# Surface area of a convex set less than that of its enclosing sphere?

Is the boundary measure ("surface area") of a convex set in $\mathbb{R}^n$ less than the boundary measure of it's enclosing hypersphere (smallest hypersphere that contains the set)?

In 2D I've found a paper that states it's true, and intuitively I think it should be true for n-D, but I'm having trouble proving it.

Edit Possible solution: approximate the boundary of the inner set with panels - line segments/triangles/tetrahedra/ect depending on the dimension. Then orthogonally project those panels onto the hypersphere. Since the set is convex, the projections of the panels don't overlap, and since the projections are orthogonal, their projections onto the sphere are larger than the original panels.

Edit 2 More general conjecture: If $X$ and $Y$ are convex sets with $X \subset Y \subset \mathbb{R}^n$, then $|\partial X| \le |\partial Y|$

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Not really sure how that helps, could you elaborate? The statement is not true for non-convex sets (eg, draw a really wiggly curve inside a circle) – Nick Alger Aug 19 '11 at 8:16
Can we assume the boundary is sufficiently smooth? – anon Aug 19 '11 at 8:45
This is not an answer; can't comment: Tried en.wikipedia.org/wiki/Coarea_formula? – user14790 Aug 19 '11 at 9:16
Slice the convex set alongside $(n-1)$-dimensional hyperplanes in a generic direction (you will probably need sard's theorem here). The result of such a slice again a convex set, and the slice of the outer circle is again circle (possibly bigger than the new outer circle, but never smaller). Now proceed by induction on $n$ and use fubini's theorem for the surface area of the outer circle from surface area of slices. – Alexander Thumm Aug 19 '11 at 9:30

Let the set be $C$ and the sphere $S$. Map $S$ onto $C$ so that $f(s)$ is the closest point to $s$ in $C$. Then $f$ is a contraction, so it decreases Hausdorff measure of any dimension.
To see that $f$ is a contraction, let $f(s_1) = c_1$ and $f(s_2) = c_2 \ne c_1$. Then $(c_1 - c_2) \cdot (s_1 - c_1) \ge 0$ (otherwise you get a closer point to $s_1$ by moving from $c_1$ a short distance in the direction of $c_2$), and similarly $(c_2 - c_1) \cdot (s_2 - c_2) \ge 0$. Add these and rearrange to get $\|c_1 - c_2\|^2 \le (c_1 - c_2) \cdot (s_1 - s_2)$, and use the Cauchy-Schwarz inequality.
This works once it is shown that $f$ is onto $C$. Convexity of $C$, which was not used above, can be used to show that $f$ is onto. – robjohn Aug 19 '11 at 20:15
for any point $c$ on the boundary of a convex set $C\subset\mathbb{R}^n$, there is a hyperplane $P$ containing $c$ and so that no points of $C$ lie on one of the sides of $P$. – robjohn Aug 19 '11 at 20:37
Consider the point $s\in S$ on the side of $P$ that doesn't contain $C$ that is on the line perpendicular to $P$ and containing $c$. $f(s)=c$. – robjohn Aug 19 '11 at 20:40
Convexity of $C$ was used in showing $f$ to be a contraction. – Robert Israel Aug 19 '11 at 23:27
Ah, yes. The existence of a point between $c_1$ and $c_2$. – robjohn Aug 19 '11 at 23:41