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We endow the space $\mathcal P_2^a (\mathbb R^d)$ of absolutely continuous probability measures with finite second moment with the Wasserstein distance $W_2$. Let $\mathcal H (\mu)$ be the relative entropy of $\mu$ w.r.t. Lebesgue measure. It is well-known that $\mathcal H$ is not continuous but only lower semi-continuous w.r.t. $W_2$.

Let $\mu : [0, T] \to \mathcal P_2^a (\mathbb R^d), t \mapsto \mu_t$ be absolutely continuous. Is $t \mapsto \mathcal H (\mu_t)$ continuous?

Thank you so much for your elaboration!


Update Let $(X, d)$ be a metric space. A map $y:[a, b] \rightarrow X$ is called absolutely continuous IFF there is $v \in L^1([a, b])$ such that $v \geq 0$ almost everywhere and $d(y(s), y(t)) \leq \int_s^t v(r) \, \mathrm{d} r$ for every $a \leq s \leq t \leq b$. A map $y:[0, \infty) \rightarrow X$ is called absolutely continuous on $[0, \infty)$ IFF $y|_{[a, b]}$ is absolutely continuous for any $0\le a<b$.

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  • $\begingroup$ What do you mean by absolute continuity of the map $t\mapsto\mu_t$? $\endgroup$
    – R W
    Commented Oct 24, 2023 at 2:31
  • $\begingroup$ @RW Please see my update. $\endgroup$
    – Akira
    Commented Oct 24, 2023 at 6:31

1 Answer 1

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$\newcommand{\R}{\mathbb R}$The answer is NO. I will provide below a counterexample in dimension $d=1$.


Preliminaries: Let's agree that the entropy is $$ H(\rho)=\int_{\mathbb R}\rho(x)\log\rho(x) dx $$ whenever $\rho=\rho(x) dx$ is absolutely continuous w.r.t. the Lebesgue measure $dx$. The counterexample to full (weak, or Wasserstein) continuity of the entropy is the typical "binary 1-0-1-0..." oscillations. I guess this is clear if you come from an information-theory background, I actually learnt this myself from here on Math Overflow. To be more precise, let $$ A_n=\bigcup_{i=0}^{n-1}\Big[2i/(2n)\,,\,(2i+1)/2n\Big) ,\qquad \rho_n=2\chi_{A_n}(x) dx $$ and $$ A=[0,1]\qquad\rho_*=\chi_{[0,1]}(x)dx. $$ You can think of this as a sequence of binary words spread across the fixed domain $x\in[0,1]$, the value $\rho_n(x)=2$ corresponding to the $1$-bit and the value $\rho_n(x)=0$ to the $0$-bit. So, we have a sequence of words $\rho_1=(1,0)$, then $\rho_2=(1,0,1,0)$, and so on, $\rho_n=\underbrace{(1,0,\dots,1,0)}_{n\text{ times}}$, each time with thinner and thinner elementary bits of width $\frac 1{2n}$. It is easy to check that $\rho_n$ converges narrowly (in duality with $C([0,1])$ test functions) to $\rho_*$, hence also in the $W_2$ distance. (Recall that, in bounded domains, convergence in any Wasserstein distance is equivalent to narrow convergence). However, one can compute explicitly $$ H(\rho_n)=n\times 2\log 2\times\frac 1{2n}=\log 2 $$ (number $n$ of bits, each of height $2$ and of width $\frac1{2n}$). This obviously does not converge to $H(\rho_*)=\int 1\log 1\,dx =0$.

The counterexample below will be constructed by suitably interpolating the $\rho_n$'s in time.


Construction of the counterexample: Pick any decreasing sequence $t_k\to 0$ with $t_1=1$, and set $h_k=t_k-t_{k+1}$. By an immediate extraction argument, there exists a subsequence $\rho_k=\rho_{n_k}$ such that $$ W_2(\rho_k,\rho_*)\leq h_k $$ and therefore by triangular inequality $$ W_2(\rho_k,\rho_{k+1})\leq h_k+h_{k+1} $$ for all $k\geq 1$. Let us now define the curve $(\rho(t))_{t\in(0,1]}$ by setting $$ t=t_k:\qquad \rho(t_k):=\rho_k $$ and $$ t\in(t_{k+1},t_k):\qquad \rho(t):=\text{the time-}h_k\text{ geodesic }W_2 \text{interpolation between }\rho_{k+1},\rho_k. $$ (I hope I don't need to make this more explicit.) By construction this curve is absolutely continuous (at least for $t>0$) with piecewise-constant metric speed $$ t\in(t_{k+1},t_k):\qquad |\dot \rho|(t) =\frac{W_2(\rho_k,\rho_{k+1})}{h_k} \leq \frac{h_k+h_{k+1}}{h_{k+1}}. $$ Moreover one also has that, for any small $t_0>0$ and with the appropiate choice of $k_0$ such that $t_0\in[t_{k_0},t_{k_0+1})$, $$ \int_{t_0}^1|\dot\rho|(t)dt \leq \sum\limits_{k=1}^{k_0}\frac{h_k+h_{k+1}}{h_k} h_k \leq \sum\limits_{k\geq 1}h_k+h_{k+1} \leq 2, $$ since by construction $\sum h_k=1$. This shows that the metric speed $t\mapsto |\dot\rho|(t)$ is globally $L^1$ up to $t_0=0^+$. By standard completeness one can extend by continuity $\rho(0)=\lim \rho(t_k)=\rho_*$ while maintaining absolute continuity (including up to $t=0$), and the thesis follows since then $$ H(\rho(t_k))=\log 2 \qquad \text{but}\qquad H(\rho(0))=0. $$

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  • $\begingroup$ Thank you very much for taking time to write such a detailed answer! $\endgroup$
    – Akira
    Commented Mar 27 at 15:14
  • $\begingroup$ Errrrr.... I just realized there was a very crucial mistake in my proof, I modified to fix it using extraction of a subsequence $\endgroup$ Commented Apr 1 at 9:47

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