## An inequality for functions on the plane

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.