Pome

Theorem 16.28.27865AACD2F96A

Theorem 16.28.27.865AACD2F96A For every $a\in \mathbb C^ n$, $q\in C(K,\mathbb R)$ and $\delta >0$,

2AF7902848CF 16.28.28
\begin{equation} F_{\delta ,q}(a)=\min _{(\mu ,c)\in \mathcal J_ a(K)} \left(\int _ K q\, \mathrm{d}\mu +\delta c\right). \end{equation}

In particular, $\mathcal J_ a(K)$ is nonempty and

\[ \min _{(\mu ,c)\in \mathcal J_ a(K)}c=V_ K^*(a). \]

Proof. On the real Banach space $C(K)$, put $T(f)=F_{1,f}(a)$. By (16.28.25), $T$ is concave, since convex combinations of competitors are competitors. Comparison of obstacles and translation by constants show that $T$ is nondecreasing and satisfies

\[ T(f+t)=T(f)+t,\qquad |T(f)-T(g)|\leq \| f-g\| _\infty . \]

Fix $f_0\in C(K)$. The directional derivative of the convex Lipschitz function $-T$ at $f_0$ is finite and sublinear. By the Hahn–Banach theorem, there is a continuous linear functional dominated by it, and since difference quotients of a convex function are monotone, this functional supports $-T$ at $f_0$. Changing the sign, we obtain a continuous linear functional $\ell $ with

\[ T(f)\leq T(f_0)+\ell (f-f_0)\qquad (f\in C(K)). \]

Testing $f=f_0+th$ with $h\geq 0$ and $t>0$, and using monotonicity, gives $\ell (h)\geq 0$. Testing with $f=f_0\pm t$ gives $\ell (1)=1$. By the Riesz representation theorem, $\ell (f)=\int _ K f\, \mathrm{d}\mu $ for a Borel probability measure $\mu $ on $K$. Set $c=T(f_0)-\int _ K f_0\, \mathrm{d}\mu $. The supporting inequality at $f=0$ gives $c\geq T(0)=V_ K^*(a)\geq 0$. Together with (16.28.25), it shows that $(\mu ,c)\in \mathcal J_ a(K)$ and

\[ T(f_0)=\int _ K f_0\, \mathrm{d}\mu +c. \]

Conversely, applying (16.28.26) to $u=F_{1,f_0}$, which lies below $f_0$ quasi-everywhere on $K$, shows that $T(f_0)\leq \int _ Kf_0\, \mathrm{d}\mu +c$ for every pair in $\mathcal J_ a(K)$. This proves (16.28.28) for $\delta =1$, with the minimum attained. The identity $F_{\delta ,q}=\delta F_{1,q/\delta }$ gives the general case, and the zero obstacle gives the last assertion. $\square$

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