Let $f\colon[0,1]\to\mathbb{R}$ the function defined by

$$f(u)=\begin{cases}-u\log(u),&\text{ if }0\leq u\leq\operatorname{e}^{-1},\\ \operatorname{e}^{-1},&\text{ if }\operatorname{e}^{-1}\leq u\leq 1-\operatorname{e}^{-1},\\ -(1-u)\log(1-u),&\text{ if }1-\operatorname{e}^{-1}\leq u\leq 1.\end{cases}$$

Then $\operatorname{stab}(f)=\mathbb{R}$ and the Legendre-Fenchel dual is the function $f^{\vee}(x)=x-\operatorname{e}^{x-1}$ for $x\leq 0$ and $f^{\vee}(x)=-\operatorname{e}^{-x-1}$ for $x\geq 0$. Then ${\operatorname{dom}}(\partial f)=(0,1)$ and ${\operatorname{dom}}(\partial f^{\vee})=\mathbb{R}$. Moreover,

$$\Pi(f)=(0,\operatorname{e}^{-1})\cup\{[\operatorname{e}^{-1},1-\operatorname{e}^{-1}]\}\cup(1-\operatorname{e}^{-1},1),\quad\Pi(f^{\vee})=\mathbb{R}.$$

The Legendre-Fenchel correspondence sends bijectively $(0,\operatorname{e}^{-1})$ to $\mathbb{R}_{>0}$ and $(1-\operatorname{e}^{-1},1)$ to $\mathbb{R}_{<0}$, and sends the element $[\operatorname{e}^{-1},1-\operatorname{e}^{-1}]$ to the point $\{0\}$. In this example, $\Pi(f)$ is not a subdivision while $\Pi(f^{\vee})$ is.