existence of the path on the space of probability density connecting two functions whose entropy are the same. Let $f,g:\mathbb{R}\rightarrow\mathbb{R}$ be a smooth probability density on $\mathbb{R}$ satisfying
$$
\int_{\mathbb{R}} f\cdot \log{f} \, dx = \int_{\mathbb{R}} g\cdot \log{g} 
\, dx<+\infty.
$$
Then, is there a path on probability density $f_t$ with $0\leq t\leq 1$ , satisfying
$$
\int_{\mathbb{R}} f_t\cdot \log{f_t} \, dx = \int_{\mathbb{R}} f\cdot \log{f} 
\, dx=\int_{\mathbb{R}} g\cdot \log{g} 
\, dx
$$
and $f_0=f$, $f_1=g$? If not, is there some counterexample for this statement?
 A: You can indeed always connect these probability density. Here is one method (but there are probably a lot of them) : start by considering the path $t \mapsto h_t := (1-t) f + t g$. This path does not preserve entropy, but by convexity the entropy remains finite along the path. Furthermore the entropy is a continuous function of $t$.
Now, given a probability density $h$ with finite entropy and $\lambda > 0$, you can define its dilatation $h_{\lambda}$ by $h^{\lambda}(x) = \lambda h(\lambda x)$. $h^{\lambda}$ also have finite entropy and more precisely :
$$H(h^{\lambda}) = \int h^{\lambda}(x) \log(h^{\lambda}(x)) dx \\
= \int h(x) \log(\lambda h(x)) dx \\
= H(h) + \log(\lambda) $$
Therefore, by choosing $\lambda(t) = \exp\left(H(f) - H(h_t)\right)$, the path $t\mapsto h_t^{\lambda(t)}$ has constant entropy and, since $\lambda(0) = \lambda(1) = 1$, it connects $f$ with $g$.
This method can be adapted to higher dimensions, but it seems to rely strongly on the fact that we are in $\mathbb{R}^d$. I wonder if there is a more general construction for the case of an arbitrary domain $\Omega$ with a reference measure $\mu$ (instead of $\mathbb{R}^d$ with reference measure Lebesgue).
