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So if we have a compact surface $S \subset \mathbb{R}^3 $ then the function $f:S\rightarrow \mathbb{R}_{\geq 0 },\ x \mapsto ||x|| $ has a maximum value say $R >0$ at $p \in S$. Then we must have the surface $S$ fully enclosed by the sphere of radius $R$ as otherwise $f(p)$ is not the maximum value. This means that the curvature $\kappa $ at $p$ must be at least $1/R \ $ for any curve on $S$ passing through $p.$

This next part is what I don't understand

So every normal curvature is at least $1/R$ so $K(p)\geq 1/R^2 >0 $.

Why must every normal curvature be at least $1/R$ and why should that then mean $K(p)\geq 1/R^2 $ ? For the final part I suspect it is because the Gaussian curvature is the product of the principal curvatures, but why should they be at least $1/R$ ?

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  • $\begingroup$ For each tangent vector $v$ at $p$ there is a curve on $S$ in that direction whose curvature is all in the normal direction, namely the intersection of $S$ with the plane spanned by $v$ and the normal to $S$ at $p$. $\endgroup$ Apr 20 at 18:26
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Because if there is a curve in $S$ through $p$ with less curvature, it wouldn't be able to stay inside the sphere centered at the origin with radius $R$. And it has to stay inside that sphere, as that sphere contains all of $S$.

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  • $\begingroup$ I understand that the curvature of any curve should be no less than $1/R$ but why should the this be true for the normal curvature, where $\kappa ^2 = \kappa _n^2 +\kappa _g^2 $? $\endgroup$
    – Ben
    Apr 20 at 16:53
  • $\begingroup$ @Anon It's true for any curve, including geodesics ($\kappa_G=0$). $\endgroup$
    – Arthur
    Apr 20 at 17:03
  • $\begingroup$ But if $1/R^2 \leq \kappa ^2 = \kappa _n^2 +\kappa _g^2 $ why does this necessarily mean $\kappa _n \geq 1/R $ ? $\endgroup$
    – Ben
    Apr 20 at 17:33
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    $\begingroup$ @Anon Because for any curve in $S$ through $p$ there is a geodesic that's tangent to it and has the same $\kappa_n$. That's what $\kappa_g$ is for: measure how far the curve is from being a geodesic (that's what the $g$ stands for). $\endgroup$
    – Arthur
    Apr 20 at 20:45

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