ScalingStacks

Proof. For any $s\in V_{1}(L)$, if $\lVert s\rVert_{\phi_{1}}\geq\lVert s\rVert_{\phi_{2}}$, let $x_{1}\in X^{\mathrm{an}}$ be a point where $\lVert s\rVert_{\phi_{1}}$ is attained. One has

$$\Big|\log\frac{\lVert s\rVert_{\phi_{1}}}{\lVert s\rVert_{\phi_{2}}}\Big|\leq\Big|\log\Big|\frac{s(x_{1})}{s(x_{1})}\Big|_{\widehat{\kappa}(x_{1})}\Big|\leq\dist(\phi_{1},\phi_{2}).$$

Otherwise, let $x_{2}\in X^{\mathrm{an}}$ be a point where $\lVert s\rVert_{\phi_{2}}$ is attained. Then

$$\Big|\log\frac{\lVert s\rVert_{\phi_{2}}}{\lVert s\rVert_{\phi_{1}}}\Big|\leq\Big|\log\Big|\frac{s(x_{2})}{s(x_{2})}\Big|_{\widehat{\kappa}(x_{2})}\Big|\leq\dist(\phi_{1},\phi_{2}).$$

Hence the desired inequality holds. ∎