Example 3.4.158DF74A8828CF

Example 3.4.15.8DF74A8828CF Let $I\subset \mathbb R$ be an open interval. Every $u\in \mathrm{SH}(I)$ other than the constant $-\infty $ is a finite, continuous, convex function. Indeed, $u$ is locally integrable by proposition 3.4.8. Given $x\in I$, choose an open interval $V\ni x$ with $\overline V\subset I$ and apply theorem 3.4.14 with $\eta =\mathbf1_ V$. On $V$ we have $u=E_1*(\eta \mu _ u)+h$ with $h$ affine, and the potential $E_1*(\eta \mu _ u)(\xi )=\frac12\int |\xi -y|\, \eta (y)\, \mathrm d\mu _ u(y)$ is finite and Lipschitz in $\xi $, since $\bigl ||\xi -y|-|\xi '-y|\bigr |\leq |\xi -\xi '|$ and $\eta \mu _ u$ is a finite measure. Hence $u$ is finite and continuous.

In dimension one the submean inequality reads $u(x)\leq \frac12(u(x-r)+u(x+r))$, and by theorem 3.4.5 it holds whenever $[x-r,x+r]\subset I$. Thus $u$ satisfies the midpoint inequality $u\bigl ((a+b)/2\bigr )\leq \bigl (u(a)+u(b)\bigr )/2$ for all $a,b\in I$. By induction on $n$, it follows that $u((1-\lambda )a+\lambda b)\leq (1-\lambda )u(a)+\lambda u(b)$ for every dyadic number $\lambda =k2^{-n}\in [0,1]$, because a dyadic number of order $n+1$ is the midpoint of two adjacent dyadic numbers of order $n$. Since $u$ is continuous and the dyadic numbers are dense in $[0,1]$, the same inequality holds for every $\lambda \in [0,1]$.

Conversely, a finite convex function $f$ on $I$ is continuous. For $p<q$ in $I$ let $\sigma (p,q)=(f(q)-f(p))/(q-p)$ be the slope of the chord of $f$ over $[p,q]$. If $p<q<r$ lie in $I$, then $q=(1-\lambda )p+\lambda r$ with $\lambda =(q-p)/(r-p)\in (0,1)$, and convexity gives $f(q)\leq (1-\lambda )f(p)+\lambda f(r)$, that is, $f(q)-f(p)\leq \lambda (f(r)-f(p))$ and $f(r)-f(q)\geq (1-\lambda )(f(r)-f(p))$. Dividing by $q-p=\lambda (r-p)$ and by $r-q=(1-\lambda )(r-p)$, respectively, we obtain

Hence $\sigma (p,q)$ increases with $q$ for fixed $p$, and with $p$ for fixed $q$. Now let $[a,b]$ be a compact subinterval of $I$, choose $a',b'\in I$ with $a'<a$ and $b<b'$, and let $a\leq s<t\leq b$. Applying these monotonicity properties and the displayed inequality to the triples $(a',s,t)$ and $(s,t,b')$, we find

so $|f(t)-f(s)|\leq \max \{ |\sigma (a',a)|,|\sigma (b,b')|\} \, (t-s)$. Thus $f$ is Lipschitz on every compact subinterval of $I$, and in particular continuous. Since $f$ satisfies the midpoint inequality, which is the submean inequality in dimension one, it is subharmonic. Hence the subharmonic functions on $I$ other than $-\infty $ are exactly the finite convex functions.

Next let $D\subset \mathbb C=\mathbb R^2$ be a domain, and let $f$ be holomorphic on $D$ and not identically zero. Then $\log |f|$ is subharmonic, and

where $\operatorname{ord}_ a(f)$ is the order of the zero of $f$ at $a$. Indeed, the zeros of $f$ form a locally finite subset of $D$, and away from them $\log |f|$ is locally the real part of a holomorphic logarithm of $f$, hence harmonic. Near a zero $a$ of order $m$ we can write $f(z)=(z-a)^ mg(z)$ with $g$ holomorphic and nonvanishing, so that $\log |f(z)|=2\pi mE_2(z-a)+\log |g(z)|$, where $\log |g|$ is harmonic and $E_2(\cdot -a)$, the fundamental potential of $\delta _ a$, is subharmonic with $\Delta E_2(\cdot -a)=\delta _ a$ (proposition 3.4.11). Hence $\log |f|$ is subharmonic near every point, and therefore on $D$. Moreover, the right-hand side of the displayed formula is a positive Radon measure, as is $\Delta \log |f|$ (theorem 3.4.12(1)). The two measures agree on a neighborhood of every point of $D$ (proposition 2.5.3), and hence on $D$: cover $D$ by countably many such neighborhoods $V_ k$, and decompose a Borel set $E\subset D$ into the disjoint sets $E\cap V_ k\setminus (V_1\cup \dots \cup V_{k-1})$. For $\log |f|^2=2\log |f|$ the coefficient $2\pi $ becomes $4\pi $.