5.13.4 The higher dimensional caseEB9F157B42F2

Proof. In positive dimension, regard a locally integrable $\varphi$ as a real distribution. If $\mathrm{dd}^{\mathrm c}\varphi\geq0$, the plurisubharmonic representative theorem supplies a plurisubharmonic function $\psi$ representing this distribution, hence equal to $\varphi$ almost everywhere. Conversely, that theorem proves local integrability and positivity of $\mathrm{dd}^{\mathrm c}\psi$ for every plurisubharmonic $\psi$; these properties are unchanged by almost-everywhere equality.

Uniqueness follows from the shrinking-ball formula for the subharmonic representative. This is essential: an arbitrary pointwise representative of a positive distribution need not itself be plurisubharmonic. In dimension zero the domain is a point, with its counting measure; local integrability is finiteness, so both conditions simply say that the value is finite.

Proof. For positive source and target dimensions, apply the holomorphic pullback theorem. Its conclusion is precisely that $\varphi\circ f$ is plurisubharmonic or identically $-\infty$ on each connected component. Here the source is connected, so there is just one alternative to consider. They are disjoint because the definition of plurisubharmonicity excludes the function identically $-\infty$.

If the source has dimension zero, it is a point: the pullback is either a finite number or $-\infty$. If the target has dimension zero, $\varphi$ is finite and constant, and so is its pullback. Thus these cases satisfy the same assertion.

Proof. Write $M(r)=\sup_{B_n(a,r)}\varphi$. This is finite: upper semicontinuity bounds $\varphi$ above on the closure of each smaller ball, and a plurisubharmonic function not identically $-\infty$ cannot equal $-\infty$ on a ball. It is increasing by inclusion of balls. In dimension zero it is constant.

Assume $n\geq1$ and fix $0<r_1<r_2<r_0$. Let $A\log r+B$ interpolate $M(r_1)$ and $M(r_2)$. On every complex line through $a$, comparison with this harmonic function on the annulus gives \[\varphi(z)\leq A\log|z-a|+B\qquad(r_1\leq|z-a|\leq r_2).\] The boundary bounds also hold on the spheres: if its value at a sphere point exceeded the interior supremum by $\delta>0$, upper semicontinuity would bound nearby values by that value plus $\delta/4$. A sufficiently small centered circle has at least one third of its length on the interior side, so its average would be strictly below the center value, contradicting the submean inequality. For $|z-a|<r_1$, the bound $\varphi(z)\leq M(r_1)$ suffices, because $A\geq0$. Taking the supremum over $|z-a|<r$, where $r_1<r<r_2$, yields \[M(r)\leq \frac{\log r_2-\log r}{\log r_2-\log r_1}M(r_1)+\frac{\log r-\log r_1}{\log r_2-\log r_1}M(r_2).\] This is convexity in $\log r$.

Proof. Convolve $g$ on smaller strips with a nonnegative smooth radial approximate identity. The regularizations $g_\varepsilon$ are smooth and subharmonic and decrease pointwise to $g$ by subharmonic regularization. They are still invariant under imaginary translations, hence have the form $g_\varepsilon(z)=f_\varepsilon(\operatorname{Re}z)$. Therefore $f_\varepsilon''=\Delta g_\varepsilon\geq0$, and each $f_\varepsilon$ is convex on its interval of definition.

Passing to the decreasing limit in the convexity inequality proves convexity of $f$ in the extended sense. This limit is finite everywhere: it is bounded above locally because $g$ is subharmonic, and if it were $-\infty$ at an interior point, convexity with a point beyond any other given point would force it to be $-\infty$ throughout $(a,b)$. That would make $g\equiv-\infty$, which is excluded. Thus $f$ is a real-valued convex function.