The Largest Eigenvalue of a Convex Function, Duality, and a Theorem of Slodkowski

Abstract

First, we provide an exposition of a theorem due to Slodkowski regarding the largest “eigenvalue” of a convex function. In his work on the Dirichlet problem, Slodkowski introduces a generalized second-order derivative which for $C^2$ functions corresponds to the largest eigenvalue of the Hessian. The theorem allows one to extend an a.e. lower bound on this largest “eigenvalue” to a bound holding everywhere. Via the Dirichlet duality theory of Harvey and Lawson, this result has been key to recent progress on the fully non-linear, elliptic Dirichlet problem. Second, using the Legendre–Fenchel transform we derive a dual characterization of this largest eigenvalue in terms of convexity of the conjugate function. This dual characterization offers further insight into the nature of this largest eigenvalue and allows for an alternative proof of a necessary bound for the theorem.

Appendix

1.1 Lipschitz Gradient

Here we show that the generalized derivative K(f, x) retains the following standard property regarding the derivative of a Lipschitz continuous function.

Proposition 4.1

Suppose $f:{\mathbb {R}}^n\rightarrow {\mathbb {R}}$ is convex and $C^{1,1}$ (i.e., f is differentiable and has Lipschitz gradient), with Lipschitz constant L. Then $K(f,x)\le L$ for all x.

Proof

Let $x_0\in {\mathbb {R}}^n$.

$$\begin{aligned} K(f,x_0) := \limsup _{\epsilon \rightarrow 0} 2\epsilon ^{-2} \text { max } \big \{ f(x_0 + \epsilon h) - f(x_0) - \epsilon \langle \nabla f(x_0), h \rangle : |h| =1\big \}, \end{aligned}$$

which can be can written as

$$\begin{aligned} K(f,x_0)= \limsup _{\epsilon \rightarrow 0} \text { max } \left\{ 2 \dfrac{f(x_0 + \epsilon h) - f(x_0) - \epsilon \langle \nabla f(x_0), h \rangle }{\epsilon ^2} : |h|= 1 \right\} . \end{aligned}$$

Differentiability lets us use the Cauchy mean value theorem. Let $\phi _1(\epsilon )= f(x_0+\epsilon h) -\epsilon \langle \nabla f(x_0),h\rangle$, and $\phi _2(\epsilon )=\epsilon ^2$. Note that

$$\begin{aligned} 2 \dfrac{f(x_0 + \epsilon h) - f(x_0) - \epsilon \langle \nabla f(x_0), h \rangle }{\epsilon ^2}= 2 \dfrac{\phi _1(\epsilon )- \phi _1(0)}{\phi _2(\epsilon )- \phi _2(0)}. \end{aligned}$$

Thus, there exists $\gamma \in (0,\epsilon )$ such that

$$\begin{aligned} 2 \dfrac{\phi _1(\epsilon )- \phi _1(0)}{\phi _2(\epsilon )- \phi _2(0)} =2 \dfrac{\phi _1'(\gamma )}{\phi _2'(\gamma )} =&\dfrac{\big \langle \nabla f(x_0+\gamma h),h\big \rangle - \big \langle \nabla f(x_0),h\big \rangle }{\gamma }\\ =&\dfrac{\big \langle \nabla f(x_0+\gamma h)- \nabla f(x_0) ,h\big \rangle }{\gamma }\\ \le&\dfrac{\big |\nabla f(x_0+\gamma h)- \nabla f(x_0)\big | }{\gamma } \le L. \end{aligned}$$

Therefore, $K(f,x_0)\le L$, and thus $\frac{1}{K(f,x_0)}$ bounds the modulus of convexity of $f^*$, for any $x_0$. $\square$

1.2 Example of a Non-$C^{1,1}$ Function with a Sphere of Support

Example 4.2

It may seem that since a bound on K(u, x) implies a sphere of support to the graph of u at (x, u(x)), that this in turn implies some kind of Lipschitz continuity of the gradient in a small neighborhood of x. Here we construct an example of a strictly convex function f that is $C^1$ and twice differentiable with $K(f,0)<\infty$, but with gradient not Lipschitz in any neighborhood of 0, to show this is not the case. Let $f:[-1,1]\rightarrow {\mathbb {R}}$ be given by $f(0)=0$, and for $x\ge 0$

$$\begin{aligned} f'(x)=\int _0^x \gamma (t)dt, \quad \text {where } \gamma (t):=n+4 \text { on } I_n \text { and } 0 \text { otherwise,} \end{aligned}$$

with $I_n= \dfrac{1}{(n+4)^2}\left[ 1-\dfrac{1}{(n+4)^2},\;1\right]$. Define $f'(-x):=-f'(x)$.

Then $f'$ is clearly increasing and so f is convex. And for $x_n =\dfrac{1}{(n+4)^2}$,

$$\begin{aligned} f'(x_n)=\int _{0}^{x_1}\gamma (t)\,dt=\sum _{k\ge n}\frac{1}{(k+4)^3}\le \int _{n+3}^{\infty }\frac{dt}{t^3}= \frac{1}{2(n+3)^2}<\frac{1}{(n+4)^2}=x_n. \end{aligned}$$

So we have $f'(x)\le x$ for all $x \in [0,1]$ and $f'(x)\ge x$ for all $x \in [-1,0]$. Since $d'(x)\ge x$ for all $x \in [0,1]$ and $d'(x)\le x$ for all $x \in [-1,0]$, it follows that the graph of d, and thus the unit circle centered at (0, 1), is always at or above the graph of f, with $f(0)=d(0)$. Therefore, f has a sphere of support at $x_0=0$.

However, there exist sequences $\{x_i\}, \{x_j\}$ such that

$$\begin{aligned} \dfrac{f'(x_i)-f'(x_j)}{x_i-x_j} \end{aligned}$$

blows up: Taking $x_i$ and $x_j$ as the endpoints of $I_n$,

$$\begin{aligned} \dfrac{f'(x_i)-f'(x_j)}{x_i-x_j}= & {} \frac{1}{x_i-x_j}\left( \int _0^{x_i} \gamma (t)dt - \int _0^{x_j} \gamma (t)dt\right) \\= & {} (n+4)^4 \int _{x_j}^{x_i} n+4 dt=n+4. \end{aligned}$$

We can make f strictly convex by adding an $x^m$ term, which does not affect any of the above analysis. The above example can be adjusted to show that $f'$ is not $\alpha$-Hölder continuous for any $\alpha$.

1.3 Osculating and Locally Supporting Spheres

Here we extend the concept of an osculating circle to a plane curve to that of an “osculating sphere” to the graph of a function in higher dimensions. The bound on the “largest eigenvalue” K(u, x) can be seen as a generalization of the relationship between the second derivative of a $C^2$ plane curve u and the radius of its osculating circle:

Let $u:{\mathbb {R}}\rightarrow {\mathbb {R}}$ be $C^{2}$. Provided $u''\ne 0$, the radius of curvature at x is defined as

$$\begin{aligned} r_{u,x}:= \dfrac{1}{\kappa }= \dfrac{\big (1 + u'^{2}\big )^{\frac{3}{2}}}{u''}, \end{aligned}$$

where $\kappa$ is the curvature of u at x, and the right-hand side is the standard formula for computing the curvature of a planar curve [2, Sect. 8]. Thus,

$$\begin{aligned} u''=\frac{\big (1+ u'^2\big )^{3/2}}{r}. \end{aligned}$$

Definition 4.3

The osculating circle, or circle of curvature, to a planar curve C at p is the circle that touches C (on the concave side) at p and whose radius is the radius of curvature of C at p.

We extend this to the graphs of $C^2$ convex functions in higher dimensions by

Definition 4.4

For a convex function $u:{\mathbb {R}}^{n}\rightarrow {\mathbb {R}}$ let the osculating sphere to the graph of u at x be the n-sphere tangent to the graph of u at x the with radius equal to that of $\frac{1}{\lambda _{max}}$.

It is easy to show that any tangent sphere at (x, u(x)) with radius less than the osculating sphere at that point is a (local) sphere of support. And any tangent sphere at (x, u(x)) with radius greater than the osculating sphere cannot be a (local) sphere of support.

1.4 Spheres of Support to a Function and Its Dual

Given a convex function u with a sphere of support at $(x_0, u(x_0))$, the conjugate function $u^*$ will not necessarily have a sphere of support at the corresponding point $(\nabla u(x_0), u^*(\nabla u(x_0))$. For example, take $u=\frac{1}{4}|x|^4$ and $u^*=\frac{3}{4}|x|^{\frac{4}{3}}$. However, for more regular and sufficiently convex functions (e.g., $C^2$ and locally strongly convex), we will have a sphere of support (locally) to both graphs at corresponding points, and the order-reversing property of ${\mathscr {L}}$ provides a simple inequality relating the radii of these spheres. We state this without proof.

Proposition 4.5

Let $u:{\mathbb {R}}^n \rightarrow {\mathbb {R}}$ be strongly convex and $C^2$ near $x_0$, and suppose u has a sphere of support of radius $r_{x_0}$. If $r_{y_0}$ is the radius of a sphere of support to $u^*$ at $y_0=\nabla u(x_0)$, then

$$\begin{aligned} r_{y_0}\le \dfrac{\left( 1+|x|^{2}\right) ^{\frac{3}{2}} \left( 1+|\nabla u(x_0)|^{2} \right) ^{\frac{3}{2}}}{r_{x_0}}. \end{aligned}$$