6
$\begingroup$

I'm trying to understand something about the Monge problem. The Monge problem is:

Let $c(x,y): \mathbb{R}^d \times \mathbb{R}^d \rightarrow \mathbb{R}^d$ and $$\mathcal{T}(\mu_1,\mu_2) = \{ T: \mathbb{R}^d \rightarrow \mathbb{R}^d | \text{ Borel maps with condition} \, \, T\#\mu_1 = \mu_2 \}$$ where $\mu_1$ and $\mu_2$ are given, compactly supported measures which are absolutely continuous with respect to Lebesgue measure. The Monge problem is to find: $$\inf_{T\in\mathcal{T}(\mu_1,\mu_2)}C_M[T] = \int_{\mathbb{R}^d}c(x,Tx) \mu_1(dx)$$

A book I am reading provides the following discussion on why finding a minimizer directly using "usual" methods (ie taking a minimizing sequence) is tough:

Usually, what one does is the following: take a minimizing sequence $T_n$, find a bound on it giving compactness in some topology (here, if the support of $\mu_2$ is compact, the maps $T_n$ take value in a common bounded set $\text{spt}\mu_2$, and so one can get compactness of $T_n$ in the weak-* $L^{\infty}$ convergence), take a limit $T_n \rightharpoonup T$, and prove that $T$ is a minimizer. This requires semicontinuity of the functional $C_M$ with respect to this convergence (which is true in many cases, for instance, if $c$ is convex in its second variable): we need $T_n \rightharpoonup T \implies \liminf_nC_M[T_n] \geq C_M[T]$, but we also need that the limit $T$ still satisfies the constraint. Yet, the nonlinearity of the pushforward condition prevents us from proving this stability when we only have weak convergence.

Loosely, this all makes sense to me: but I'm getting turned around working the details.

  1. Explicitly: Suppose $\text{spt}\mu_2$ is compact, why does $T_n \rightarrow T$ weak-* in $L^{\infty}$.

  2. Explicitly: what are the sufficient conditions on $C_M[T]$ to ensure the implication in the above statement $$T_n \rightharpoonup T \implies \liminf_nC_M[T_n] \geq C_M[T]$$ Do I just need to say $C_M[T]$ is weakly lower semicontinuous and bounded from below? (i.e. I think this is the same as convexity)

  3. Suppose I show that a minimizer of $C_M[T]$ exists. How do I see that minimizing sequence isn't preserving the pushforward condition in the limit? Apprently "non-linearity" is preventing this -- but I don't see how.

EDIT:

For 1&2: Okay then combining all your comments:

The statement: Let $C_M[T]: L^{\infty} \rightarrow \mathbb{R}$ be weak-* lower semicontinous and bounded from below. Then $C_M[T]$ has a minimizer.

Pf: $C_M[T]$ is bounded from below so $\inf C_M[T]$ exists. Let $T_n \in L^{\infty}$ be a minimizing sequence and let $T$ denote the minimizer. Consider a closed ball in the weak-* topology of positive radius centered at $T$. This ball is compact in weak-* topology by banach-aloglu - so there exists a subsequence of $T_n$, call it $T_{n_k} \rightarrow T$. This gives us $$\liminf C_M[T_{n_k}] \leq C_M[T_i] \quad \forall T_i \in L^{\infty}$$ -- in particular $\liminf C_M[T_{n_k}] \leq C_M[T]$. Weak-* LSC gives the other inequality, so $C_M[T]$ achieves its min.


for 3

It doesn't make sense to write: $$(aT_1+bT_2)_{\#}\mu = a(T{_1}_{\#}\mu) + b(T{_2}_{\#}\mu)$$ Let $a=b=1$, $T_1 = 2x$, $T_2 = x^2$, $\mu$ to be leb msr, and consider a set $A = [0,-1]$. Then $$(x^2 + 2x)_{\#}\mu\big([0,-1]\big) = \mu \bigg((x^2+2x)^{-1}\big([0,-1]\big)\bigg)$$ But $(x^2+2x)^{-1}[0,-1]$ can be $[0,1]$ or $[1,2]$ so LHS of such a statement is not well defined. So in what sense is the push forward non linear?

$\endgroup$

1 Answer 1

3
$\begingroup$
  1. It is not true that $T_n\to T$ in any sense. However, since $L^\infty=(L^1)^*$, the unit ball in $L^\infty$ is weak* compact by Banach-Alaogu theorem. Hence, one can extract from $T_n$ a weak* convergent subsequence.
  2. Yes, this is a standard notion of lower semi-continuity from calculus of variations.
  3. For $d=1$, let $\mu_1$ be the restriction of Lebesgue measure to $[0,1]$, and $\mu_2$ the sum of $1/2$-point masses at $-1$ and $1$. Let $T_n(t):=\text{sgn}\sin(2\pi n t)$. Then, clearly, $T_n\sharp\mu_1=\mu_2$ for all $n$. But $T_n\to T\equiv 0$ in the weak* topology and $T$ maps the Lebesgue measure to the point mass at $0$.
$\endgroup$
3
  • $\begingroup$ My follow ups are too long, so I edited the orginal post -- for context see question edits. 1. Is this statement/proof correct? 2. So compact support of $\mu_2$ wasn't essential? That statement in the highlighted passage made me first investigate using Uniform Boundedness Principle - was this a misleading statement then? Or is there another of showing a minimizer using this statement? $\endgroup$
    – yoshi
    Jun 7, 2019 at 18:12
  • $\begingroup$ 3. Okay, so if I find some $C_M[T]$ such that $T_n$ is given as you state, I can use your observation to show the pushforward is not preserved. The passage states "non-linearity" culprit. In what way is the pushforward non-linear? (again for context, see above edit) $\endgroup$
    – yoshi
    Jun 7, 2019 at 18:13
  • 1
    $\begingroup$ No, your proof is wrong, you cannot prove the existence of a minimiser starting with "let $T$ denote the minimiser". Rather, you have to assume that all $T_n$ belong to some ball in $L^\infty$, this is where you use that $\mu_2$ has compact support. As for nonlinearity, it just means that if for two maps you have $T_{1,2}\sharp \mu_1=\mu_2$, then $T_1+T_2$ usually does not have this property. Since the weak* topology is defined by linear functionals, it is usually difficult to prove that a non-linear (and non-convex) condition is weak*-closed - and indeed in our case it is not. $\endgroup$
    – Kostya_I
    Jun 7, 2019 at 18:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.