Why is it difficult to solve the Monge problem directly?

Question

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.

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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?

Answer 1

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 $0$.