A proof of the Michael-Simon-Sobolev inequality

The Michael-Simon-Sobolev inequality states that

Theorem 1 Let {M^m} ({m\ge 2}) be a smooth immersed submanifold in {\mathbb R^n} and {f} be a non-negative smooth function with compact support, then

\displaystyle  C_m \|f\|_{\frac{m}{m-1}}\le \|\nabla f\|_{1}+ \|fH\|_{1}. \ \ \ \ \ (1)

Here {\|f\|_p= (\int_M |f|^p)^{\frac 1p}} is the {L^p} norm of {f} on {M}, {H} is the mean curvature vector of {M} and {C_m} depends on {m} only.  

In our convention, the mean curvature vector of {\mathbb S^2\subset \mathbb R^3} is {-2x} where {x} is the position.

Remark 1

  1. Of course, by completion the above inequality also holds on {W^{1,1}_{}(M)} (completion of {C^\infty_0} w.r.t. {W^{1,1}} norm).
  2. When {M} is {\mathbb R^m} (or subset of it), then it becomes the basic ordinary Sobolev inequality

    \displaystyle  C_m \|f\|_{\frac{m}{m-1}}\le \|\nabla f\|_{1}. \ \ \ \ \ (2)

  3. The term {\|fH|_1} can’t be dropped in general when {M} has non-zero curvature. E.g. when {M} is the sphere of radius {r} in {\mathbb R^{m+1}}, choose {f=1}, then LHS is of order {r^{m-1}}, whereas {\|\nabla f\|_1=0}. As the mean curvature of the sphere is of order {1/r}, the {\|fH\|_1} term is also of order {r^{m-1}}, so this inequality is quite “reasonable”.

 

The interesting thing about the inequality is that {C} depends on {m} only (not even depends on {M})! (Of course, the constant in the ordinary Sobolev inequality (2) also doesn’t depend on {\Omega}, even if we restrict our domain to {\Omega \subset \mathbb R^m}, the reason is that such a function {f} can always be extended (trivially) as a function compactly supported in {\mathbb R^m}. )

I haven’t really read the proof of the most general form (1) of this inequality, but I come across a cute proof of it when {m=2} (i.e. {M} is a surface), which I am going to give here. As a bonus, {C_m } can be explicitly given in this case. (Actually Michael-Simon’s paper also gives an explicit constant. )

Proof: (Proof of Theorem 1 for {m=2})

For an {\mathbb R^n}-valued vector field {\phi} which is defined on {M}, we define

\displaystyle \mathrm{div}_M \phi= \sum_{i=1}^m \langle \overline \nabla _{i} \phi, e_i\rangle

where \{e_i\}_{i=1}^m is a local orthonormal frame on M and \overline \nabla is the connection on \mathbb R^n.
Suppose {N^m} is a smooth submanifold in {\mathbb R^n} with boundary {\Sigma} with \nu being the unit outward normal of \Sigma w.r.t. N, then as {\Delta x=H} (here for the proof),

\displaystyle  \begin{array}{rcl} \int_N \phi\cdot H = \int_N \phi\cdot \Delta x &= &\int_\Sigma \phi \cdot \langle \nabla x, \nu\rangle- \int_N \langle \nabla \phi, \nabla x\rangle\\ &= &\int_\Sigma \phi \cdot \langle \nabla x, \nu\rangle- \int_N \mathrm{div}_N \phi\\ &= &\int_\Sigma \phi \cdot \nu- \int_N \mathrm{div}_N \phi. \end{array} \ \ \ \ \ (3)

Remark 2 In particular, if {N} has no boundary, this becomes

\displaystyle  \int_N \phi\cdot H = - \int_N\mathrm{div}_N \phi.

This becomes zero if {\phi} is a tangent vector field (as {\phi\perp H}), thus this formula generalizes the ordinary divergence theorem.  

Now, assume {M} contains {0} for the time being and let {\phi= \frac {f(x)x}{|x|^m}}.

Easy computations give {\overline \nabla _{e_i} (|x|^{-k})= -k |x|^{-k-2} \langle x, e_i\rangle} and

\displaystyle  \mathrm{div}_M (\frac x{|x|^k})= \frac m {|x|^k} - \frac {k |x^T|^2} {|x|^{k+2} } = \frac{m-k}{|x|^k }+ \frac{k |x^\perp|^2}{|x|^{k+2}}. \ \ \ \ \ (4)

In particular,

\displaystyle  \mathrm{div}_M( \frac x { |x|^m})= \frac m {|x|^{m+2}} |x^\perp|^2\ge 0 .

Let {B_r} to be the geodesic ball around {0} of radius {r}, {\Sigma_r=\partial B_r} and {M_r= M\setminus B_r}. Then applying (3) on {M\setminus B_r},

\displaystyle  \int_{M_r} \phi\cdot H = \int_{\Sigma _r} \phi\cdot \nu - \int_{M_r} \mathrm{div}_M \phi.

As {r\rightarrow 0}, we have {\nu(x)= - \frac x{|x|}+o(1)} and {\mathrm{Area}(\Sigma_r)= \omega r^{m-1} + O(r^m)}, where {\omega= \omega_{m-1}} is the area of the standard {(m-1)}-dimensional sphere, so taking {r\rightarrow 0} in the above, we have

\displaystyle  \begin{array}{rcl}  - \omega f(0)-\int_{M } f \frac x{|x|^m}\cdot H= \int_{M } \mathrm{div}_M \phi &= &\int_M (\nabla f \cdot \frac x{|x|^m} + f \mathrm{div}_M (\frac {x}{|x|^m}))\\ &\ge& \int_M \nabla f \cdot \frac x{|x|^m} . \end{array}

So

\displaystyle  \omega f(0) \le \int_M (\frac{|\nabla f(x)|}{|x|^{m-1}} + \frac{f(x)|H(x)|}{|x|^{m-1}})dx. \ \ \ \ \ (5)

Translating the origin to {y} on {M}, we have

\displaystyle  \omega f(y) \le \int_M (\frac{|\nabla f(x)|}{|x-y|^{m-1}} + \frac{f(x)|H(x)|}{|x-y|^{m-1}})dx.

Multiply {f(y)^{\alpha-1}} to both sides and integrating, we have

\displaystyle  \begin{array}{rcl} \omega \int_M f(y)^\alpha dy &\le &\int_M \left(\int_M (\frac{|\nabla f(x)|}{|x-y|^{m-1}} + \frac{f(x)|H(x)|}{|x-y|^{m-1}})dx \right)f(y)^{\alpha-1}dy\nonumber\\ &= &\int_M (|\nabla f(x)|+ f(x)|H(x)|)(\int_M \frac{f(y)^{\alpha-1}}{|x-y|^{m-1}}dy )dx. \end{array} \ \ \ \ \ (6)

Now, we take {\alpha = m=2}. In this case {\omega=2\pi}. We estimate {\int_M \frac{f(y)^{}}{|x-y|^{}}dy} as follows. We apply (3) to {\phi= \frac{f(y)y}{|y|}}, then as { \mathrm{div}_M (\frac y{|y|})\ge \frac 1{|y|}} by (4), we have

\displaystyle  \begin{array}{rcl}  - \int_M f \frac y{|y|}\cdot H = \int_M \mathrm{div}_M (\frac{fy}{|y|}) &= &\int_M (\frac {y\cdot \nabla f }{|y|} + f \mathrm{div}_M (\frac y{|y|}))\\ &\ge &\int_M (\frac {y\cdot \nabla f }{|y|} + \frac f{|y|}). \end{array}

So

\displaystyle \int_M \frac{f}{|y|}\le \int_M (f(y)|H(y)|+ |\nabla f(y)|)dy.

By translating the origin from {0} to {x}, we have

\displaystyle \int_M \frac{f}{|x-y|}\le \int_M (f(y)|H(y)|+ |\nabla f(y)|)dy.

So (6) becomes

\displaystyle  2\pi \int_M f^2 \le (\int_M |\nabla f|+f|H|)^2.

In particular the constant {C_2} can be taken to be {\sqrt{2\pi}}. \Box

Remark 3 The constant {\sqrt{2\pi}} above is not optimal. Suppose not, by translating {M}, we can assume {M} contains {0}, but then the proof shows that {x^\perp =0} for all {x\in M}. So {x=x^T} for all {x}. However, the choice of the origin is arbitrary in the proof, i.e. for any points {x, y\in M}, the vector {x-y} is tangential to both {T_xM} and {T_yM}. We then conclude that {M} is a domain in a {2}-plane {\mathbb R^2}. Also, for fixed {y\in M}, by tracing the proof, we see that {\nabla f(x)} is a multiple of {x-y} for all {x\in M}. But this is impossible unless {f} is zero, for if {\nabla f(x)\ne 0}, for any {y\in M} which does not lie on {\{x+t\nabla f(x)\}_{t\in \mathbb R}}, clearly {x-y} and {\nabla f(x)} are not parallel.  

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One Response to A proof of the Michael-Simon-Sobolev inequality

  1. Anonymous says:

    I have used this inequality in my mphil thesis

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