Proposition 21.2.8909102168A21

Proof. Define $f(t)=U(t)$, identifying $t\in V$ with $t+\mathrm i0$; by invariance, $U(t+\mathrm is)=f(t)$. By theorem 5.9.1, $U$ is locally integrable, so the set $\{ U=-\infty \} =\{ f=-\infty \} +\mathrm i\mathbb R^ n$ is a Lebesgue null set in $\mathbb C^ n$. Applying Fubini's theorem on products of compact boxes in $t$ and in $s$, we see that $\{ f=-\infty \} $ is a null set in $V$.

For $\varepsilon >0$ let $V_\varepsilon =\{ t\in V:\operatorname {dist}(t,\mathbb R^ n\setminus V)>\varepsilon \} $. A point $t$ lies in $V_\varepsilon $ if and only if the closed ball $\overline{B(t,\varepsilon )}$ lies in $V$, so $V_\varepsilon $ is the intersection of the convex sets $V-y$, $|y|\leq \varepsilon $, and is convex. The points of the tube at distance more than $\varepsilon $ from its complement form the tube $V_\varepsilon +\mathrm i\mathbb R^ n$. On this set let $U_\varepsilon =U*\rho _\varepsilon $, where $\rho _\varepsilon $ is the radial mollifier of section 2.5 on $\mathbb C^ n=\mathbb R^{2n}$. The function $U_\varepsilon $ is smooth, and its complex Hessian matrix $(\partial ^2U_\varepsilon /\partial w_ j\partial \bar w_ k)$ is positive semidefinite. Indeed, for $\xi \in \mathbb C^ n$, the function $\sum _{j,k}(\partial ^2U_\varepsilon /\partial w_ j\partial \bar w_ k)\xi _ j\bar\xi _ k$ is, by proposition 2.5.3(1), the mollification of the distribution $\sum _{j,k}(\partial ^2U/\partial w_ j\partial \bar w_ k)\xi _ j\bar\xi _ k$, which is positive by theorem 5.9.1 and the matrix criterion of section 5.4, so it is nonnegative by proposition 2.5.3(3).

Moreover, $U_\varepsilon \downarrow U$ as $\varepsilon \downarrow 0$ by (3.4.13), because $U$ is subharmonic and locally integrable (theorem 5.9.1). Since imaginary translations commute with convolution, $U_\varepsilon $ is invariant under them; write $U_\varepsilon (t+\mathrm is)=f_\varepsilon (t)$. For a function of $t=\operatorname{Re}w$ alone we have $\partial /\partial w_ j=\frac12\partial /\partial t_ j=\partial /\partial \bar w_ j$, hence

The left side is a positive semidefinite matrix, so $f_\varepsilon $ has positive semidefinite real Hessian. Hence the restriction of $f_\varepsilon $ to every segment in the convex set $V_\varepsilon $ has nonnegative second derivative, and $f_\varepsilon $ is convex on $V_\varepsilon $.

Suppose that $f(t_0)=-\infty $ for some $t_0\in V$, and choose $r\in (0,1)$ with $\overline{B(t_0,3r)}\subset V$. The upper semicontinuous function $U$ is bounded above, say by $C$, on the compact set $\overline{B(t_0,3r)}+\mathrm i[-1,1]^ n$. Since $U_\varepsilon (w)$ is an average of $U$ over the ball $\overline{B(w,\varepsilon )}$, we get $U_\varepsilon \leq C$ on $B(t_0,2r)$ for $\varepsilon <r$; moreover $B(t_0,2r)\subset V_\varepsilon $ for such $\varepsilon $. For $t\in B(t_0,r)$, the point $2t-t_0$ lies in $B(t_0,2r)$, and $t$ is the midpoint of $t_0$ and $2t-t_0$. Convexity of $f_\varepsilon $ therefore gives

As $\varepsilon \downarrow 0$, $f_\varepsilon (t_0)\downarrow f(t_0)=-\infty $ and $f_\varepsilon (t)\downarrow f(t)$, so $f=-\infty $ on $B(t_0,r)$. This contradicts the fact that $\{ f=-\infty \} $ is a null set. Hence $f$ is finite. Finally, every segment $[a,b]\subset V$ is compact and therefore lies in $V_\varepsilon $ for small $\varepsilon $; the convexity inequalities for $f_\varepsilon $ on $[a,b]$ pass to the pointwise limit, so $f$ is convex.

Conversely, let $f$ be a finite convex function on $V$. By proposition 21.2.4, $f$ is continuous and has a subgradient $p\in \mathbb R^ n$ at every point. Let $w_0=t_0+\mathrm is_0$ be a point of the tube, $a\in \mathbb C^ n\setminus \{ 0\} $, and $r>0$ such that the circle $\{ w_0+r\mathrm e^{\mathrm i\theta }a:\theta \in [0,2\pi ]\} $ lies in the tube. If $p$ is a subgradient of $f$ at $t_0$, then

The last term has mean zero over $\theta \in [0,2\pi ]$, so the mean of the left side over the circle is at least $f(t_0)=f(\operatorname{Re}w_0)$. Since $w_0$, $a$ and $r$ are arbitrary, the restriction of the continuous function $f(\operatorname{Re}w)$ to every affine complex line satisfies the submean inequality of definition 3.4.2 at every point, and it is therefore subharmonic. As $f(\operatorname{Re}w)$ is finite and continuous and the tube is a domain, definition 5.8.1 shows that $f(\operatorname{Re}w)$ is plurisubharmonic. It is invariant under imaginary translations by construction. $\square$