Lecture 6 ing

lecture 6

Lemma 6.1 Suppose \(\varphi\) is lower-semi continuous and is a proper convex function.

(1). if \(\varphi\) is strictly convex in a neighborhood of some \(x \in \rm{Dom}(\varphi)\), then \(\varphi^\ast\) is differentiable on \(\partial \varphi (x)\), and \(\nabla \varphi^\ast(y) = x, \forall y \in \partial \varphi(x).\)

(2). if \(\varphi\) is strictly convex and differentiable, then \(\varphi^\ast\) is also strictly convex and differentiable. And \(\nabla \varphi\) is an one-to-one mapping, satisfying \((\nabla \varphi )^{-1} = \nabla \varphi^\ast.\)

(3). under the condition of (2), if moreover \(\varphi\) is superlinear, then \(\nabla \varphi ( \boldsymbol{R}^n) = \boldsymbol{R}^n\), in which case \(\nabla \varphi\) is a bijection \(\boldsymbol{R}^n \to \boldsymbol{R}^n\), and \(\nabla \varphi^\ast\) is its inverse.

The general idea about this problem is that:

if \(\varphi\) is not strictly convex in a neighborhood of \(x\), then there exist a neighborhood \(U(x, \delta)\) in which \(\varphi\) is linear, thus \(U(x, \delta) \subset \partial \varphi^\ast ( \partial \varphi(x) )\)

Proof. (1). Suppose \(\varphi\) is strictly convex in a neighborhood of some \(x \in \rm{Dom}(\varphi).\) Then forall \(y \in \partial \varphi(x)\), we want to prove that \(\varphi^\ast\) is differentiable at \(y,\) i.e. \(\partial \varphi^\ast(y)\) is a singleton set.

Firstly, \(y \in \partial \varphi(x) \implies x \in \partial \varphi^\ast(y).\)

If \(\partial \varphi^\ast (y)\) is not a singleton set, then there exist \(x' \in \partial \varphi^\ast (y)\) such that \(x' \neq x.\)

By the theorem—, we have

\[ \langle x',y \rangle =\varphi(x') + \varphi^\ast (y) ; \quad \langle x,y \rangle = \varphi(x) + \varphi^\ast (y). \\ \implies \varphi(x') - \varphi(x) = \langle x'-x, y \rangle \]

Since \(y \in \partial \varphi(x) \implies \varphi(x') \geq \varphi(x) + \langle x'-x, y \rangle,\) from the convex property of \(\varphi\), we have

\[ \forall z \in [x,x'], \quad \varphi(z) = \varphi(x) + \langle z-x, y \rangle. \\ i.e. \quad \varphi(\alpha x + (1-\alpha)x') =\alpha \varphi(x) + (1-\alpha) \varphi(x'). \]

which is a contradiction to the assumption that \(\varphi\) is strictly convex in a neighborhood of \(x.\)

(2). if \(\varphi\) is strictly convex and differentiable, from (1) we have \(\varphi^\ast\) is differentiable on \(\nabla \varphi (\rm{Dom}(\varphi) ),\) and \(\nabla \varphi^\ast(\nabla \varphi(x)) = x.\)

Next, we want to prove that \(\nabla \varphi (\rm{Dom}(\varphi)) = \rm{Dom} (\varphi^\ast).\)

Since \(\nabla \varphi (\rm{Dom}(\varphi)) \subset \rm{Dom} (\varphi^\ast),\) we only need to prove the inverse direction.

Suppose \(y \in \rm{Dom}(\varphi^\ast),\) then there exists \(x \in \partial \varphi^\ast(y),\) which implies \(x \in \rm{Dom}(\varphi)\) i.e. \(y \in \nabla \varphi(\rm{Dom}(\varphi)).\)

\(int(\rm(Dom))\) (existence of x) ?

\(\varphi^\ast\) strict convexity

if \(\varphi^\ast\) is not strictly convex, then there exist \(y, y' \in \rm{Dom}(\varphi^\ast)\) s.t. the restriction of \(\varphi^\ast\) on \([y,y']\) is an affine transformation \(f\).

Randomly choose \(z_0 \in (y, y')\), and let \(\xi = \nabla \varphi^\ast(z_0) \in \rm{Dom}(\varphi).\) Then we have:

\[ \varphi^\ast(z) = \varphi^\ast(z_0) + \langle z-z_0, \xi \rangle, \quad \forall z \in [y,y']. \]

Since \(\xi \in \partial \varphi^\ast (z_0)\), we obtain

\[ \begin{align} \forall \omega \in \boldsymbol{R}^n, z \in [y,y'] \quad \varphi^\ast(\omega) &\geq \varphi^\ast(z_0) + \langle \omega-z_0, \xi \rangle \\ &= \varphi^\ast(z) + \langle \omega-z, \xi \rangle. \end{align} \]

So \(\xi \in \partial \varphi^\ast (z) , \forall z \in [y,y'].\)

unfinished

6.1 Back to the study of the dual Monge-Kantorovich problem

Definition 6.1 (Existence of an optimal pair of convex conjugate functions) Let \(\mu, \nu\) be to probability measures on \(\boldsymbol{R}^n\) with finite second order moments. Let \(\tilde{\Phi}\) defined in previous section. Then there exists a pair of lower semi-continuous proper conjugate convex functions \((\varphi, \varphi^\ast)\) on \(\mathbb{R}^n\) such that

\[ \inf_{\tilde{\Phi}} J = J(\varphi, \varphi^\ast) \]

Strategy of proof:

Suppose \((\varphi_k, \psi_k)\) is the minimizing sequence: \(\lim_k J(\varphi_k, \psi_k) = \inf_{\tilde{\Phi}} J\).

And we may assume \((\varphi_k, \psi_k)\) are pairs of conjugate convex functions. (assume the second order moment is finite for the moment)

Extract a subsequence \((\varphi_k, \psi_k)\) such that \(\varphi_k \overset{L^1(\rm{d} \mu)}{\to} \varphi, \psi_k \overset{L^1(\rm{d} \nu)}{\to} \psi\) and \((\varphi, \psi) \in \tilde{\Phi}, J(\varphi, \psi) \leq \liminf J(\varphi_k, \psi_k)\).

Problem

The convergence connot be true in full generality. But \(J\) has invariance under addition of constant: \(J(\varphi, \psi) = J(\varphi + a, \psi + a)\), so we can translate \((\varphi_k, \psi_k)\) properly, s.t. they are stable on a neighborhood.