This is the sequel of the post about Bean’s theorem on convex curve.

I will state the smooth version of Bean’s theorem in the general dimension.

Theorem 1 (Bean’s theorem, 2011, Smooth version)Suppose is a ((n-1)-dimensional) smooth convex hypersurface contained in the interior of another smooth compact hypersurface in , then the area of is not greater than the area of .

First of all let us make some remarks. By a smooth convex hypersurface we mean is compact, connected and such that for all , we can find a (outward) normal vector such that all , . i.e. lies on the non-positive side of .

Since (and also ) is a compact hypersurface in , it is orientable (see also here). Thus we can find a global outward normal field of , and thus we can globally define its second fundamental form without any ambiguity, once is fixed (here is only the ordinary directional derivative of the vector field in in direction ). In our convention, the second fundamental form of the unit sphere with respect to its outward normal is .

We claim that

Lemma 2If is convex, then its second fundamental form with respect to the outward normal is always non-negative definite.

*Proof:* Since is convex, it is locally convex in the sense that at any point , can be locally represented by a graph of function with and . (What does that mean precisely, and how is related to ? )

Since , by Taylor’s theorem, at (corresponding to the point ), for any tangent vector

But it is easy to see that the hessian of is equal to the second fundamental form:

Remark: Under some additional assumptions, the converse is also true (but much harder) by a result of Sacksteder.

Lemma 3For any hypersurface in , suppose is a unit vector field defined on a neighborhood of such that is a normal vector field to . Then at each point of , we have

where is the mean curvature of w.r.t. (i.e. sum of eigenvalues of the second fundamental form w.r.t. , which is symmetric). In our notation, the mean curvature of the unit sphere in is 2.

*Proof:* Since , differentiating it gives and thus at

where forms an orthonormal basis of .

Now for convex and , we define the surface by

By our previously result (Lemma 2 in here), is a regular surface (i.e. without any self-intersection). We claim

*Proof:* Let be a local parametrization of around , let us for simplicity denote simply by . Then for fixed

gives a natural parametrization for . It is easy to see that is also the normal vector of at and that . Let us compute the second fundamental form of (denote by etc) at :

Thus the second fundamental form of is also non-negative and so its mean curvature is non-negative.

*Proof:*(Proof of Bean’s theorem)

Extend the outward normal of to the region bounded between and by parallel translating along its outward direction. Then by our previous result (Lemma 2 in here), this defines a vector field on , extending .

It is easy to see that at each point , is the normal to the surface containing where . Thus by Lemma 4 and Lemma 3,

Now, we apply divergence theorem to in ,

where is the unit outward normal of . But clearly , and , so

Is it related to Bishop volume comparison theorem?

(I didn’t even read your article nor remember what’s Bishop volume comparison theorem)

No. At least I don’t see the relation. The Bishop comparison theorem compares the volume of geodesic balls when we have a comparison of the Ricci curvatures between a certain Riemannian manifold and a space form.