Let us first prove uniqueness. If $\overline{L}$ and $\overline{L}^{\prime}$ are two metrics on $L$, let $\varphi$ be the continuous function such that $\left\|{\cdot}\right\|^{\prime}=e^{-\varphi}\left\|{\cdot}\right\|$. Assuming that $\varepsilon$ is an isometry for these two metrics, one obtains the following equation

for any $x\in\mathrm{X}$. Since $\mathrm{X}$ is compact, $\varphi$ is bounded and this equation implies that $\left\|{\varphi}\right\|_{\infty}\leq\frac{1}{d}\left\|{\varphi}\right\|_{\infty}$. Since $d\geq 2$, one concludes that $\varphi\equiv 0$.

For the existence, one begins with any continuous metric $\overline{L}_{0}$ on $L$. Let us then consider the sequence of metrics $(\overline{L}_{n})$ on $L$ induced by the pull-backs on $L^{d}=\varepsilon f^{*}L$, $L^{d^{2}}=(\varepsilon f^{*})^{2}L$, etc., hence on $L$. Since $d\geq 2$, a similar contraction argument as the one used for uniqueness shows that this is a Cauchy sequence of metrics on $L$ ; consequently, it converges to a continuous metric on $L$. If $\overline{L}_{0}$ is chosen to be semi-positive, which we may if $L$ is ample, then all al of the metrized line bundles $\overline{L}_{n}$ are semi-positive, hence the canonical metric is semi-positive.

Concretely, in the non-archimedean case, one begins with a model $(\mathfrak{X}_{0},\mathfrak{L}_{0},e)$ such that $\mathfrak{L}_{0}$ is numerically effective. Then one considers the map $f\colon X\rightarrow\mathfrak{X}_{0}$ and the normalization $\mathfrak{X}_{1}$ of $\mathfrak{X}_{0}$ in $X$ ; this is a projective model $\mathfrak{X}_{1}$, equiped with a finite morphism $f_{1}\colon\mathfrak{X}_{1}\rightarrow\mathfrak{X}_{0}$ extending $f$. Moreover, $\mathfrak{L}_{1}=f_{1}^{*}\mathfrak{L}_{0}$ is a model of $f^{*}L^{e}$ which is identified with $L^{ed}$ via the fixed isomorphism $\varepsilon$. Iterating this construction defines a sequence $(\mathfrak{X}_{n},\mathfrak{L}_{n},ed^{n})$ of models of $(X,L)$, with finite morphisms $f_{n}\colon\mathfrak{X}_{n}\rightarrow\mathfrak{X}_{n-1}$ such that $f_{n}^{*}\mathfrak{L}_{n-1}=\mathfrak{L}_{n}$. The metric on $L$ defined by any of these models is semi-positive, hence so is their uniform limit. ∎