# 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?

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.

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

$$\newcommand{\R}{\mathbb R}$$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 $$H(\rho(t_k))=\log 2 \qquad \text{but}\qquad H(\rho(0))=0.$$