I am reading these notes on non-linear electrodynamics (NED). On page 8, below equation (5.1) the author states that the modified electromagnetism parameter $\gamma$ should be non-negative in order to ensure causality and unitarity. How can we see that? For completeness, here's the Lagrangian
$$L_{ModMax} = -\frac{\cosh \gamma}{4}F_{\mu\nu}F^{\mu\nu} + \frac{\sinh \gamma}{4}\sqrt{(F_{\mu\nu}F^{\mu\nu})^2 + (F_{\mu\nu}\tilde{F}^{\mu\nu})^2} \tag{5.1}$$
In the very next sentence below eq. (5.1) Ref. 1 writes:
These values of $\gamma$ also ensure that the Lagrangian density is a convex function of the electric field $E^i$.
And on p. 5:
Other restrictions on the form of the NED action are imposed by the physical requirements of causality and unitarity of the theory (see e.g. [35–39]). These are related to the requirement that the Lagrangian density is a convex function of the electric field strength $E^i$, i.e. that the Hessian $3 \times 3$ matrix $${\cal L}_{ij} ~=~\frac{\partial^2{\cal L}}{\partial E^i\partial E^j} \tag{3.6}$$ has only non-negative eigenvalues for all values of ${\bf E}$ and ${\bf B}$.
This is because the electric field $E^i$ is the momentum field to the gauge field $A_i$, so that the Hessian is the kinetic term. See also e.g. this related Phys.SE post.
On one hand, for $\gamma\geq 0$ the Lagrangian density (5.1) is a non-negative linear combination of convex functions and therefore also convex.
On the other hand, to actually check that the Lagrangian density (5.1) is not a convex function for $\gamma<0$, calculate e.g. $$ \left. {\cal L}_{xx}\right|_{{\bf E}=(0,1+\epsilon,0),{\bf B}=(1,0,0)} \quad\sim\quad \frac{\sinh\gamma}{|\epsilon|} \quad\text{for}\quad \epsilon~\to~ 0,$$ and note that it would be negative.
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