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.
> 
> 
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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?



Why is it difficult to solve the Monge problem directly?
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 $0$.


