Level sets of convex functions

Question

Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ be a convex function. For $t\in\mathbb{R}$, consider the corresponding level set $$f^{-1}\{t\}=\{(x,y)\in\mathbb{R}^2: f(x,y)=t\}.$$

For the application I have in mind, it will suffice to assume that all the level sets of $f$ are compact curves. In that case, denoting arc length measure on the plane by $\ell(\cdot)$, when does an estimate of the form $$\Big|\ell(f^{-1}\{t\})-\ell(f^{-1}\{t'\})\Big|\leq C |t-t'|^{\delta}$$ hold for some $C<\infty$ and $\delta>0$? And how does the optimal exponent $\delta$ depend on $f$?

The intuition is that the curves $f^{-1}\{t\}$ and $f^{-1}\{t'\}$ should be "close to each other" whenever $t$ is close to $t'$. In which other ways can this be made rigorous?

Thank you.

Answer 1

Here is an approach in terms of "intuitive calculus":

Let $$\gamma_v:\quad s\mapsto z(s)=\bigl(x(s),y(s)\bigr)\qquad(0\leq s\leq L_v)$$ be the arc length parametrization of the level set $f^{-1}(v)$. For each $s$ the tangent vector $\dot\gamma_v(s)$ is orthogonal to the gradient $\nabla f$ at $z(s)$. If we orient $\gamma_v$ such that $\nabla f$ points to the outside we can therefore write $\nabla f\bigl(z(s)\bigr)$ in the form $$\nabla f\bigl(z(s)\bigr)=\rho\bigl(z(s)\bigr)\bigl(\dot y(s),-\dot x(s)\bigr)\ ,\qquad\rho(z):=\bigl|\nabla f(z)\bigr|\ .$$ For an infinitesimal $\epsilon>0$ the level set $\gamma_{v+\epsilon}:=f^{-1}(v+\epsilon)$ has the parametrization $$\gamma_{v+\epsilon}:\quad s\mapsto\Bigl(x(s)+\epsilon{ \dot y(s)\over\rho(z(s))},\ y(s)-\epsilon{ \dot x(s)\over\rho(z(s))}\Bigr)\qquad(0\leq s\leq L_v)\ ,$$ where we have neglected higher order terms in $\epsilon$. Thus we get $$\dot\gamma_{v+\epsilon}(s)\doteq \Bigl(\dot x(s)+\epsilon{ \ddot y\rho-\dot y\dot\rho\over\rho^2},\ \dot y(s)-\epsilon{ \ddot x\rho-\dot x\dot\rho\over\rho^2}\Bigr)$$ and therefore $$\bigl|\dot\gamma_{v+\epsilon}\bigr|\doteq1+2\epsilon{\dot x\ddot y-\ddot x\dot y\over\rho}$$ or $$\bigl|\dot\gamma_{v+\epsilon}(s)\bigr|\doteq 1+\epsilon{\kappa(s)\over\rho(z(s))}\ ,$$ where we have denoted the curvature of $\gamma_v$ by $\kappa$. From this we deduce the following formula for the derivative of $L_v$ with respect to $v$: $${dL_v\over dv}=\int_0^{L_v}{\kappa(s)\over\bigl|\nabla f(z(s))\bigr|}\ ds\ .\qquad(*)$$ Now that we have this formula it seems intuitively pretty obvious: When we let $v$ increase then the length $L_v$ increases most in parts of $\gamma_v$ where the curvature is large, and does not change along straight parts of $\gamma_v$.

The integrand in formula $(*)$ can be expressed in terms of the partial derivatives of $f$. One obtains (see, e.g., Bronstein-Semendjajew) $${dL_v\over dv}=\int_0^{L_v}{f_{xx}f_y^2-2f_{xy}f_xf_y+f_{yy}f_x^2\over (f_x^2+f_y^2)^4}\Biggr|_{z(s)}\ ds\ .$$

Answer 2

What follows is the work I've done towards a solution, which is unfortunately far too large for a comment. I can make no guarantee that it goes anywhere.

I'll assume $f$ is $C^1$ and strictly convex. Assuming the level sets are compact curves, they must also be simple since a convex function has at most one local extreme. Consider values $t$ in some neighborhood $U$ of $t_0$ (which we will make as small as needed). Let $\gamma_t:[0,1]\to \mathbb R^2$ be a parametrization of $f^{-1}\{t\}$ and note that $\gamma_t$ is locally the graph of a $C^1$ function of $(t,x)$ or $(t,y)$ for fixed $t$. Thanks to compactness and various theorems in calculus, we have a finite set of $a_i,b_i\in [0,1]$ and $C^1$ functions $$\begin{align} f_{1}&:U\times (a_1,b_1)\to \mathbb R,\\ &\vdots\\ f_{n}&:U\times (a_n,b_n)\to \mathbb R\\ f_{n+1}&:U\times (a_{n+1},b_{n+1})\to \mathbb R\\ &\vdots\\ f_{n+m}&:U\times (a_{n+m},b_{n+m})\to \mathbb R \end{align}$$ such that each $a_i<b_i$, the collection $(a_i,b_i)$ covers $(0,1)$, and there exist $C^1$ functions $x_i,y_i$ which are strictly increasing in $s$ such that $$t\in U,s\in (a_i,b_i)\implies \gamma_t(s)=\begin{cases} (x_i(t,s),f_i(t,x_i(t,s))) &\text{if } 1\leq i\leq n\\ (f_i(t,y_i(t,s)),y_i(t,s)) &\text{if } n<i\leq n+m \end{cases}$$ For notational convenience, assume $$0=a_1< a_2< b_1<a_3<b_2<a_4<b_3<\cdots < b_{n+m}=1$$ i.e. that the intervals are in order, nicely trimmed and non-redundant, which saves us from having to through out/trim intervals and rearrange junk. Furthermore, we can assume that $x_i,y_i$ are affine functions in $s$, at least where their domain does not overlap with that of $x_j,y_j$ for $j>i$, by choosing an appropriate parametrization for $\gamma_t$ (arc length is parametrization-independent). Let $L(t)=\ell(f^{-1}\{t\})$ and $a_{n+m+1}=b_{n+m}=1$. Note that $$\begin{align} L(t)=\ell(\mathrm{im}(\gamma_t)) &= \int_0^1\left\|\frac{d\gamma_t}{ds}\right\|ds\\ &=\sum_{i=1}^{n}\int_{a_i}^{a_{i+1}} \frac{dx_i}{ds}\sqrt{1+\left(\frac{df_i}{dx_i}\right)^2}ds+\sum_{i=n+1}^{n+m}\int_{a_i}^{a_{i+1}} \frac{dy_i}{ds}\sqrt{\left(\frac{df_i}{dy_i}\right)^2+1}ds \end{align}$$ thus $$\begin{align} \frac{dL}{dt} &= \frac{d}{dt}\left(\sum_{i=1}^{n}\int_{a_i}^{a_{i+1}} \frac{dx_i}{ds}\sqrt{1+\left(\frac{df_i}{dx_i}\right)^2}ds+\sum_{i=n+1}^{n+m}\int_{a_i}^{a_{i+1}} \frac{dy_i}{ds}\sqrt{\left(\frac{df_i}{dy_i}\right)^2+1}ds\right)\\ &= \sum_{i=1}^{n}\int_{a_i}^{a_{i+1}} \frac{d}{dt}\left(\frac{dx_i}{ds}\sqrt{1+\left(\frac{df_i}{dx_i}\right)^2}\right)ds+\sum_{i=n+1}^{n+m}\int_{a_i}^{a_{i+1}} \frac{d}{dt}\left(\frac{dy_i}{ds}\sqrt{\left(\frac{df_i}{dy_i}\right)^2+1}\right)ds \end{align}$$ which we can analyze piece-by-piece. Looking at the first integrand, note that $x_i$ is affine in $s$ in the region we care about so $\frac{dx_i}{ds}$ is a positive $C^1$ function $c(t)$. Thus we have $$\begin{align} \frac{d}{dt}\left(\frac{dx_i}{ds}\sqrt{1+\left(\frac{df_i}{dx_i}\right)^2}\right) &= \frac{d^2x_i}{ds dt}\sqrt{1+\left(\frac{df_i}{dx_i}\right)^2}+\frac{dx_i}{ds}\frac{df_i}{dx_i}\frac{d^2f_i}{dx_i dt}\left(1+\left(\frac{df_i}{dx_i}\right)^2\right)^{-1/2}\\ &= \frac{dc}{dt}\sqrt{1+\left(\frac{df_i}{dx_i}\right)^2}+c(t)\frac{df_i}{dx_i}\frac{d^2f_i}{dx_i dt}\left(1+\left(\frac{df_i}{dx_i}\right)^2\right)^{-1/2}\\ \end{align}$$ which maybe we can sort of understand. Additionally, since the region contained in $f^{-1}\{t\}$ is the intersection of a convex set with a plane, it is convex as well thus $f_i$ (which defines part of its boundary) is concave as a function of $x_i$ so $\frac{df_i}{dx_i}$ is decreasing. We can go back and change some of what we did earlier, moving around our choices of $f_i$ to avoid the points on the curve tangent to the axes, which means we miss a set of points of measure $0$ (which doesn't affect the integral) so $\frac{df_i}{dx_i}$ will have constant sign $\sigma_i$, but this could cause some problems. From here, the best approach is probably to try to understand each of the derivatives. The hardest are those with respect to $t$.

Answer 3

Suppose $f$ is strictly convex and has its minimum at the origin. We can use polar coordinates $(r,\theta)$, in which the element of length is $ds = \sqrt{dr^2 + r^2 d\theta^2}$. Now along the level curve, $\dfrac{dr}{d\theta} = - \dfrac{f_\theta}{f_r}$ (where subscripts refer to partial derivatives wrt $r$ and $\theta$). The level curve $f(r,\theta) = t$ can be written as $r = R(t,\theta)$, and its length is $$ L = \int_0^{2\pi} \sqrt{\dfrac{f_\theta^2}{f_r^2} + r^2} \ d\theta = \int_0^{2\pi} \sqrt{\dfrac{f_\theta^2(R(t,\theta),\theta)}{f_r^2(R(t,\theta),\theta)} + R(t,\theta)^2}\ d\theta$$ We'd like to differentiate the integrand with respect to $t$ at fixed $\theta$. Note that $$\dfrac{\partial r}{\partial t} = \dfrac{1}{f_r}$$ I get $$ \dfrac{\partial}{\partial t} \sqrt{\dfrac{f_\theta^2}{f_r^2} + r^2} = \frac{1}{f_r} \dfrac{\partial }{\partial r} \sqrt{\dfrac{f_\theta^2}{f_r^2} + r^2} = \left({f_\theta^2}+ r^2 f_r^2\right)^{-1/2} \left( \dfrac{f_\theta}{f_r^2} f_{r\theta} - \dfrac{f_\theta^2}{f_r^3} f_{rr} + r \right) $$ and thus we should have $$ \dfrac{dL}{dt} = \int_0^{2\pi} \left({f_\theta^2}+ r^2 f_r^2\right)^{-1/2} \left( \dfrac{f_\theta}{f_r^2} f_{r\theta} - \dfrac{f_\theta^2}{f_r^3} f_{rr} + r \right)\ d\theta$$ I don't know if it's possible to get useful bounds out of this, but perhaps it's a start.