ScalingStacks

Proof. For any $x\in X^{\mathrm{an}}$, let $e(x)\in L^{\mathrm{an}}(x)\setminus\{0\}$. Let $s,s^{\prime}\in V_{1}(L)$ and $\lambda,\lambda^{\prime}\in\hat{\kappa}(x)$ be elements such that

$$s(x)=\lambda\cdot e(x),\ \lVert e(x)\rVert_{X|x}=\lvert\lambda\rvert^{-1}\lVert s\rVert,$$

$$s^{\prime}(x)=\lambda^{\prime}\cdot e(x),\ \lVert e(x)\rVert_{X|x}=\lvert\lambda^{\prime}\rvert^{-1}\lVert s^{\prime}\rVert^{\prime}.$$

If $\lVert e(x)\rVert_{X|x}>\lVert e(x)\rVert^{\prime}_{X|x}$, one has

$$\begin{split}\dist(\lVert e(x)\rVert_{X|x},\lVert e(x)\rVert^{\prime}_{X|x})&=\Big|\log\frac{\lvert\lambda\rvert^{-1}\lVert s\rVert}{\lvert\lambda^{\prime}\rvert^{-1}\lVert s^{\prime}\rVert^{\prime}}\Big|\\ &\leq\Big|\log\frac{\lvert\lambda\rvert^{-1}\lVert s\rVert}{\lvert\lambda\rvert^{-1}\lVert s\rVert^{\prime}}\Big|\leq\dist(\lVert\mathord{\cdot}\rVert,\lVert\mathord{\cdot}\rVert^{\prime}).\end{split}$$

Otherwise, one has

$$\begin{split}\dist(\lVert e(x)\rVert_{X|x},\lVert e(x)\rVert^{\prime}_{X|x})&=\Big|\log\frac{\lvert\lambda\rvert^{-1}\lVert s\rVert}{\lvert\lambda^{\prime}\rvert^{-1}\lVert s^{\prime}\rVert^{\prime}}\Big|\\ &\leq\Big|\log\frac{\lvert\lambda^{\prime}\rvert^{-1}\lVert s^{\prime}\rVert}{\lvert\lambda^{\prime}\rvert^{-1}\lVert s^{\prime}\rVert^{\prime}}\Big|\leq\dist(\lVert\mathord{\cdot}\rVert,\lVert\mathord{\cdot}\rVert^{\prime}).\end{split}$$

Varying $x$ and taking the supremum, one gets the desired inequality. ∎