Freiman inequality for projective space?

Question

This question is suggested by some results in a paper I am writing. I would like to write it down there but want to make sure that it is not known or at least MO-hard.

Freiman's inequality states that for a set $A$ of vectors that span $\mathbb R^d$, we have: $$|A+A| \geq (d+1)|A| -\binom{d+1}{2} $$

My question is what happens in $\mathbb{RP}^d$? To be really concrete let $L(A)$ be the set of lines given by the vectors in $A$. Is this true that if $A$ is a set of vectors that span $\mathbb R^{d+1}$, we still have:

$$|L(A+A)| \geq (d+1)|L(A)| -\binom{d+1}{2} ?$$

If not, can any inequality be proved?

Remark: there seems to be some confusion in the comments, so let's give one example. Let $d=1$ and $A=\{(1,0), (-1,0), (0,1), (0,-1)\}$. Then $|L(A)|=2$ (the x- and y-axis) and $|L(A+A)|=4$ (the axes plus the diagonal lines). Also, it is clear that this projective version implies the original one. It is tempting to try to deduce it from the original version, but I can't see a way to do it.

Answer 1

As stated, the answer is no, Freiman's inequality no longer holds. The counter example is $A$ being the vertices of an equilateral triangle on the unit circle. I found this by looking at the proof of Freiman's inequality (given as Lemma 5.13 of Tao-Vu book on Additive Combinatorics) and see where the proof fails.

In my particular setting, the set $A$ satisfies some additional convexity conditions, so perhaps the result may hold for such situations.

UPDATE: it turned out that the proposed inequality held under one extra assumption: that the origin is outside the convex hull of $A$. That will rule out examples such as above, and still implies the classical inequality. The proof follows the same line as the original proof, with some extra care needed.