An inequality for functions on the plane

Posted on 29/09/2017 by KKK

I accidentally came across a curious inequality for functions of two variables. I would like to know if this inequality is a special case of a more general result but I was unable to find a reference. It would also be interesting if applications can be found for this inequality.

Theorem 1 For any nontrivial {C^2} function {f=f(x,y)} on {\mathbb R^2} such that {[c_1,c_2]\subset f(\mathbb R^2)} and {f^{-1}([c_1,c_2])} is compact, we have

\displaystyle \begin{array}{rl} \displaystyle \int_{\{c_1\le f\le c_2\}} \frac{|f_{xx}{f_y}^2-2f_{xy}f_xf_y+f_{yy}{f_x}^2|}{|\nabla f|^2} dxdy \ge 2 \pi (c_2-c_1). \ \ \ \ \ (1)\end{array}

Here we define the integrand to be zero if {\nabla f=0}.

In particular, if {f} is compactly supported, then

\displaystyle \begin{array}{rl} \displaystyle \int_{\mathbb R^2} \frac{|f_{xx}{f_y}^2-2f_{xy}f_xf_y+f_{yy}{f_x}^2|}{|\nabla f|^2} dxdy \ge 2 \pi \left(\max f-\min f\right). \end{array}

Proof: By Sard’s theorem, for almost all {c}, {f^{-1}(c)} is a smooth level curve which must be closed (but may be disconnected). Fix such {c} for the moment. It is easy to compute that the curvature of the level curve is given by

\displaystyle \begin{array}{rl} \displaystyle k=& \displaystyle -\frac{(\nabla ^2f )(T, T)}{|\nabla f|} \end{array}

where {T=\frac{(f_y, -f_x)}{|\nabla f|}} is the unit tangent and {|\nabla f|=\sqrt{{f_x}^2+{f_y}^2}}.

Expanding, we have

\displaystyle \begin{array}{rl} \displaystyle k=-\frac{f_{xx}{f_y}^2-2f_{xy}f_xf_y+f_{yy}{f_x}^2}{|\nabla f|^3}. \end{array}

On the other hand, by the coarea formula we have

\displaystyle \begin{array}{rl} \displaystyle \int_{\{c_1\le f\le c_2\}} |\nabla ^2f (T,T)| dV =& \displaystyle \int_{\{c_1\le f\le c_2\}} \frac{|f_{xx}f_y^2-2f_{xy}f_xf_y+f_{yy}f_x^2|}{|\nabla f|^2} dV\\ =& \displaystyle \int_{t=c_1}^{c_2}\int_{\{f=t\}} \frac{|f_{xx}f_y^2-2f_{xy}f_xf_y+f_{yy}f_x^2|}{|\nabla f|^3} ds dt\\ =& \displaystyle \int_{t=c_1}^{c_2}\int_{\{f=t\}} |k| ds dt\\ \ge& \displaystyle \int_{t=c_1}^{c_2}2 \pi dt\\ =& \displaystyle 2 \pi (c_2-c_1) \end{array}

where we have used Fenchel’s inequality ([dC] p.199) in line 4. \Box

Remark 1 We record some simple observations here.

By the equality case of the Fenchel’s inequality, it can be seen that the equality holds if and only if for almost every {c\in [c_1, c_2]}, the level curve {\{f=c\}} is a (connected) closed convex curve {k\ge 0} or {k\le 0}. Clearly in this case, by choosing a suitable sign, we can remove the absolute value in the integral.

Also, if {f} is {C^2} and all its critical points are non-degenerate (i.e. Morse function), then by Morse’s theory, if there is a critical point in {f^{-1} (c_1, c_2)} with index {1}, then the inequality (1) is strict.

1. Higher dimension and a more intrinsic version

We now investigate the higher dimensional analogue. By the result of this paper (Theorem 1) of Chen, it can be similarly shown that an inequality for a certain integral involving the Hessian of {f} when restricted to the tangent spaces of the level sets holds. However, the expression is not as explicit. Indeed, it is not hard to see that under the same assumptions,

\displaystyle \begin{array}{rl} \displaystyle \int_{\{c_1\le f\le c_2\}}\frac{\left |{\det}_{n-1}(\nabla ^2f\mid_{\nabla f^\perp})\right|}{|\nabla f|^{m-1}}dV\ge s_{n-1}(c_2-c_1) \ \ \ \ \ (2)\end{array}

where {m=n-1} and {s_{n-1}} is the area of the unit {(n-1)}-sphere. Here {{\det}_{n-1}(\nabla ^2f\mid_{\nabla f^\perp})} is defined as follows: choose an orthonormal basis {\{e_i\}_{i=1}^{n-1}} for {\nabla f^\perp} ({\nabla f\ne 0} on {f^{-1}(c) } for almost all {c\in [c_1,c_2]}), then we define {{\det}_{n-1}(\nabla ^2f\mid_{\nabla f^\perp}):= \det \left(A_{ij}\right)} where {A_{ij}= \nabla ^2f (e_i, e_j)}. This is easily seen to be well-defined.

Question: I wonder if notions like this has been studied by others, and whether {\det_{n-1}\left((\nabla ^2f )|_{(\nabla f)^\perp}\right)} has a more computable form.

Now we prove a more intrinsic version of the inequality which does not involve {\det_{n-1}}. Let {m=n-1}, then by AM-GM inequality,

\displaystyle \begin{array}{rl} \displaystyle \left|{\det} _{n-1} (\nabla ^2f \mid_{\nabla f^\perp})\right| \le& \displaystyle \frac{1}{m^{\frac{m}{2}}}\left| (\nabla ^2f )\mid_{\nabla f^\perp} \right|^m\\ \le& \displaystyle \frac{1}{m^{\frac{m}{2}}}\left[\left| \nabla ^2f \right|^2-\left( \nabla ^2f(\nu, \nu) \right)^2\right]^{\frac{m}{2}}\\ =& \displaystyle \frac{1}{m^{\frac{m}{2}}}\left[\left| \nabla ^2f \right|^2-\left(\frac{\nabla ^2f(\nabla f, \nabla f)}{|\nabla f|^2}\right)^2\right]^{\frac{m}{2}} \end{array}

where {\nu=\frac{\nabla f}{|\nabla f|}} is the unit normal to the level surfaces.

So (2) can be expressed more intrinsically as the following:

Theorem 2 For any nontrivial {C^2} function {f} on {\mathbb R^n} such that {[c_1,c_2]\subset f(\mathbb R^n)} and {f^{-1}([c_1,c_2])} is compact, we have

\displaystyle \begin{array}{rl} \displaystyle \int_{\{c_1\le f\le c_2\}}\frac{1}{|\nabla f|^{m-1}}\left[\left| \nabla ^2f \right|^2-\left(\frac{\nabla ^2f(\nabla f, \nabla f)}{|\nabla f|^2}\right)^2\right]^{\frac{m}{2}}dV \ge m^{\frac{m}{2}} s_{n-1}(c_2-c_1). \end{array}

Here we define the integrand to be zero if {\nabla f=0} and recall that {s_{n-1}} is the area of the unit {(n-1)}-sphere in {\mathbb R^n} and {m=n-1}.

In particular, if {f} is compactly supported,

\displaystyle \begin{array}{rl} \displaystyle \int_{\mathbb R^n}\frac{1}{|\nabla f|^{m-1}}\left[\left| \nabla ^2f \right|^2-\left(\frac{\nabla ^2f(\nabla f, \nabla f)}{|\nabla f|^2}\right)^2\right]^{\frac{m}{2}}dV \ge m^{\frac{m}{2}} s_{n-1}(\max f-\min f). \end{array}

Let us also record a simple observation. The equality holds if and only if each level set is convex and connected, and {|\nabla f|} is constant along the non-degenerate level surfaces and {\nabla ^2f \mid_{\nabla f^\perp}} is umbilical, i.e. {\nabla ^2f(p)(v, w)=h(p)\langle v, w\rangle } for some function {h(p)} and for {v, w} tangential to the level surfaces. But these two conditions imply that the level surfaces {f^{-1}(c)} are all umbilical, and they are either planes or hyperspheres. But since all level surfaces are complete and compact by our assumptions, so they must be hyperspheres.

In particular, if {f=f(r)=f(|x|)} is a monotone radial function, then the equality holds. Indeed, by direct computation, {\nabla f=f'\partial _r} and {\nabla ^2f=f''dr^2+\frac{f'}{r}g|_{S(r)}} where {S(r)} is the sphere of radius {r} in {\mathbb R^n} and {g} is the Euclidean metric. By direct computation, the integrand becomes {\frac{m^{\frac{m}{2}}|f'|}{r^m}}. Since the Euclidean volume form is given by {r^m d\mathrm{vol}_{\mathbb S^{m}}dr} and {f'} has a definite sign, it is easy to see that the equality holds.

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