PomeBD1A1A3DE2D3As $\nu $ runs through the supporting probabilities at $F$, the integral $-\int _ XG\, \mathrm{d}\nu $ ranges exactly between the two one-sided derivatives of $B_\varphi $ at $F$ in the direction $G$. Consequently, differentiability is equivalent to uniqueness of the supporting probability. For $F,G\in C(X)$, write $D^+B_\varphi (F)[G]$ and $D^-B_\varphi (F)[G]$ for the limits of the difference quotient $(B_\varphi (F+tG)-B_\varphi (F))/t$ as $t\downarrow 0$ and $t\uparrow 0$, respectively.
Proposition 11.34.5.792D6FEF2F3C For $F,G\in C(X)$, the one-sided derivatives exist, and
Consequently $B_\varphi $ is Gâteaux differentiable at $F$ if and only if $\mathcal A_\varphi (F)$ is a singleton. Its derivative is then $G\mapsto -\int _ XG\, \mathrm{d}\nu $.
Proof. The one-sided limits exist and are finite by concavity and the Lipschitz bound. For $t>0$, every supporting probability $\nu $ at $F$ gives
Conversely, choose $\nu _ t\in \mathcal A_\varphi (F+tG)$ and apply its supporting inequality with the test function $F$; this bounds the same quotient below by $-\int _ XG\, \mathrm{d}\nu _ t$. Along a sequence $t\downarrow 0$, the measures $\nu _ t$ have a weak limit point, which belongs to $\mathcal A_\varphi (F)$ by proposition 11.34.3. The two bounds identify the right derivative with minus the maximum. Applying this identity to $-G$ gives the formula for the left derivative.
If $\mathcal A_\varphi (F)$ is a singleton $\{ \nu \} $, the two derivatives coincide for every $G$ and equal $-\int _ XG\, \mathrm{d}\nu $, a continuous linear functional of $G$. Conversely, if the two one-sided derivatives agree for every $G$, all supporting probabilities have the same integral against every continuous function, and the Riesz representation theorem shows that they coincide. $\square$
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