Is the Hessian matrix of thermodynamic potentials regarding extensive variables positive definite?

We learned that the Hessian matrix of thermodynamic potentials regarding extensive variables is positive definite (as a stability condition).

If we take the grand potential in its natural variables,
$$Omega(T,V,mu),qquad dOmega=-SdT-pdV-Ndmu$$
the only extensive variable is $V$, so it should follow that
$$left(frac{partial^2Omega}{partial V^2}right)_{T,mu}>0impliesleft(frac{-partial p}{partial V}right)_{T,mu}>0impliesleft(frac{partial p}{partial V}right)_{T,mu}<0$$

However, with help of the Gibbs-Duhem equation, one can also follow that
$$left(frac{partial p}{partial V}right)_{T,mu}=0,$$
which apparently is a contradiction. So is the Hessian matrix really only semidefinite?

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The derivation with the Gibbs-Duhem equation works as follows:
$$0=U-TS+pV-mu N=Omega+pViffOmega(T,V,mu)=-p(T,V,mu)V$$
As $dOmega=-SdT-pdV-Ndmu$ one gets:
$$left(frac{partialOmega}{partial V}right)_{T,mu}=-poverset{!}{=}left(frac{partial(-pV)}{partial V}right)_{T,mu}=-p-Vleft(frac{partial p}{partial V}right)_{T,mu}$$
Which implies
$$left(frac{partial p}{partial V}right)_{T,mu}=0$$