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I have a question which I believe to be pretty basic.

Let $\Gamma$ be some convex body, bounded inside a $L_2$ sphere of radius 1 $B(0,1)$.

Is it true that the surface area of $\Gamma$ is smaller than the surface area of the sphere?

I'm guessing that the answer involves finding a continuous deformation from $\Gamma$ to the sphere for which the area is monotonous, but I'm incapable of finding it

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The nearest-point projection from the sphere to the convex set decreases distances.

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    $\begingroup$ This is a very elegant answer. Thank you. $\endgroup$
    – Guillaume Dehaene
    Commented Mar 14, 2017 at 14:02
  • $\begingroup$ Why is the metric projection surjective? $\endgroup$
    – user60933
    Commented Apr 14, 2017 at 15:33
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    $\begingroup$ @user60933, the normal of any supporting hyperplane at a point $p$ of the compact set will meet the sphere in a point whose image under the nearest-point projection is $p$. $\endgroup$
    – Mikhail Katz
    Commented Apr 17, 2017 at 21:42
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This is a consequence of the standard inequalities on the "intrinsic volumes" (or "quermaß integrals", the two are synonymous up to a constant) of convex sets: for every $i$, the $i$-th intrinsic volume $V_i(K)$ of a convex set is monotone in $K$, and for a $d$-dimensional convex set, $V_{d-1}(K)$ is twice the $(d-1)$-dimensional surface area of $K$ (while $V_d(K)$ is the $d$-dimensional volume; the quick definition in general is that $V_i(K)$ is the coefficient of $r^{d-i}$ in the polynomial giving the volume of the distance $r$ ball $K+B(0,r)$ around $K$, divided by the volume of $B(0,1)$ in dimension $d-i$).

See, for example, J. R. Sangwine-Yager's chapter "Mixed Volumes" (specifically §3) in the Handbook of Convex Geometry (vol. A), edited by Gruber & Wills (North-Holland 1993).

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Yes, this is a general fact. If $K_1\subset K_2$ are convex, then the surface area of $K_1$ is at most that of $K_2$.

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