21.2 Closed convex functions and supporting slopes5430067C9C99

Proof. Fix a Euclidean inner product $y\cdot w$ on $W$, with norm $|y|$.

(1) Choose $y_ k\in C$ with $|z-y_ k|\to \inf _{w\in C}|z-w|$. The sequence $(y_ k)$ is bounded, so a subsequence converges to a point $y$, which lies in $C$ because $C$ is closed. Thus $y$ minimizes the distance to $z$ on $C$, and $y\neq z$. For $w\in C$ and $0\leq t\leq 1$ the point $y+t(w-y)$ lies in $C$, so the function $t\mapsto |z-y-t(w-y)|^2$ on $[0,1]$ is minimal at $t=0$, and its derivative there, $-2(z-y)\cdot (w-y)$, is nonnegative. Hence $\ell (w)=(z-y)\cdot w$ satisfies $\ell (w)\leq \ell (y)=\ell (z)-|z-y|^2<\ell (z)$ for every $w\in C$.

(2) Since $z$ is a boundary point of $C$, there are points $z_ k\notin C$ with $z_ k\to z$. By (1) there are unit vectors $a_ k$ with $a_ k\cdot w<a_ k\cdot z_ k$ for all $w\in C$. A subsequence of $(a_ k)$ converges to a unit vector $a$, and in the limit $a\cdot w\leq a\cdot z$ for all $w\in C$. As $z\in C$, the functional $\ell (w)=a\cdot w$ has the required property.

(3) Let $x=\sum _{i=1}^ q\lambda _ ix_ i$ with $x_ i\in S$, $\lambda _ i>0$ and $\sum _ i\lambda _ i=1$. If the points $x_ i$ are affinely dependent, the vectors $(x_ i,1)\in W\times \mathbb R$ are linearly dependent, say $\sum _ i\mu _ i(x_ i,1)=0$ with not all $\mu _ i$ zero. Since $\sum _ i\mu _ i=0$, some $\mu _ i$ is positive. Replacing each $\lambda _ i$ by $\lambda _ i-s\mu _ i$, where $s=\min \{ \lambda _ i/\mu _ i:\mu _ i>0\} $, preserves $\sum _ i\lambda _ i$ and $\sum _ i\lambda _ ix_ i$, keeps all coefficients nonnegative and makes at least one of them zero. We discard the points with zero coefficient and repeat. Eventually the remaining points are affinely independent, and an affinely independent subset of $W$ has at most $n+1$ elements.

(4) Let $A$ be the affine span of $C$, of dimension $k$, and write $B_ A(x,r)$ for the open ball of $A$ with center $x$ and radius $r$. If $k=0$, then $C$ is a point, equal to its relative interior and to its closure, and its relative boundary is empty; so assume $k\geq 1$. A maximal affinely independent subset of $C$ spans an affine subspace containing $C$, hence equal to $A$, so it consists of $k+1$ points $x_0',\ldots ,x_ k'$. Every point $y\in A$ has unique barycentric coordinates $\lambda _ i(y)$, which are affine functions of $y$ with $\sum _ i\lambda _ i(y)=1$ and $y=\sum _ i\lambda _ i(y)x_ i'$. At the barycenter $c$ of the simplex with vertices $x_ i'$ all of them equal $1/(k+1)$, so they are positive on a ball $B_ A(c,r)$; every point of this ball is then a convex combination of the $x_ i'$ and lies in $C$. Hence $c\in \operatorname {ri}C$.

Now let $x_0\in \operatorname {ri}C$, choose $r>0$ with $B_ A(x_0,r)\subset C$, and let $y\in \overline C$ and $0\leq t<1$. If $y\in C$, convexity gives $B_ A((1-t)x_0+ty,(1-t)r)=(1-t)B_ A(x_0,r)+ty\subset C$. In general choose $y_ k\in C$ with $y_ k\to y$; then $B_ A((1-t)x_0+ty_ k,(1-t)r)\subset C$, and for large $k$ this ball contains $B_ A((1-t)x_0+ty,(1-t)r/2)$. Thus $(1-t)x_0+ty\in \operatorname {ri}C$. Letting $t\uparrow 1$, we see that every point of $\overline C$ is a limit of points of $\operatorname {ri}C$, so $\overline{\operatorname {ri}C}=\overline C$. Next, $\operatorname {ri}C\subset \operatorname {ri}\overline C$, since both are interiors in the same affine space $A$. Conversely, if $x\in \operatorname {ri}\overline C$, then $y=x+\varepsilon (x-c)$ lies in $\overline C$ for small $\varepsilon >0$, and $x=(1-t)c+ty$ with $t=1/(1+\varepsilon )<1$; hence $x\in \operatorname {ri}C$.

For the measure assertion, translate so that $c=0$ and identify $A$ with $\mathbb R^ k$ by a linear isomorphism, which changes Lebesgue measure on $A$ only by a constant factor. Let $Z=\overline C\setminus \operatorname {ri}C$ be the relative boundary; it is closed in $A$, because $\operatorname {ri}C$ is open in $A$, hence Borel, and $0\notin Z$. We claim that the dilates $sZ$, $s>0$, are pairwise disjoint. Indeed, suppose that $sz=s'z'$ with $z,z'\in Z$ and $0<s<s'$. Then $z'=(s/s')z$ is of the form $(1-t)\cdot 0+tz$ with $z\in \overline C$ and $0<t<1$, so $z'\in \operatorname {ri}C$ by what we have just shown, which is impossible. Now fix $R>0$ and put $Z_ R=Z\cap B(0,R)$. The sets $sZ_ R$ with $s$ rational in $[\frac12,1]$ are pairwise disjoint subsets of $B(0,R)$, and each has measure $s^ k|Z_ R|\geq 2^{-k}|Z_ R|$. Since infinitely many of them fit into a set of finite measure, $|Z_ R|=0$. Letting $R\to \infty $, we obtain $|Z|=0$. $\square$

Proof. We first show that $F$ is bounded, hence Lipschitz, near each point of $\operatorname {ri}(\operatorname {dom}F)$, and find a subgradient there by separation; then we prove (1)–(4). Let $A$ be the affine span of $\operatorname {dom}F$, of dimension $k$, and let $x\in \operatorname {ri}(\operatorname {dom}F)$. Choose affine coordinates on $A$ centered at $x$, with the associated Euclidean norm $|\cdot |$; since all norms on $A$ are equivalent, local Lipschitz continuity does not depend on this choice. For small $\delta >0$ the cube $Q$ of points of $A$ whose coordinates are at most $2\delta $ in absolute value lies in $\operatorname {dom}F$. Every point of $Q$ is a convex combination of the $2^ k$ vertices of $Q$; this follows by induction on $k$, since writing the last coordinate of a point as a convex combination of $\pm 2\delta $ exhibits the point as a convex combination of two points of the opposite faces of $Q$, which are cubes of dimension $k-1$. By the finite form of Jensen's inequality, $F\leq C$ on $Q$, where $C$ is the largest value of $F$ at a vertex. If $y\in Q$, then also $2x-y\in Q$, and $F(x)\leq \frac12(F(y)+F(2x-y))$ gives $F(y)\geq 2F(x)-C$. Thus $F$ is bounded on $Q$, and by (21.2.2) it is Lipschitz on the cube of half-side $\delta $ centered at $x$.

Next we construct a subgradient at $x$. Since $\operatorname {dom}F\subset A$, the epigraph of $F$ lies in $A\times \mathbb R$; let $E$ be its closure, a closed convex subset of $A\times \mathbb R$. Because $F$ is continuous at $x$ as a function on $A$, the points $(x,F(x)-\varepsilon )$, $\varepsilon >0$, do not belong to $E$, so $(x,F(x))$ is a boundary point of $E$ in $A\times \mathbb R$. By lemma 21.2.3(2), applied in $A\times \mathbb R$ with origin $(x,F(x))$, there are a linear functional $a$ on the linear subspace $A_0=\{ y-y':y,y'\in A\} $ parallel to $A$ and a real number $c$, not both zero, such that

Letting $s\to +\infty $ shows that $c\geq 0$. If $c=0$, then $a\neq 0$ and $\langle a,y-x\rangle \geq 0$ for all $y\in \operatorname {dom}F$; but $\operatorname {dom}F$ contains a ball of $A$ about $x$, and this ball contains points $y$ with $\langle a,y-x\rangle <0$. Hence $c>0$, and taking $s=F(y)$ we obtain $F(y)\geq F(x)+\langle p_ A,y-x\rangle $ for $y\in \operatorname {dom}F$, where $p_ A=-a/c$. Extend $p_ A$ to a linear functional $p\in M_{\mathbb R}$. The inequality $F(y)\geq F(x)+\langle p,y-x\rangle $ holds for $y\in \operatorname {dom}F$, because then $y-x\in A_0$, and also for $y\notin \operatorname {dom}F$, where $F(y)=+\infty $. Thus $p\in \partial F(x)$.

From now on let $F$ be a finite convex function on the open convex set $\Omega $, so that $A=N_{\mathbb R}$ and $\operatorname {ri}(\operatorname {dom}F)=\Omega $. Fix a basis $e_1,\ldots ,e_ n$ of $N_{\mathbb R}$, let $K\subset \Omega $ be compact, and choose $r>0$ such that the set $K_ r=\{ y+\sum _ is_ ie_ i:y\in K,\ |s_ i|\leq r\} $ lies in $\Omega $. If $x\in K$ and $p\in \partial F(x)$, the subgradient inequality at $x$, evaluated at the points $x\pm re_ i$, gives $\pm r\langle p,e_ i\rangle \leq F(x\pm re_ i)-F(x)$, hence $|\langle p,e_ i\rangle |\leq 2r^{-1}\max _{K_ r}|F|$, where $F$ is bounded on the compact set $K_ r$ because it is continuous. This is the uniform bound in (1). If $x_ k\to x$ in $\Omega $, $p_ k\in \partial F(x_ k)$ and $p_ k\to p$, then letting $k\to \infty $ in $F(y)\geq F(x_ k)+\langle p_ k,y-x_ k\rangle $ and using that $F$ is continuous at $x$, we get $p\in \partial F(x)$. Thus the graph is closed, and each $\partial F(x)$ is closed and bounded, hence compact.

For (2) and (4) we use the following slope inequalities for a convex function $\phi $ on an interval: if $s_1<s_2<s_3$, then

Indeed, $s_2=\lambda s_1+(1-\lambda )s_3$ with $\lambda =(s_3-s_2)/(s_3-s_1)\in (0,1)$, and the convexity inequality $\phi (s_2)\leq \lambda \phi (s_1)+(1-\lambda )\phi (s_3)$ is equivalent to each of the two inequalities. Now fix $x$ and $v$, and apply (21.2.5) to $\phi (t)=F(x+tv)$, which is convex on the open interval of those $t$ with $x+tv\in \Omega $: with $s_1=0$ it shows that the difference quotient in (2) is nondecreasing in $t>0$, so its limit as $t\downarrow 0$ exists in $[-\infty ,+\infty )$. Each $p\in \partial F(x)$ gives $F(x+tv)-F(x)\geq t\langle p,v\rangle $; hence $F'(x;v)\geq \max _{p\in \partial F(x)}\langle p,v\rangle $, the maximum being attained by compactness. Conversely, choose $p_ t\in \partial F(x+tv)$. The subgradient inequality at $x+tv$, evaluated at $x$, gives $F(x+tv)-F(x)\leq t\langle p_ t,v\rangle $. By (1) the vectors $p_ t$ stay bounded as $t\downarrow 0$, and each of their cluster points lies in $\partial F(x)$. Therefore $F'(x;v)\leq \varlimsup _{t\downarrow 0}\langle p_ t,v\rangle \leq \max _{p\in \partial F(x)}\langle p,v\rangle $.

For (3), suppose that $\partial F(x)=\{ p\} $, and choose $p_ h\in \partial F(x+h)$. By (1), every cluster point of $p_ h$ as $h\to 0$ lies in $\partial F(x)$, so $p_ h\to p$. The subgradient inequalities at $x$ and at $x+h$ give

so $F$ is differentiable at $x$, with $\nabla F(x)=p$. Conversely, if $F$ is differentiable at $x$ and $p\in \partial F(x)$, dividing $F(x+tv)-F(x)\geq t\langle p,v\rangle $ by $t$ and letting $t\downarrow 0$ gives $\langle \nabla F(x),v\rangle \geq \langle p,v\rangle $ for every $v$, hence $p=\nabla F(x)$. As $\partial F(x)\neq \varnothing $, it equals $\{ \nabla F(x)\} $.

For (4), fix $i$, and write $D_ i^+F(x)=F'(x;e_ i)$ and $D_ i^-F(x)=-F'(x;-e_ i)$ for the one-sided partial derivatives; they are finite by (2). Let $\phi $ be the restriction of $F$ to the intersection of $\Omega $ with a line parallel to $e_ i$, parametrized as $\phi (s)=F(y+se_ i)$ on an open interval. By (21.2.5), $\phi $ has one-sided derivatives $\phi '_\pm $ at every point, $\phi '_-(s)\leq \phi '_+(s)$, and $\phi '_+(s)\leq \phi '_-(s')$ whenever $s<s'$, because both are bounded by the slope of the chord from $s$ to $s'$. Hence the open intervals $(\phi '_-(s),\phi '_+(s))$ are pairwise disjoint; each nonempty one contains a rational number, so only countably many of them are nonempty, and $\phi '_-=\phi '_+$ outside a countable set. Next, $D_ i^+F$ is the pointwise limit on $\Omega $ of the Borel functions equal to $k(F(x+e_ i/k)-F(x))$ where $x+e_ i/k\in \Omega $ and to $0$ elsewhere, so it is a Borel function, and similarly for $D_ i^-F$. Thus the set $Z_ i\subset \Omega $ where $D_ i^+F\neq D_ i^-F$ is a Borel set that meets each line parallel to $e_ i$ in a countable set, and Fubini's theorem, in the coordinates of the basis $e_1,\ldots ,e_ n$, shows that $Z_ i$ is a null set. At a point $x\notin Z_1\cup \cdots \cup Z_ n$, (2) gives

for every $i$, so all subgradients at $x$ coincide, and $F$ is differentiable at $x$ by (3). Conversely, if $F$ is differentiable at $x$, then $D_ i^+F(x)=D_ i^-F(x)=\langle \nabla F(x),e_ i\rangle $ for every $i$. Therefore $\Omega _ F=\Omega \setminus (Z_1\cup \cdots \cup Z_ n)$ is a Borel set with null complement, and the coordinates $\langle \nabla F,e_ i\rangle =D_ i^+F$ of $\nabla F$ are Borel functions on $\Omega _ F$. $\square$

Proof. By proposition 21.2.4, $F$ has a subgradient $p$ at some point $x_1\in \operatorname {ri}(\operatorname {dom}F)$, so $F$ lies above the affine function $\ell (y)=F(x_1)+\langle p,y-x_1\rangle $. The closure $E$ of $\operatorname {epi}F$ is convex, and it lies above the graph of $\ell $ because $\ell $ is continuous. Hence, for each $x$, the set $\{ r:(x,r)\in E\} $ is closed and bounded below by $\ell (x)$, and with each of its points $r$ it contains $[r,+\infty )$, because $\operatorname {epi}F$ has this property. So this set is either empty or a closed half-line $[\widetilde F(x),+\infty )$ with $\widetilde F(x)\geq \ell (x)>-\infty $. Setting $\widetilde F(x)=+\infty $ in the first case, we see that $E$ is the epigraph of a convex function $\widetilde F$. It is lower semicontinuous because its epigraph is closed, and $\widetilde F\leq F$, so $\operatorname {dom}\widetilde F\supset \operatorname {dom}F\neq \varnothing $. If $G\leq F$ is lower semicontinuous, then $\operatorname {epi}G$ is closed and contains $\operatorname {epi}F$, hence also $E$, and so $G\leq \widetilde F$. Thus $\widetilde F=\overline F$, which is therefore closed, proper and convex.

Let $x\in \operatorname {ri}(\operatorname {dom}F)$. If $(y_ k,r_ k)\in \operatorname {epi}F$ converge to $(x,r)$, then $y_ k$ lies in the affine span of $\operatorname {dom}F$, on which $F$ is continuous at $x$; hence $r\geq \lim _ kF(y_ k)=F(x)$. This shows $\overline F(x)\geq F(x)$, and so $\overline F(x)=F(x)$.

Now let $F$ be closed, fix $x_0\in \operatorname {ri}(\operatorname {dom}F)$, and let $x\in \overline{\operatorname {dom}F}$. By lemma 21.2.3(4), the points $(1-t)x_0+tx$ with $0\leq t<1$ lie in $\operatorname {ri}(\operatorname {dom}F)$, so the convex function $\phi (t)=F((1-t)x_0+tx)$ is finite on $[0,1)$. Convexity gives $\phi (t)\leq (1-t)\phi (0)+t\phi (1)$, hence $\varlimsup _{t\uparrow 1}\phi (t)\leq \phi (1)$, while lower semicontinuity gives $\varliminf _{t\uparrow 1}\phi (t)\geq \phi (1)$. Therefore

also when $F(x)=+\infty $. Since $F=+\infty $ outside $\overline{\operatorname {dom}F}$, which is the closure of $\operatorname {ri}(\operatorname {dom}F)$, the function $F$ is determined by its restriction to $\operatorname {ri}(\operatorname {dom}F)$.

Finally, let $(F_ j)$ be a sequence of finite convex functions on $\Omega $ converging to a finite limit at every point of a dense set $S\subset \Omega $. We show that near each point of $\Omega $ the functions $F_ j$ are uniformly bounded, hence uniformly Lipschitz, so that convergence on $S$ propagates to uniform convergence. Fix a basis of $N_{\mathbb R}$, write $Q(y,\rho )$ for the closed cube of half-side $\rho $ centered at $y$ in the corresponding coordinates, and let $x\in \Omega $ and $\rho >0$ with $Q(x,2\rho )\subset \Omega $. For each vertex $y$ of $Q(x,\rho )$ choose a point of $S$ whose coordinates differ from those of $y$ by less than $\rho /2$, and let $S_0\subset S\cap Q(x,2\rho )$ be the finite set of these points. Its convex hull $H$ is compact as the image of a simplex under a linear map, and it contains $Q(x,\rho /2)$. Indeed, let $\ell $ be a nonzero linear functional, $\ell (z)=\sum _ ia_ iz_ i$ in the chosen coordinates, and put $\| a\| =\sum _ i|a_ i|$. At a vertex $y^*$ of $Q(x,\rho )$ where $\ell $ is maximal, $\ell (y^*)=\ell (x)+\rho \| a\| $, and the point $s\in S_0$ chosen for $y^*$ satisfies $\ell (s)>\ell (y^*)-\frac\rho 2\| a\| =\max _{Q(x,\rho /2)}\ell $. Hence no point of $Q(x,\rho /2)$ can be separated from $H$ as in lemma 21.2.3(1).

The sequences $(F_ j(s))_ j$, $s\in S_0$, converge, so they are bounded above by some $C$, and the finite form of Jensen's inequality gives $F_ j\leq C$ on $H$ for all $j$. Choose $s_0\in S\cap Q(x,\rho /8)$. Then $Q(s_0,3\rho /8)\subset Q(x,\rho /2)$, and for $y\in Q(s_0,3\rho /8)$ the reflected point $2s_0-y$ lies in the same cube, so that $F_ j(y)\geq 2F_ j(s_0)-C\geq 2\inf _ kF_ k(s_0)-C$. Hence $|F_ j|\leq B$ on $Q(x,\rho /4)\subset Q(s_0,3\rho /8)$ for all $j$, with a constant $B$. By (21.2.2), applied with $\delta =\rho /8$ and $\omega \leq 2B$, the $F_ j$ are Lipschitz on $Q(x,\rho /8)$ with the common constant $L=16B/\rho $, for the Euclidean norm $|\cdot |$ of the chosen coordinates.

Given $\varepsilon >0$, choose a finite set $S_\varepsilon \subset S\cap Q(x,\rho /8)$ such that every point of $Q(x,\rho /8)$ lies within distance $\varepsilon /L$ of $S_\varepsilon $. To this end, cover the compact cube $Q(x,\rho /8)$ by finitely many balls of radius $\varepsilon /(2L)$ centered in the cube, and pick a point of the dense set $S$ in the intersection of each ball with the interior of the cube, a nonempty open set. For all sufficiently large $j,k$ we have $|F_ j-F_ k|<\varepsilon $ on $S_\varepsilon $, hence $|F_ j-F_ k|<3\varepsilon $ on $Q(x,\rho /8)$. Thus $(F_ j)$ is uniformly Cauchy on $Q(x,\rho /8)$. Finitely many such cubes cover any given compact subset of $\Omega $, so $(F_ j)$ converges locally uniformly to a finite continuous function, which is convex because the convexity inequalities pass to pointwise limits. $\square$

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. $\square$

Proof. Consider the holomorphic map

Since $\log |E_ j(w)|^2=\operatorname{Re}w_ j$, it maps the tube $V+\mathrm i\mathbb R^ n$ onto $\Omega _ V$; in particular, $\Omega _ V$ is connected. By proposition 5.9.3, $\varphi \circ E$ is either plurisubharmonic or identically $-\infty $ on the tube, and the second alternative is impossible because $E$ is surjective and $\varphi $ is nontrivial. Since $E(w+\mathrm is')=(\mathrm e^{\mathrm is_1'/2}E_1(w),\ldots ,\mathrm e^{\mathrm is_ n'/2}E_ n(w))$, the rotation invariance of $\varphi $ makes $\varphi \circ E$ invariant under imaginary translations. Proposition 21.2.8 now gives a finite convex function $f$ with $\varphi (E(w))=f(\operatorname{Re}w)$, which is the asserted formula.

Conversely, let $f$ be finite and convex. Then $f(\operatorname{Re}w)$ is plurisubharmonic on the tube by proposition 21.2.8. Near each point of $\Omega _ V$, the map $E$ has the holomorphic local inverse $z\mapsto (2\log z_1,\ldots ,2\log z_ n)$, defined with local branches of the logarithm, and $f(\log |z_1|^2,\ldots ,\log |z_ n|^2)$ is the composition of $f(\operatorname{Re}w)$ with this inverse. By proposition 5.9.3 it is plurisubharmonic near that point, since it is finite. Plurisubharmonicity is a local property, so the function is plurisubharmonic on $\Omega _ V$, and it is $(S^1)^ n$-invariant because it depends only on $|z_1|,\ldots ,|z_ n|$. The domains $(\Delta ^*)^ n$ and $(\mathbb C^*)^ n$ are $\Omega _ V$ for $V=(-\infty ,0)^ n$ and $V=\mathbb R^ n$. $\square$