Concavity of magnetization for Potts model For the Ising model the magnetization $ \langle \sigma_x \rangle_{\beta,h} $ is concave in the variable $h$. This means that
\begin{align*}
\frac{ \partial^2 \langle \sigma_x \rangle_{\beta,h} }{\partial^2 h} \leq 0. 
\end{align*}
See for example $[1]$.
As far as I know, this does not hold for Potts models, but is it known to be false?
What about random cluster models for general $1 \leq q \leq \infty$?
Reference:

*

*R. B. Griffiths, C. A. Hurst and S. Sherman, Concavity of Magnetization of an Ising Ferromagnet in a Positive External Field, J. Math. Phys. 11, 790 (1970).

 A: The question is bit ambiguous, since the Potts spins are not really scalar quantities (sure, you can map the $q$ states to $\{1,\dots,q\}$, but this does not properly reflect the permutation symmetry of the model). A better representation is as the vertices of the $(q-1)$-simplex, but then you have to say what you precisely mean by the GHS inequality in this model (in particular, what plays the role of the magnetic field and of the magnetization).
I'll interpret the question in the simplest way possible: the magnetic field acts as $-h\sum_i \delta_{\sigma_i,1}$ and the "magnetization" is given by $\langle\delta_{\sigma,1}\rangle_h$.
In this setting, the GHS inequality does not extend to the $q$-state Potts model with $q>2$. If I haven't made a stupid mistake, the simplest counterexample is a $q$-state Potts model with a single spin $\sigma\in\{1,\dots,q\}$ and free boundary condition, that is, the Hamiltonian reads
$$
\mathcal{H(\sigma)} = -h \delta_{\sigma, 1}. 
$$
In this case, $\langle\delta_{\sigma,1}\rangle_h = \operatorname{Prob}_{\,h}(\sigma=1) = \frac{e^h}{e^h + (q-1)}$.
One easily verifies that, when $q>2$,
$\frac{\mathrm{d}^2}{\mathrm{d}h^2} \langle\delta_{\sigma,1}\rangle_h$ does not have a constant sign when $h>0$.
