I'm trying solve the following problem:
Find a sequence $T_n$ of transport maps from $\mathbb{I}_{[0,1]}dx$ to $\mathbb{I}_{[0,1]}dx$ such that $T_n$ converges weakly to $\frac{1}{2}$. Deduce that the weak limit of transport maps need not be a transport map. Hint: take $T_n$ to be a piecewise linear sawtooth function.
I was told to use the following sequence
$$ T_n(x)=\sum_{j=0}^{2^{n-1}-1}2^{n-1}\left(x-\frac{j}{2^{n-1}}\right)\mathbb{I}_{\Big[\frac{j}{2^{n-1}},\frac{j+1}{2^{n-1}}\Big)}(x)\text{ and }T_n(1)=1. $$
Could anyone please help me with this exercise? That is
- a). How do I prove that for each $n\in\mathbb{N}, T_n$ is a pushforward for Lebesgue measure over** $[0,1]$? I.E.
$$ T_{\#\mathbb{I}_{[0,1]}}=\mathbb{I}_{[0,1]}. $$
- b). How do I prove the weak convergence to 1/2?
For (a), under the $\pi-\lambda$ Theorem, it suffices to prove $\mathcal{L}_{[0,1]}(T_n^{-1}([a,b]))=\mathcal{L}_{[0,1]}([a,b])$ for all intervals $[a,b]\subset[0,1]$, $n\in \mathbb{N}$ (here $\mathcal{L}_{[0,1]}$ is the Lebesgue measure restricted to [0,1]). Based on the definition of your $T_n$, this should be obvious by observing that the inverse image of an interval under $T_n$ is a disjoint union of smaller intervals with a total length equal to $[a,b]$. Maybe it is a good idea to plot the function for a few small $n$.
For (b), first note the integral of $T_n$ is $\frac{1}{2}$ for all $n\in\mathbb{N}$. Then since $L^2([0,1])$ is a reflexive Banach space, any norm-closed ball is weakly sequentially compact, which means that given any subsequence $\{T_{n_j}\}$ of $\{T_n\}$, we have a further subsequence $\{T_{n_{j_k}}\}$ that weakly converges to a function $f$ in $L^2([0,1])$. By choosing the test function to be 1, one can show that $f$ is always $\frac{1}{2}$, which shows that the original sequence converges weakly to $\frac{1}{2}$ itself, thus concluding the proof.