5.13 The definition of plurisubharmonic functions92A4FA45DFE1

Proof. Suppose first that $\varphi\in L^1_{\mathrm{loc}}(\Omega)$ and $\Delta\varphi\geq0$. Apply the distributional characterization of subharmonic functions to the distribution defined by $\varphi$. It gives a subharmonic representative $\psi$ of this distribution. Equality of the distributions defined by two locally integrable functions is equality almost everywhere, so $\varphi=\psi$ almost everywhere.

Conversely, a subharmonic function on the connected domain $\Omega$ is locally integrable and has positive distributional Laplacian by the same theorem. Replacing it on a null set changes neither its distribution nor its local integrability. Finally, two subharmonic representatives that agree almost everywhere agree at every point: their values are the limits of their averages over shrinking balls. This also proves uniqueness. Multiplying the usual planar Laplacian by a fixed positive normalization constant does not change positivity.

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.

Proof. Work on one connected component $Y_0$ of $Y$. In coordinate neighborhoods of $Y$ and $X$, holomorphic pullback says that $\varphi\circ f$ is either plurisubharmonic or identically $-\infty$. A plurisubharmonic function cannot be identically $-\infty$ on a nonempty open subset of a connected domain, by its local integrability. Thus a neighborhood on which the pullback is identically $-\infty$ cannot overlap a neighborhood on which it is plurisubharmonic. The two types form disjoint open subsets of $Y_0$; connectedness leaves only one type. If the first type never occurs on any component, the pullback is plurisubharmonic on $Y$. Zero-dimensional charts are covered by the finite-value convention.

Proof. Restrict a locally integrable $\varphi$ with $\mathrm{dd}^{\mathrm c}\varphi\geq0$ to each coordinate chart and apply the local representative theorem. The resulting plurisubharmonic representatives agree almost everywhere on overlaps, hence everywhere there by the uniqueness part of the same theorem. They therefore glue to a plurisubharmonic function on $X$ representing $\varphi$.

Conversely, local integrability and positivity hold in every chart for a plurisubharmonic function and are preserved by almost-everywhere equality. The same overlap argument proves uniqueness globally. On a zero-dimensional component all assertions reduce to finiteness at each point.

Proof. Apply Josefson's theorem, proved earlier in the collection to $E$. It constructs a function $V\in\mathrm{PSH}(\mathbb C^n)$ with $E\subset\{V=-\infty\}$, in fact with the additional bound $V(z)\leq\log^+|z|$. Taking $\varphi=V$ gives the assertion. The earlier construction is a proof of this stronger statement, so no additional external theorem is being invoked here. In dimension zero a pluripolar set is empty and a finite constant suffices.