Let $K=\mathbb{Q}_p$ and $G$ a split reductive group over $K$ with split maximal torus $T$. Furthermore, $X$ is a smooth projective variety over $K$ with a free $G$-action. Let $U \subset X$ be a $T$-invariant affine subspace, i.e. we have $U=\mathrm{Spec}(A)$ with $A=K[T_0,\ldots, T_N]/\mathfrak{a}$ for some $N \in \mathbb{N}$ and radical ideal $ \mathfrak{a}\subset K[T_0,\ldots, T_N].$

Let $m \in \mathbb{N}$ and $c\in K$ with $\lvert c \rvert_K>1$. We define $$A_m:=K\langle c^{-m}T_0,\ldots, c^{-m}T_N \rangle/(\mathfrak{a})$$ and $U_m:=\mathrm{Sp}A_m$. Hence $A \subset A_m$ and $U \supset U_m$.

Then, $\mathcal{O}_X(U)$ is $K$-vector space equipped with a $T$-action. Hence, we have a decomposition into weight spaces $$ \mathcal{O}_X(U)= \bigoplus_{\lambda \in X^*(T)} \mathcal{O}_X(U)_\lambda. $$ Additionally, we equip $\mathcal{O}_X(U)$ with a norm $\lvert \, \, \rvert_m$ such that
$$ \lvert f \rvert_m =\sup_{x \in U_m} \{\lvert f(x)\rvert_K \} $$ for $f \in \mathcal{O}_X(U).$ Note that it is the supremum norm from $A_m$ and that $A$ is dense in $A_m$.

Let $f \in \mathcal{O}_X(U)$ and $f = \sum f_\lambda$ with $f_\lambda \in \mathcal{O}_X(U)_\lambda$. By the properties of the absolute value on $K$, we have $$ \lvert f \rvert_m \leq \sup_{\lambda \in X^*(T)} \vert f_\lambda \rvert_m. $$

But is it true that $$ \lvert f \rvert_m= \sup_{\lambda \in X^*(T)} \vert f_\lambda \rvert_m $$ holds?