ScalingStacks

Proof. By assumption, there exists a familly of norms $\{\lVert\mathord{\cdot}\rVert_{n}\}_{n\in\mathbb{N}}$ such that uniformly for $x\in X^{\mathrm{an}}$,

$$\quad\lim_{\begin{subarray}{c}n\to\infty\end{subarray}}\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})(\ast)(x)=\lvert\ast\rvert_{\phi}(x),$$

so for any $\epsilon>0$, there exists $N_{0}\in\mathbb{N}$ such that for any $n\geq N_{0}$

$$\dist(n\phi,\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n}))\leq n\epsilon,$$

hence by Lemma 3.9 and 3.10,

$$\dist(\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n\phi}),\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})}))\leq n\epsilon.$$

By Proposition 3.11, one has $\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})}))=\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})$, so

$$\dist(\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n\phi}),\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n}))\leq n\epsilon,$$

and

$$\dist(\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n\phi}),\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n}))\leq\epsilon.$$

Taking limit for $n\to\infty$ and then for $\epsilon\to 0$, one has

$$\dist(\mathcal{P}(\vvvert\mathord{\cdot}\vvvert_{\phi},\phi)=0,$$

so the two metrics are equal. ∎