Consider the function

$$f_{\operatorname{FS}}\colon\mathbb{R}^{n}\longrightarrow\mathbb{R},\quad u\longmapsto-\frac{1}{2}\log\Big(1+\sum_{i=1}^{n}\operatorname{e}^{-2u_{i}}\Big).$$

Let $\Delta^{n}=\{(x_{1},\dots,x_{n})\subset\mathbb{R}^{n}\mid x_{i}\geq 0,\sum x_{i}\leq 1\}$ be the standard simplex of $\mathbb{R}^{n}$. For $(x_{1},\dots,x_{n})\in\Delta^{n}$, write $x_{0}=1-\sum_{i=1}^{n}x_{i}$ and set

(3.54)

$$\varepsilon_{n}\colon\Delta^{n}\longrightarrow\mathbb{R},\quad x\longmapsto-\sum_{i=0}^{m}x_{i}\log(x_{i}).$$

We have $\displaystyle\nabla f_{{\operatorname{FS}}}(u)=\frac{1}{1+\sum_{i=1}^{n}\operatorname{e}^{-2u_{i}}}\left(\operatorname{e}^{-2u_{1}},\dots,\operatorname{e}^{-2u_{n}}\right)$ and so

$\displaystyle\frac{1}{2}\varepsilon_{n}(\nabla f_{{\operatorname{FS}}}(u))$

$\displaystyle=\frac{\sum_{i=1}^{n}\operatorname{e}^{-2u_{i}}u_{i}}{1+\sum_{i=1}^{n}\operatorname{e}^{-2u_{i}}}+\frac{1}{2}\log\Big(1+\sum_{i=1}^{n}\operatorname{e}^{-2u_{i}}\Big)=\langle\nabla f_{\operatorname{FS}}(u),u\rangle-f_{\operatorname{FS}}(u),$

which shows that $\operatorname{stab}(f_{{\operatorname{FS}}})=\Delta^{n}$ and that $f_{\operatorname{FS}}^{\vee}=\frac{1}{2}\varepsilon_{n}$.