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Suppose we have two compact convex bodies one contained in the other in $\mathbb{R}^n$, $C\subset D\subset \mathbb{R}^n$. Does it follow that the ($n-1$ dimensional) surface area of $C$ is less than $D$? If so is there a natural sequence of $n-k$ dimensional quantities ($g_0=$Volume, $g_1=$Surface area,...) such that whenever $C\subset D\subset \mathbb{R}^n$ (where $C$ and $D$ are again compact) and $n>k$ we have $g_k(C)>g_k(D)$?

zyx
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Josh F
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  • Yes, because higher dimensional Crofton formulas exist. Adding the integral-geometry tag. – zyx May 13 '16 at 04:59

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Short answer: yes. The equivalent 2D-claim is easy to understand: assume that $A\subset B$ are two convex bodies and $P,Q$ are two distinct points on the boundary of $B$, such that the segment $PQ$ does not meet $A$. If we make a cut along the $PQ$ segment, the perimeter of $B$ strictly decreases. Moreover, we may find a sequence of cuts such that $B$ "repeatedly cutted" converges to $A$, hence the perimeter of the original $B$ set is greater than the perimeter than the $A$ set.
In the 3D context we just have to perform cuts along planes not meeting the innermost convex body and replace the word perimeter with area of the boundary: the same argument applies.

Jack D'Aurizio
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  • Thanks, that makes a lot of sense. I'm still curious whether this is a co-dimension analog of the volume case, but you've made the surface area case crystal clear. – Josh F Aug 12 '16 at 05:09