ScalingStacks

Proof. Let $\underline{s}\in\widehat{V}_{{\scriptscriptstyle\bullet}}(L,\phi)$ be an element in $\text{rad}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L,\phi))$, then by Proposition 3.1, one has

$$0=\vvvert\underline{s}\vvvert_{\phi;\mathrm{sp}}=\vvvert s\vvvert_{\phi},$$

so

$$\forall n\in\mathbb{N},\quad\lVert s_{n}\rVert_{n\phi}=0.$$

By the assumption, all componets $s_{n}$ are zero sections. So $\underline{s}=\underline{0}$, hence $\widehat{V}_{{\scriptscriptstyle\bullet}}(L,\phi)$ is semi-simple. Same arguments works for $\widehat{V}_{{\scriptscriptstyle\bullet}}(L|_{Y},\phi|_{Y})$.

Let $\underline{t}\in\text{rad}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\phi_{X|Y}))$. Then $t_{n}\in\text{rad}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\phi_{X|Y}))$ for every $n\in\mathbb{N}$. For any $m\in\mathbb{N}$, there exists $s_{nm}\in V_{nm}(L)$ such that $s_{nm}|_{Y}=t_{n}^{m}$ and

$$\lim_{\begin{subarray}{c}m\to\infty\end{subarray}}\lVert s_{nm}\rVert_{nm\phi}^{\frac{1}{m}}=0.$$

As $\lVert t_{n}^{m}\rVert_{nm\phi|_{Y}}\leq\lVert s_{nm}\rVert_{nm\phi}$, one has that

$$\forall n\in\mathbb{N},\quad\lVert t_{n}\rVert_{n\phi|_{Y}}=0$$

so $\underline{t}=\underline{0}$. Hence $\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\phi_{X|Y})$ is semi-simple. ∎