In fact, let $\underline{s}=(s_{n})_{n\in\mathbb{N}}$ be an element of $V_{{\scriptscriptstyle\bullet}}(L)$ and $m\in\mathbb{N}$, let $n_{0}\in\mathbb{N}$ be the smallest integer for which $\vvvert\underline{s}\vvvert_{\phi}=\lVert s_{n_{0}}\rVert_{n_{0}\phi}$. By the ultrametricity of $\vvvert\mathord{\cdot}\vvvert_{\phi}$ and the power-multiplicativity of $\{\lVert\mathord{\cdot}\rVert_{n\phi}\}_{n\in\mathbb{N}}$, one has

$$\vvvert\underline{s}^{m}\vvvert_{\phi}\leqslant\lVert(s_{n_{0}})^{m}\rVert_{mn_{0}\phi}=(\lVert s_{n_{0}}\rVert_{n_{0}\phi})^{m}=\vvvert\underline{s}\vvvert_{\phi}^{m}.$$

By the choice of $n_{0}$, one has

$$\forall(j_{0},\dots,j_{l})\in\mathbb{N}^{l+1},\sum_{i\in\{0,\dots,l\}}i\cdot j_{i}=mn_{0},\quad\Big\lVert\prod_{i\in\{0,\dots,l\}}(s_{i})^{j_{i}}\Big\rVert_{mn_{0}\phi}\leq\lVert(s_{n0})^{m}\rVert_{mn_{0}\phi}$$

and the equality holds if and only if $(j_{0},\dots,j_{l})=(0,\dots,0,n_{0},0,\dots,0)$ where $n_{0}$ is on the $m$-th place, so by the definition of $\vvvert\mathord{\cdot}\vvvert$ and its ultra-metricity, one get

$$\vvvert\underline{s}^{m}\vvvert_{\phi}\geqslant\lVert(\underline{s}^{m})_{mn_{0}}\rVert_{mn_{0}\phi}\geq\lVert(s_{n0})^{m}\rVert_{mn_{0}\phi}=\vvvert\underline{s}\vvvert_{\phi}^{m},$$

hence there is an equality. ∎

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. ∎