Size of a compact set in space from its shadows

Question

Say we have a finite number of disjoint balls in the space forming the set $A.$ We have some bound $R^2\pi$ for the areas of all its projections to planes. Is it true that the size of the set $A$ in space is bounded by $\frac{4}{3}R^3\pi$?

Answer 1

Ok, I got the following by myself. I only need to assume that the set $A$ is bounded and closed in space, that is $$A \subset C$$ for some ball $C > 0.$ Now, via scaling let $$c^34/3\pi = \mu(A) = \int_{B(0,1)} 1_{A}\mu$$ and $A \subset B(0,1).$ So $1_A$ is the indicator function of $A$. Let $e$ be an unit vector. So we need to show a bound $$\mu_{2}(A|e^{\bot}) \geq c^2\pi$$ for one of the projections, but I only get a bound $$\mu_{2}(A|e^{\bot}) \geq c^3\frac{4}{3}.$$ Now, we assume some uniform density $\gamma$ on $B(0,1)$. This is a scaled Lebesgue-measure in space so it projects to measure $\gamma^{2}\mu$ to a plane. We write $$c^34/3\pi =\mu(A) = \int_{B(0,1)}\gamma\mu$$ and we now should have $$\gamma^{2/3}\mu_{2}(B(0,1)|e^{\bot}) = \mu_{2}(B(0,\gamma^{1/3})|e^{\bot}) = c^{2}\pi.$$ So that $c^3 = \gamma.$ It was not easy to me to show that any kind of projection bound holds. But if we have $$\mu_{2}(A|e^{\bot}) < \frac{4c^3}{3},$$ then there exist some $\gamma_1 < \gamma = c^3$ s.t $$\mu_2(A|e^{\bot}) = \gamma_1\pi < \frac{4c^3}{3}.$$ It's clear that we have $A \cap H^e \subset (A \cap (A|e),$ where $H^e$ is a subspace orthogonal to $e.$ But now we have $A \subset \bigcup(A \cap H^e)$ and for some $H^e$ that $$\mu_2(A \cap H^e) = \gamma_1\pi.$$ Let $\delta > 0$ be small. Then we have for $(A \cap H^{e}_{\delta})$ that $$\mu(A) \leq \mu(A \cap \bigcup H^e_{\delta}) \leq \sum_{i=1}^m \mu(A \cap H^{e_i}_{\delta}) \leq \sum_{i=1}^{m} \pi\gamma_1\delta < \gamma\frac{4}{3}\pi = \mu(A),$$ which is a contradiction. Above I used compactness of $A$ in order to get a finite sum, where we should have $m \leq \frac{\pi}{\delta}.$ Now, $m \leq \frac{\pi}{\delta}$ is the right bound because $$A \subset B(0,1) \subset \bigcup_{i=1}^{\frac{\pi}{\delta}}B_2(0,1) \cap H^{e_i}.$$ So we have $$\mu_{2}(A|e^{\bot}) \geq \frac{4c^3}{3}\,$$ for some projection $(A|e^{\bot})$. This does not imply my original question. But it's clear that my proof for analogical bound works in every dimension. In other words I proved that small projections imply small volume. But what happens if the largest projections equal the largest sections? Then we would have a that a small sections imply small volume. My proof works for sections also. (EDIT) However, if we have an annulus $S_2$ of thickness $r$ in the unit ball, then I got that$$\frac{4}{3}\pi(1-(1-r)^3) = 4\pi\int_{1-r}^1 r^2 dr = \int_S = \frac{4}{3}\gamma\pi.$$ Now, an annulus has a slice of size $2\pi\int_{1-r}^1 r dr = \pi*r(2-r).$ Which means that there is a maximum slice of the annulus smaller than $\gamma^{2/3}\pi.$ So this is a counter example.