# Projected Area vs. Surface Area of a 3D Set

(In what follows, I'm making up the nomenclature as I go along, so please pardon anything nonstandard.)

Suppose I have a set of points $A \in \mathcal{R}^3$ which is compact, convex, and simply connected. Let's call its surface area $S(A)$. Given a unit vector $\hat{n}$ in some direction, I define the area of the projection of $A$ onto a plane with normal $\hat{n}$ as $\mathcal{P}_{\hat{n}}(A)$. I can imagine varying $\hat{n}$ to find the direction that maximizes $\mathcal{P}_{\hat{n}}(A)$. Clearly (?) this maximum projected area will be less than $S(A)$. In fact, my intuition tells me that there should exist some inequality of the form $$\max_{\hat{n}}\, \mathcal{P}_{\hat{n}}(A) \le C\, S(A)\qquad\qquad \text{(1)}$$ for some general $C \in (0, 1]$ which is independent of $A$.

For example, for the special case where $A$ is a sphere, we can see that $S(A) = 4\pi R^2$ and $\mathcal{P}_{\hat{n}}(A) = \pi R^2$ for any $\hat{n}$, so that Eq. (1) above becomes an equality with $C = 1/4$. But what I want is a general inequality with a (hopefully optimal) value of $C$ which is independent of $A$ for suitably "nice" sets $A$.

Is my intuition correct, does a general inequality like Eq. (1) exist? If so, what is $C$? If not, can someone provide a counterexample?

Thanks very much!

• Convex implies simply connected — indeed, it implies contractibility — so I'm a bit puzzled by the inclusion of both assumptions. Is there something missing or extra?
– user21467
Oct 20, 2014 at 16:53
• @StevenTaschuk No need to be puzzled: I'm a physicist, not a mathematician, so I just listed all of the "niceness" conditions I could think of without worrying about whether they were redundant... although in retrospect, I can see that convexity implies simply-connected. Oct 20, 2014 at 17:01
• Oh, okay. No matter. I imagine the best constant is $C=\frac12$, with equality for two-dimensional convex sets (or, if you prefer, arbitrarily close to equality for sufficiently thin pancakes).
– user21467
Oct 20, 2014 at 17:05
• That seems plausible, but is there a proof? (I've learned to be wary of tricky counterexamples.) Oct 20, 2014 at 17:16

Yes, with $C=\frac12$ (which is sharp, as may be seen by considering a two-dimensional convex set, or an arbitrarily thin pancake if you prefer nondegenerate examples).

To prove the inequality with $\frac12$, first consider the case when the convex set is symmetric under reflection in the plane onto which it's being projected. Then the projection of the whole convex set is the same as the projection of just the half of it on one side of the plane, and its area is smaller than that of the half of the surface on that side because orthogonal projection is a contraction.

If the convex set isn't symmetric under reflection in the plane onto which it's being projected, apply Steiner symmetrization wrt that plane, which doesn't change the projection onto that plane but reduces surface area (see, e.g., Gruber's Convex and Discrete Geometry, Springer 2007, Proposition 9.1(vii)).

• You might also find the Cauchy surface area formula interesting, though I don't think it helps with your precise question. See arxiv.org/abs/1109.0595
– user21467
Oct 20, 2014 at 17:56
• Oh, and if you're able to apply an affine transformation to your convex set to make it less flat, you can get a bound of order $\frac c{\sqrt n}$ (where $n$ is the dimension). See Giannopoulos and Papadimitrakis, "Isotropic surface area measures", Mathematika 46 (1999), pp.1–13, doi:10.1112/S0025579300007518
– user21467
Oct 20, 2014 at 18:08
• Thanks for your responses, which answer my question. For anyone who stumbles upon this post in the future, and who isn't familiar with Steiner symmetrization (as I wasn't), here is a link to some notes which explain it: math.utah.edu/~treiberg/Steiner/SteinerSlides.pdf Oct 21, 2014 at 15:57