Example 8.9.9.5F56B4D9CC35 On the unit ball in $\mathbb C^3$, put $\beta _ j=\mathrm{dd}^{\mathrm{c}}|z_ j|^2$, $\beta =\sum _{j=1}^3\beta _ j$ and
\[ u=-\tfrac 12\sum _{j=1}^3|z_ j|^2, \qquad v=-\tfrac 45|z_1|^2. \]
Both $\beta +\mathrm{dd}^{\mathrm{c}}u$ and $\beta +\mathrm{dd}^{\mathrm{c}}v$ are strictly positive, but
\[ (\beta +\mathrm{dd}^{\mathrm{c}}v)^2\wedge \beta > (\beta +\mathrm{dd}^{\mathrm{c}}u)^2\wedge \beta , \qquad (\mathrm{dd}^{\mathrm{c}}v)^2\wedge \beta <(\mathrm{dd}^{\mathrm{c}}u)^2\wedge \beta . \]
Thus shifted Hessian order does not imply unshifted Hessian order, even for smooth potentials with positive perturbed forms.
Proof.
The eigenvalues of the two perturbed forms relative to $\beta $ are $(\tfrac 12,\tfrac 12,\tfrac 12)$ and $(\tfrac 15,1,1)$. For a diagonal form with relative eigenvalues $\lambda _1,\lambda _2, \lambda _3$, its second Hessian is $(\lambda _1\lambda _2+\lambda _1\lambda _3+\lambda _2\lambda _3)\beta ^3/3$. Consequently the shifted measures of $u,v$ are $\tfrac 14\beta ^3$ and $\tfrac 7{15}\beta ^3$, while their unshifted expressions are $\tfrac 14\beta ^3$ and $0$. Their unshifted traces are negative, so neither original function belongs to the local unshifted class. Adding $\sum _ j|z_ j|^2$ gives an exact correction in this constant closed background. An exact correction $\mathrm{dd}^{\mathrm{c}}\rho =\omega $ is impossible where $\mathrm d\omega \ne 0$.
$\square$
Pome
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