtain some bound on $d(A-B,A-B)$ in terms of $d(A,B)$.
EDIT (this edit has been added after Thomas Blooms's hint:
Dear, Prof. Bloom! I tried to follow your hint(of course, if I understand it correctly) and this is what I obtained so far: so our goal is to obtain some upper bound for $d(A-B,A-B)$ in terms of $d(A,B)$. Assume that $|A|\geq |B|$, then by Ruzsa's covering lemma we have that $B\subseteq A-A+X$ for some additive set $X$ suh that $|X|\leq \frac{|A-B|}{|A|}$.
Hence $A-A+B-B\subseteq 3A-3A+X-X$ and $$d(A-B,A-B)=\log \dfrac{|A-A+B-B|}{|A-B|}\leq \log\dfrac{|3A-3A||X|^2}{|A-B|}\leq \log \dfrac{\delta[A]^{5(3+3-1)}|A||A-B|^2}{|A|^2|A-B|}=$$ $$=\log \dfrac{\delta[A]^{25}|A-B|}{|A|}=25d(A,A)+\log \dfrac{|A-B|}{|A|}\leq 25d(A,A)+\log \dfrac{|A-B|}{|A|^{1/2}|B|^{1/2}}=25d(A,A)+d(A,B)\leq $$ $$\leq51d(A,B).$$
If $|A|\leq |B|$, then we need to cover $A$ by at most $\frac{|A-B|}{|B|}$ translates of $B-B$. The same reasoning shows us that $d(A-B,A-B)\leq 51 d(A,B)$. I guess this constant can be improved but that was not my final goal. Did I understand you correctly?