Let $\psi$ be a positive singular metric of $L$ on $X$. Then $\psi-h$ is a globally well defined $\omega$-psh function. It is u.s.c. hence bounded from above on $X$: we let $C_{\psi}$ denotes its maximum. Then $\psi-h-C_{\psi}\leq 0$ on $X$, hence $\psi-h\leq V_{X,\omega}^{*}+C_{\psi}$, which yields $\psi\leq h_{min}+C_{\psi}$. ∎

The core of the proof consists in showing that

$$V_{K,\omega}(x)=\sup\left\{\frac{1}{N}\log||s||_{Nh}(x)\,/\,N\geq 1,s\in\Gamma(X,L^{N})\text{ and }\sup_{K}||s||_{Nh}\leq 1\right\}.$$

Note that for any of the sections $s$ involved in the supremum, $\varphi:=N^{-1}\log||s||_{Nh}$ belongs to $PSH(X,\omega)$ and satisfies $\varphi\leq 0$ on $K$. Therefore $\varphi\leq V_{K,\omega}$.

Conversely fix $x_{0}\in X$ and $a<V_{K,\omega}(x_{0})$. Fix $\varphi\in PSH(X,\omega)$ such that $\sup_{K}\varphi\leq 0$ and $\varphi(x_{0})>a$. Regularizing $\varphi$ (see Appendix) and translating, we can assume $\varphi\in PSH(X,\omega)\cap{\mathcal{C}}^{\infty}(X)$, $\sup_{K}\varphi<0$ and $\varphi(x_{0})>a$. Fix $\varepsilon>0$. Let $B=B(x_{0},r)$ be a small ball on which $\varphi>a$. We choose $B$ so small that the oscillation of $h$ is smaller than $\varepsilon$ on $B$. Let $\chi$ be a test function with compact support in $B$ and such that $\chi\equiv 1$ in $B(x_{0},r/2)$. We can assume w.l.o.g. that $B\subset{\mathcal{U}}_{\alpha_{0}}$ for some $\alpha_{0}$ but $B\cap{\mathcal{U}}_{\beta}=\emptyset$ for all $\beta\neq\alpha_{0}$. This insures that $\chi$ is a smooth section of $L^{N}$ for all $N\geq 1$.

Let $\psi_{1}$ be a smooth positive metric of $L^{N_{1}}\otimes K_{X}^{*}$ on $X$ (this is possible if $N_{1}$ is chosen large enough since $L$ is positive). Let $\psi_{2}$ be a positive metric of $L^{N_{2}}$ on $X$ which is smooth in $X\setminus\{x_{0}\}$ and with Lelong number $\nu(\psi_{2},x_{0})\geq n=\dim_{\mathbb{C}}X$ (this is again possible if $N_{2}$ is large enough, since $L$ is ample). Observe that $\overline{\partial}\chi$ is a smooth $\overline{\partial}$-closed $(0,1)$-form with values in $L^{N}$ (for all $N\geq 1$). Alternatively it is a smooth $\overline{\partial}$-closed $(n,1)$-form with values in $L^{N}\otimes K_{X}^{*}$. Applying Hörmander’s $L^{2}$-estimates (see e.g. [15], chapter VIII) with weight $\psi_{N}:=(N-N_{1}-N_{2})(\varphi+h)+\psi_{1}+\psi_{2}$, we find a smooth section $f$ of $L^{N}$ such that $\overline{\partial}f=\overline{\partial}\chi$ and

$$\int_{X}|f|^{2}e^{-2(N-N_{1}-N_{2})(\varphi+h)-2\psi_{1}-2\psi_{2}}dV_{\omega}\leq C_{1}\int_{X}|\overline{\partial}\chi|^{2}e^{-2\psi_{N}}dV_{\omega}.$$

Note that $\overline{\partial}\chi$ has support in $B\setminus B(x_{0},r/2)$ where $\psi_{N}$ is smooth so that both integrals are finite. Since $\nu(\psi_{2},x_{0})\geq n$, this forces $f(x_{0})=0$. The second integral is actually bounded from above by $C_{2}e^{-2N(a-\varepsilon)}$, where $C_{2}$ is independent of $N$, since $-\varphi<-a$ on $B$ and the oscillation of $h$ is smaller than $\varepsilon$ on $B$. Therefore $s:=\chi-f\in\Gamma(X,L^{N})$ satisfies $s(x_{0})=1$ and

$$\int_{X}|s|^{2}e^{-2N(\varphi+h)}dV_{\omega}\leq C_{3}e^{-2N(a-\varepsilon)},$$

where $C_{3}$ is independent of $N$. Now $\varphi<0$ in a neighborhood of $K$, so the mean-value inequality applied to the subharmonic functions $|s_{\alpha}|^{2}$ yields for all $x$ in $K$,

	$\displaystyle|s|^{2}e^{-2Nh}(x)$	$\displaystyle\leq$	$\displaystyle C_{\delta}\int_{B(x,\delta)}|s|^{2}(y)e^{-2N[\varphi+h](y)}e^{2N[h(y)-h(x)+\varphi(y)]}d\lambda(y)$
		$\displaystyle\leq$	$\displaystyle C_{4}e^{-2N(a-\varepsilon)}$

if $\delta$ is so small that $|\sup_{B(x,\delta)}\varphi|>0$ is bigger than the oscillation of $h$ on $B(x,\delta)$. Therefore $S:=C_{4}^{-1/2}e^{N(a-\varepsilon)}s\in\Gamma(X,L^{N})$ satisfies $\sup_{K}||S||_{Nh}\leq 1$ and $N^{-1}\log||S||_{Nh}(x_{0})\geq a-\varepsilon-\frac{\log C_{4}}{2N}$. Letting $N\rightarrow+\infty$, $\varepsilon\rightarrow 0$ and $a\rightarrow V_{K,\omega}(x_{0})$ completes the proof of the equality.

To conclude observe that by rescaling one gets

	$\displaystyle-\log T_{\omega}(K)=\sup_{X}V_{K,\omega}$
		$\displaystyle=$	$\displaystyle\!\!\!\!\sup\left\{\frac{1}{N}\sup_{X}\log||S||_{Nh}\,/\,N\geq 1,S\in\Gamma(X,L^{N})\text{ and }\sup_{K}||S||_{Nh}=1\right\}$
		$\displaystyle=$	$\displaystyle\!\!\!\!\sup\left\{-\frac{1}{N}\sup_{K}\log||S||_{Nh}\,/\,N\geq 1,S\in\Gamma(X,L^{N})\text{ and }\sup_{X}||S||_{Nh}=1\right\}$
		$\displaystyle=$	$\displaystyle-\log T_{\omega}^{\prime}(K).$

∎

Let $K,K_{0}$ be as in the theorem. Observe that $K_{0}$ is a circled subset of $\partial\mathbb{B}^{n+1}$: if $z\in K_{0}$ then $e^{i\theta}z\in K_{0}$, $\forall\theta\in[0,2\pi]$. For such compacts, the polynomial hull $\widehat{K_{0}}$ coincides with the ”homogeneous polynomial hull”,

$$\widehat{K_{0}}^{h}:=\{x\in\mathbb{C}^{n+1}\,/\,|P(x)|\leq\sup_{F}|P|,\,\forall P\text{ homogeneous polynomial}\}.$$

Indeed one inclusion $\widehat{K_{0}}\subset\widehat{K_{0}}^{h}$ is clear, so assume $z_{0}\in\widehat{K_{0}}^{h}$. Let $P=\sum_{j=0}^{d}P_{j}$ be a polynomial of degree $d$ decomposed into its homogenous components. Observe that $P_{j}(x)=(2\pi)^{-1}\int_{0}^{2\pi}P(e^{i\theta}x)e^{-ij\theta}d\theta$. Therefore $\sup_{K_{0}}|P_{j}|\leq\sup_{K_{0}}|P|$ since $K_{0}$ is circled. Fix $t\in]0,1[$. Then

$$|P(tz_{0})|\leq\sum_{j=0}^{d}t^{j}|P_{j}(z_{0})|\leq\frac{1}{1-t}\sup_{K_{0}}|P|.$$

We infer $tz_{0}\in\widehat{K_{0}}$. Letting $t\rightarrow 1^{-}$ and using that $K_{0}$ is closed we get $z_{0}\in\widehat{K_{0}}$, whence $\widehat{K_{0}}=\widehat{K_{0}}^{h}$.

Fix now $z\in\mathbb{C}^{n+1}$ such that $||z||\leq T_{\omega}(K)$. Let $P$ be a homogeneous polynomial of degree $d$. Then

(2)

$$|P(z)|=||z||^{d}\left|P\left(\frac{z}{||z||}\right)\right|\leq T_{\omega}(K)^{d}\sup_{\partial\mathbb{B}^{n+1}}|P|$$

Now set $\psi(z)=d^{-1}\log|P(z)|-\log||z||$ and $\varphi=\psi-\sup_{K}\psi$. Then $\varphi\in PSH(X,\omega)$ with $\sup_{K}\varphi\leq 0$ hence $\varphi\leq V_{K,\omega}$. Therefore

$$T_{\omega}(K)^{d}\leq\exp(-d\sup_{\mathbb{C}\mathbb{P}^{n}}\varphi)=\frac{\sup_{K_{0}}|P|}{\sup_{\partial\mathbb{B}^{n+1}}|P|}.$$

Together with $(2)$ this yields $|P(z)|\leq\sup_{K_{0}}|P|$ hence $z\in\widehat{K_{0}}^{h}=\widehat{K_{0}}$. Thus $K_{0}$ contains the ball centered at the origin of radius $T_{\omega}(K)$.

Conversely since $T_{\omega}(K)=T_{\omega}^{\prime}(K)$ (theorem 4.1), one can find homogenous polynomials $P_{j}$ of degree $d_{j}$ such that $\sup_{\partial\mathbb{B}^{n+1}}|P_{j}|^{-1/d_{j}}\cdot\sup_{K_{0}}|P_{j}|^{1/d_{j}}\rightarrow T_{\omega}(K).$ Assume $r\mathbb{B}^{n+1}\subset\widehat{K_{0}}$. Then

$$r^{d_{j}}\sup_{\partial\mathbb{B}^{n+1}}|P_{j}|=\sup_{r\mathbb{B}^{n+1}}|P_{j}|\leq\sup_{K_{0}}|P_{j}|$$

yields $r\leq T_{\omega}(K)$. ∎