Proof. Using continuity, we may restrict to the case $\underline {y} \in \mathbf {Z}^d_{\geq 0}$ . By writing $\underline {y}$ as a sum of $1_k$ ’s and applying the identity (1) inductively, we may write

$$ \begin{align*}\Delta_{\underline{y}} = \sum_j{a_j \mathrm{sh}_{\underline{z}_j}} \Delta_{i_j}\end{align*} $$

for some $a_j \in \mathbf {Z}$ , $\underline {z}_j \in \mathbf {Z}^d$ and $1 \leq i_j \leq d$ (the $i_j$ may repeat). Raising this to the n-th power, we obtain a writing of $\Delta _{\underline {y}}$ as a $\mathbf {Z}$ -linear combination up to shifts of $\Delta ^{\underline {m}}$ ’s where for each $\underline {m}$ we have $\sum {m_i}=n$ . For each such $\underline {m}$ , choose the maximal coordinate $m_i$ . For every $g:\mathbf {Z}_p \rightarrow M$ one has $\mathrm {val}_M^{\mathrm {op}}(\Delta ^{\underline {m}}(g))\geq \mathrm {val}_M^{\mathrm {op}}(\Delta _i^{m_i}(g))$ ; applying this observation to shifts of f and using that $m_i\geq \lceil n/d \rceil $ , this concludes the proof.

Proof. We follow the proof of Bojanic [Reference BojanicBoj74], see also Theorem 1.13 of [Reference Berger and RozensztajnBR22].

First, suppose f has an expansion of the form $f(\underline {x}) = \sum {a_{\underline {n}}}\binom {x}{\underline {n}}$ with $a_{\underline {n}} \rightarrow 0$ . Then f is continuous since it is the uniform limit of continuous functions. We now explain why do we have in this case the equality

$$ \begin{align*}\mathrm{val}_M^{\mathrm{op}}(f) = \inf_{\underline{n}\in \mathbf{Z}_{\geq 0}}\mathrm{val}_M(a_{\underline{n}}).\end{align*} $$

Indeed, for each $\underline {n}$ , the element $a_{\underline {n}}=\Delta ^{\underline {n}}(f)(\underline {0})$ is a sum of elements in $\mathrm {Im}(f)$ , which shows $\inf _{\underline {x}}\mathrm {val}_M(f(\underline {x})) \leq \inf _{\underline {n}}\mathrm {val}_M(a_{\underline {n}}).$ The inequality in other direction is clear given the existence of the expansion.

Conversely, suppose f is continuous. We set $a_{\underline {n}} = \Delta ^{\underline {n}}(f)(0)$ . It suffices to show that $a_{\underline {n}} \rightarrow 0$ ; indeed, with that given, the function f and the one given by $g(\underline {x}):=\sum _{\underline {n}}a_{\underline {n}}\binom {x}{\underline {n}}$ agree on $\mathbf {Z}_{\geq 0}^d$ , hence are equal.

Since $\mathbf {Z}_p^d$ is compact, the function f is uniformly continuous. Fix $s \geq 0$ ; then there exists $t \geq 0$ such that if $\mathrm {val}(\underline {x}-\underline {y}) \geq t$ then $\mathrm {val}_M(f(\underline {x})-f(\underline {y})) \geq s\mathrm {val}(p) + \mathrm {val}_{M}^{\mathrm {op}}(f)$ . The same statement, with the same t, holds for any $\Delta ^{\underline {n}}(f)$ . To conclude the proof, it suffices to show that if $|\underline {n}|_{\infty }$ is larger than $sp^t$ then $\mathrm {val}(a_{\underline {n}})\geq s\mathrm {val}(p)+\mathrm {val}_M^{\mathrm {op}}(f)$ .

Now given a function g we have

$$ \begin{align*}\Delta^{p^t1_k}(g)(\underline{x}) = g(\underline{x}+p^t1_i)-g(\underline{x}) + p\cdot y(\underline{x})\end{align*} $$

for some element $y(\underline {x})$ with $\mathrm {val}_M(y(\underline {x})) \geq \mathrm {val}_M^{\mathrm {op}}(g)$ . Taking g to be a function of the form $\Delta ^{\underline {n}}(f)$ , we obtain $\mathrm {val}(g(\underline {x}+p^t1_i)-g(\underline {x})) \geq s\mathrm {val}(p) + \mathrm {val}_{M}^{\mathrm {op}}(f)$ and so for every $\underline {n}$ and k we have the inequality

(2) $$ \begin{align} \text{ }\mathrm{val}_{M}^{\mathrm{op}}(\Delta^{p^t1_k}(\Delta^{\underline{n}}(f))) \geq \min(s\mathrm{val}(p) + \mathrm{val}_{M}^{\mathrm{op}}(f), \mathrm{val}(p) + \mathrm{val}_M^{\mathrm{op}}(\Delta^{\underline{n}}(f))). \end{align} $$

Now if $|\underline {n}|_{\infty }> sp^t$ then the k-th coordinate of $\underline {n}$ is larger than $sp^t$ for some k. Applying (2) consecutively to

$$ \begin{align*}\Delta^{\underline{n}-sp^t1_k}(f),\Delta^{\underline{n}-(s-1)p^t1_k}(f),...,\Delta^{\underline{n}-p^t 1_k}(f)\end{align*} $$

shows $\mathrm {val}_{M}^{\mathrm {op}}(\Delta ^{\underline {n}}(f)),$ hence also $a_{\underline {n}}(f)$ , has valuation at least $s\mathrm {val}(p) + \mathrm {val}_{M}^{\mathrm {op}}(f)$ . This finishes the proof.

Proof. To show 1 implies 2, we may assume $n=p^k$ . Write $f(\underline {x}) =\sum a_{\underline {n}} \binom {\underline {x}}{\underline {n}}$ (as we may according to 2.2), then

$$ \begin{align*}\Delta_i^{p^k}(f)(\underline{x}) = \sum a_{\underline{n}} \Delta_i^{p^k}(\binom{\underline{x}}{\underline{n}}) = \sum_{\underline{n}:n_i\geq p^k} a_{\underline{n}}\binom{\underline{x}}{\underline{n}-1_ip^k},\end{align*} $$

for which each term has $\mathrm {val}\geq p^\lambda \cdot p^k + \mu $ by assumption.

Conversely, choose i with $n_i = |\underline {n}|_\infty $ . Let $\underline {n}'$ be $\underline {n}$ with the i-th coordinate removed. Then

$$ \begin{align*}a_{\underline{n}}(f) = \Delta^{\underline{n}}(f)(\underline{0}) =(\Delta_i^{n_i}(\Delta^{\underline{n}'}(f))(\underline{0})).\end{align*} $$

As $\Delta ^{\underline {n}'}(f)$ is a $\mathbf {Z}$ -linear combination of shifts of f, we obtain that $a_{\underline {n}}(f)$ is a $\mathbf {Z}$ -linear combination of $\Delta _i^{n_i}(f)(t_j)$ for various $t_j$ . Each of these has valuation $\geq p^{\lfloor \log _p(n_i) \rfloor } + \mu =p^{\lfloor \log _p(|\underline {n}|_\infty ) \rfloor } + \mu ,$ as required.

Proof. For $\mathcal {C}^{\mathrm {an}\text {-}\lambda ,\mu }(\mathbf {Z}_p^d,M)$ the statement follows from Proposition 2.3. We now show the statement holds for $\mathcal {C}^{\lambda \text {-}\mathrm {an}}(\mathbf {Z}_p^d,M)$ . Suppose it were known for shifts by elements of $\mathbf {Z}^d$ . Let $\underline {z} \in \mathbf {Z}_p^d$ and let $f\in \mathcal {C}^{\lambda \text {-}\mathrm {an}}(\mathbf {Z}_p^d,M)$ . As f is uniformly continuous, we can find some m so that $\mathrm {val}_M(f(\underline {x}+\underline {y})-f(\underline {x})> 1$ if $\mathrm {val}(\underline {x}-\underline {y})> m$ . Choosing $\underline {z}_0 \in \mathbf {Z}^d$ with $\mathrm {val}(\underline {z}-\underline {z}_0)> m$ , we see that $\mathrm {val}_M^{\mathrm {op}}(\mathrm {sh}_{\underline {z}}(f)-\mathrm {sh}_{\underline {z}_0}(f))> 1$ and $\mathrm {sh}_{\underline {z}_0}(f) \in \mathcal {C}^{\lambda \text {-}\mathrm {an}}(\mathbf {Z}_p^d,M)$ which implies $\mathrm {sh}_{\underline {z}}(f) \in \mathcal {C}^{\lambda \text {-}\mathrm {an}}(\mathbf {Z}_p^d,M)$ . This argument also shows the action of $\mathbf {Z}_p^d$ on $\mathcal {C}^{\lambda \text {-}\mathrm {an}}(\mathbf {Z}_p^d,M)$ is continuous (provided we show it exists). We thus reduce to the case where $\underline {z} \in \mathbf {Z}^d$ , which allows us to reduce further to the case $\underline {z}=1_i$ for $1 \leq i \leq d$ . Finally, we have $a_{\underline {n}}(\mathrm {sh}_{1_i}(f))=a_{\underline {n}}(f)+a_{\underline {n}+1_i}(f)$ which shows $\mathrm {sh}_{1_i}(f) \in \mathcal {C}^{\lambda \text {-}\mathrm {an}}(\mathbf {Z}_p^d,M)$ .

It remains to show each element of $\mathbf {Z}_p^d$ acts as an isometry on $\mathcal {C}^{\lambda \text {-}\mathrm {an}}(\mathbf {Z}_p^d,M)$ . We may reduce again to showing that $\mathrm {sh}_{1_i}$ acts as an isometry. To show this, take $f \in \mathcal {C}^{\lambda \text {-}\mathrm {an}}(\mathbf {Z}_p^d,M)$ and let $\underline {n}_0$ be maximal with $\mathrm {val}_\lambda (f) = \mathrm {val}_M(a_{\underline {n}_0}(f))-\lfloor p^\lambda \underline {n}_0 \rfloor )$ . Then $\mathrm {val}_M(a_{\underline {n}_0+1_i}(f))> \mathrm {val}_M(a_{\underline {n}_0}(f))$ , and so the formula $a_{\underline {n}}(\mathrm {sh}_{1_i}(f))=a_{\underline {n}}(f)+a_{\underline {n}+1_i}(f)$ shows $\mathrm {val}_{\lambda }(\mathrm {sh}_{1_i}(f)) \leq \mathrm {val}_\lambda (f).$ Now by using the uniform continuity again, we know that for $p^k$ for k sufficiently large we have $\mathrm {val}_{\lambda }(\mathrm {sh}_{p^k 1_i}(f)) = \mathrm {val}_\lambda (f)$ . Hence the inequalities

$$ \begin{align*}\mathrm{val}_{\lambda}(f) = \mathrm{val}_{\lambda}(\mathrm{sh}_{p^k 1_i}(f)) \leq \mathrm{val}_{\lambda}(\mathrm{sh}_{(p^k-1) 1_i}(f)) \leq... \leq \mathrm{val}_{\lambda}(\mathrm{sh}_{1_i}(f)) \leq \mathrm{val}_{\lambda}(f)\end{align*} $$

are all forced to be equalities, which shows $\mathrm {val}_{\lambda }(\mathrm {sh}_{1_i}(f)) = \mathrm {val}_\lambda (f),$ as required.

Proof. Let $l\geq 0$ be arbitrary. We have the Vandermonde identity

$$ \begin{align*}\Delta_{p^l 1_i}(\binom{\underline{x}}{\underline{n}}) = \sum_{j=1}^{n_i}\binom{\underline{x}}{\underline{n}-j1_i}\binom{p^l}{j}.\end{align*} $$

Writing $f(\underline {x}) = \sum _{\underline {n}} a_{\underline {n}} \binom {\underline {x}}{\underline {n}},$ the identity above shows that

$$ \begin{align*}\Delta_{p^l 1_i}(f) = \sum_{\underline{n}} (\sum_{j=1}^\infty a_{\underline{n} +j 1_i} \binom{p^l}{j}) \binom{\underline{x}}{\underline{n}},\end{align*} $$

and hence

$$ \begin{align*}\Delta^{\underline{n}}(\Delta_{p^l 1_i}(f)(0)) = \sum_{j=1}^\infty a_{\underline{n}+j1_i} \binom{p^l}{j}.\end{align*} $$

From this we get the lower bound

$$ \begin{align*}\mathrm{val}_M(\Delta^{\underline{n}}(\Delta_{p^l 1_i}(f)(0))) \geq \inf_{j\geq 1}(\mathrm{val}(a_{\underline{n}+j1_i}) + \mathrm{val}(\binom{p^l}{j})).\end{align*} $$

On the other hand, by definition we have the inequality

$$ \begin{align*}\mathrm{val}_M(a_{\underline{n}+j1_i}) \geq \sum_{i=1}^d\lfloor p^{\lambda}(n_i + j\delta_{ij})\rfloor + \mathrm{val}_{\lambda}(f).\end{align*} $$

Putting this all together, we get

$$ \begin{align*}\mathrm{val}_{\lambda}(\Delta_{p^l 1_i}(f)) &= \inf_{\underline{n}}(\mathrm{val}_M(\Delta^{\underline{n}}(\Delta_{p^l 1_i}(f)(0))) -\lfloor p^{\lambda} \underline{n} \rfloor)\\&\geq \mathrm{val}_{\lambda}(f)+ \inf_{j\geq 1}(p^\lambda j -1 + \mathrm{val}(\binom{p^l}{j})),\end{align*} $$

and so any choice of l which satisfies $\inf _{j\geq 1}(p^\lambda j -1 + \mathrm {val}(\binom {p^l}{j}))> c$ works.

Proof. Choose any $c> p^{\lambda '} + 1$ and $l\geq 0$ so that the previous lemma applies to c and $\lambda $ . Let $f \in \mathcal {C}^{\lambda \text {-an}}(\mathbf {Z}_p^d,M)$ a function. We wish to show its restriction lies in $\mathcal {C}^{\lambda '\text {-an}}(p^l\mathbf {Z}_p^d,M)$ . We may assume f lies in $\mathcal {C}^{\lambda \text {-an}}(\mathbf {Z}_p^d,M)^+.$ Let $\Delta _{p^l}^{\underline {m}}$ denote the composition $\Delta ^{m_1}_{p^l 1_1} \circ ... \circ \Delta ^{m_d}_{p^l 1_d}.$ We need to show that

$$ \begin{align*}\mathrm{val}(\Delta_{p^l}^{\underline{m}}(f)(0)) - p^{\lambda'}|\underline{m}| \rightarrow \infty.\end{align*} $$

Applying the previous lemma successively to $f \in \mathcal {C}^{\lambda \text {-an}}(\mathbf {Z}_p^d,M)^+$ we get the inequality

$$ \begin{align*}\mathrm{val}_\lambda(\Delta_{p^l}^{\underline{m}}(f)) \geq |\underline{m}|c.\end{align*} $$

By the general inequality $\mathrm {val}_M(g(0)) \geq \mathrm {val}_\lambda (g)$ we deduce

$$ \begin{align*}\mathrm{val}_M(\Delta_{p^l}^{\underline{m}}(f)(0)) \geq |\underline{m}|c \geq p^{\lambda'} |\underline{m}| + |\underline{m}|\end{align*} $$

which gives the desired lower bound for the $\mathrm {val}_M(\Delta _{p^l}^{\underline {m}}(f)(0))$ .

Proof. The cofinality of $\{\mathcal {C}^{\mathrm {an}\text {-}\lambda ,\mu }(\mathbf {Z}_p^d,M)\}_{\lambda ,\mu }$ and $\{\mathcal {C}^{\lambda \text {-}\mathrm {an}}(\mathbf {Z}_p^d,M)\}_\lambda $ follows from the inequality $|\underline {n}|_\infty \leq |\underline {n}| \leq |\underline {n}|_\infty /d$ for $\underline {n} \in \mathbf {Z}^d_{\geq 0}$ by condition 1 of Proposition 2.3. The cofinality with respect to a different basis follows from Lemma 2.1 and condition 2 of Proposition 2.3.

Proof. Let $f \in \mathcal {C}^{\mathrm {an}\text {-}\lambda ,\mu }(G_0,M)$ . By Lemma 2.1 and condition 2 of Proposition 2.3 we have that f restricts to $ \mathcal {C}^{\mathrm {an}\text {-}\lambda ',\mu '}(G_0',M)$ for some $\lambda ',\mu '$ , from which the lemma follows.

Proof. This is similar to argument of Lemma 2.5 of [Reference Berger and ColmezBC16] which we produce here for the convenience of the reader. We may assume $G=G_0$ is a uniform subgroup and identify it analytically with $\mathbf {Z}_p^d$ . Recall that locally analytic functions are those for which the valuation of $a_{\underline {n}}$ grows at least linearly with $|\underline {n}|.$ For (i), note that for $\underline {n},\underline {m}\in \mathbf {Z}^d_{\geq 0}$ we have

(3) $$ \begin{align} \binom{\underline{x}}{\underline{n}}\cdot\binom{\underline{x}}{\underline{m}}=\sum_{\underline{k}\leq\underline{n}+\underline{m}}c_{\underline{k}}\binom{\underline{x}}{\underline{k}} \end{align} $$

for some $c_{\underline {k}} \in \mathbf {Z}.$ This proves $R^{\mathrm {la}}$ is a ring. For (ii), write $\mathrm {orb}_r(\underline {x})= r+\sum _{\underline {n} \neq 0}a_{\underline {n}}\binom {\underline {x}}{\underline {n}}$ . We have the identity

$$ \begin{align*}\mathrm{orb}_{1/r}(\underline{x})=\frac{1}{r}\sum_{j \geq 0}(-1)^j(\sum_{\underline{n}\neq 0}a_{\underline{n}}\binom{\underline{x}}{\underline{n}}/r)^j,\end{align*} $$

which is still locally analytic.

Proof. Again, we reproduce the argument of 2.3 of [Reference Berger and ColmezBC16] for the convenience of the reader. Let $a_{ij}(g)$ be the coordinates of $g\mapsto \mathrm {Mat}(g)$ . Then $g(m_i)=\sum _{j}a_{ij}(g)m_j,$ so each $m_i$ is locally analytic. This proves the inclusion $\oplus _i R^{\mathrm {la}} \cdot m_i \subset M^{\mathrm {la}}$ . Conversely, let $m \in M^{\mathrm {la}}$ . Write $m=\sum _i r_i m_i$ with $r_i \in R.$ By assumption, we may write $g(m) = \sum _i f_i(g)m_i$ with each $f_i$ locally analytic. Applying g to $m = \sum _i r_i m_i$ , we get the identity $g(r_i) = \sum _{j}b_{ij}(g)f_j(g),$ where $b_{ij}(g)$ are the coordinates of $\mathrm {Mat}(g)^{-t}$ . These are polynomials in the $a_{ij}(g)$ multiplied by $\det (g)$ , hence are locally analytic by the previous lemma.

Proof. As $\Gamma \cong \mathbf {Z}_p$ we know that $\mathrm {H}^{i}(G,M) = \mathrm {H}^{i}(G,M^{\mathrm {la}}) = 0$ for $i \geq 2$ (see V, 2.2.3.3 of [Reference LazardLaz65]), while the equality for $i=0$ is obvious. So the only nontrivial case is $i=1$ . Choosing $\gamma $ for a generator of G, we have an exact sequence

$$ \begin{align*}0\rightarrow M/\mathrm{H}^{0}(G,M) \xrightarrow{\gamma-1} M \rightarrow \mathrm{H}^{1}(G,M)\rightarrow 0.\end{align*} $$

Taking locally analytic vectors (regarding $\mathrm {H}^{0}(G,M)$ and $\mathrm {H}^{1}(G,M)$ as having a trivial G-action) we obtain an exact sequence

$$ \begin{align*}0\rightarrow (M/\mathrm{H}^{0}(G,M))^{\mathrm{la}} \xrightarrow{\gamma-1} M^{\mathrm{la}} \rightarrow \mathrm{H}^{1}(G,M)\rightarrow \mathrm{R}_{\mathrm{la}}^1(M/\mathrm{H}^{0}(G,M)).\end{align*} $$

First, we claim $\mathrm {R}_{\mathrm {la}}^1(M/\mathrm {H}^{0}(G,M))=0$ . This is because there is an exact sequence

$$ \begin{align*}\mathrm{R}_{\mathrm{la}}^1(M) \rightarrow \mathrm{R}_{\mathrm{la}}^1(M/\mathrm{H}^{0}(G,M)) \rightarrow \mathrm{R}_{\mathrm{la}}^2(\mathrm{H}^{0}(G,M))\end{align*} $$

and both of its outer terms vanish (the $\mathrm {R}_{\mathrm {la}}^1$ term because of the theorem, the $\mathrm {R}_{\mathrm {la}}^2$ term because $\Gamma \cong \mathbf {Z}_p$ ).

On the other hand, we claim that $(M/\mathrm {H}^{0}(G,M))^{\mathrm {la}} = M^{\mathrm {la}}/\mathrm {H}^{0}(G,M).$ This reduces to showing $\mathrm {R}^1_{\mathrm {la}}(\mathrm {H}^{0}(G,M)) = 0$ ; for this vanishing we need to show that $\gamma -1$ acts surjectively on $\mathcal {C}^{\mathrm {la}}(\Gamma ,\mathrm {H}^{0}(G,M))$ . Now if $\sum _n a_n \binom {x}{n} \in \mathcal {C}^{\mathrm {la}}(\Gamma ,\mathrm {H}^{0}(G,M))$ with the coordinate x corresponding to $\gamma $ , then $(\gamma -1)(\sum _n a_n \binom {x}{n}) = \sum _n a_n \binom {x}{n-1}$ , from which the surjectivity is clearly seen.

Putting the two claims together, we obtain an exact sequence

$$ \begin{align*}0\rightarrow M^{\mathrm{la}}/\mathrm{H}^{0}(G,M) \xrightarrow{\gamma-1} M^{\mathrm{la}} \rightarrow \mathrm{H}^{1}(G,M)\rightarrow 0\end{align*} $$

which shows $\mathrm {H}^{0}(G,M)$ is the cokernel of $M^{\mathrm {la}}\xrightarrow {\gamma -1} M^{\mathrm {la}}$ , that is to say naturally isomorphic to $\mathrm {H}^{1}(G,M^{\mathrm {la}})$ , as required.

Proof. Suppose $\{e_i\}_{1\leq i \leq n}$ and $\{f_j\}_{1\leq j \leq m}$ are respective bases of $N^+$ and $M^+$ such that $\mathrm {val}(\mathrm {Mat}_{\{e_i\}}(\gamma )-1),\mathrm {val}(\mathrm {Mat}_{\{f_j\}}(\gamma )-1)> c_1+2c_2+2c_3$ for $\gamma \in G'.$ Using the identity

$$ \begin{align*}(\gamma-1)(e_i\otimes f_j)-e_i\otimes f_j=(\gamma-1)(e_i)\otimes\gamma(f_j)+e_i\otimes (\gamma-1)(f_j),\end{align*} $$

we see that $\{e_i \otimes f_j\}_{1\leq i \leq n, 1\leq j \leq m}$ is a basis of $N^+ \otimes _{\widetilde {\Lambda }^+} M^+$ with

$$ \begin{align*}\mathrm{val}(\mathrm{Mat}(\gamma)-1)> c_1+2c_2+2c_3.\end{align*} $$

Hence, $N^+ \otimes _{\widetilde {\Lambda }^+} M^+$ lies in $\mathrm {Mod}^{G_0}_{\widetilde {\Lambda }^+}(G')$ . Next, we show that

$$ \begin{align*}D^+_{H',n}(M^+) \otimes_{\Lambda^+_{H',n}} D^+_{H',n}(M^+) =D^+_{H',n}(M^+ \otimes_{\widetilde{\Lambda}^+} N^+).\end{align*} $$

Both the left- and right-hand side are objects of $\mathrm {Mod}^{G_0}_{\Lambda _{H',n}^+}(G')$ , so by virtue of Proposition 3.7 it is enough to show they are equal after tensoring with $\widetilde {\Lambda }^+$ . To show this, we compute

$$ \begin{align*} \widetilde{\Lambda}^+\otimes_{\Lambda^+_{H',n}}(D^+_{H',n}(M^+) \otimes_{\Lambda^+_{H',n}} D^+_{H',n}(M^+)) &= M^+ \otimes_{\Lambda^+_{H',n}} D^+_{H',n}(M^+) \\ &=M^+ \otimes_{\widetilde{\Lambda}^+}(\widetilde{\Lambda}^+\otimes_{\Lambda^+_{H',n}} D^+_{H',n}(M^+))\\ &= M^+ \otimes_{\widetilde{\Lambda}^+}N^+\\ &= \widetilde{\Lambda}^+\otimes_{\Lambda^+_{H',n}}D^+_{H',n}(M^+ \otimes_{\widetilde{\Lambda}^+} N^+). \end{align*} $$

This completes the proof.

Proof. We know that $D_{H',n}^+(M)^+$ has a basis of $c_3$ -fixed elements. Using the identity

$$ \begin{align*}(\gamma-1)(ab)=(\gamma-1)(a)b + \gamma(a)(\gamma-1)(b)\end{align*} $$

we see that $(\gamma -1)(D_{H',n}^+(M)^+) \subset \varpi ^{s}D_{H',n}^+(M)^+$ for $s = \mathrm {min}(c_3,t)$ where $t = t(G',n)$ is as in (TS4). If $x \in \varpi ^n\cdot D_{H',n}^+(M)^+$ then writing $x = \varpi ^n y$ and using the same identity as before we conclude that $(\gamma -1)(x) \in \varpi ^{n+s}\cdot D_{H',n}^+(M)^+.$ This concludes the proof.

Proof. Use the expansion $\gamma ^{x}(m)=\sum _{n\geq 0}(\gamma -1)^{n}(m)\binom {x}{n}$ for $x\in \mathbf {Z}_p$ .

Proof of part 1 of Theorem 3.1.

For $G'$ small enough and n large enough the descent to $D_{H',n}$ applies (this can always be achieved because of continuity). We need to show that the natural map

$$ \begin{align*}\widetilde{\Lambda} \otimes_{{\Lambda}^{\mathrm{la}}} M^{\mathrm{la}} \rightarrow M\end{align*} $$

is an isomorphism. We shall that show this map is an isomorphism for $G'$ small enough when $M^{\mathrm {la}}$ is replaced with $M^{G'\text {-la}}$ , which suffices (the transition maps along different $G'$ ’s are then forced to be isomorphisms). By Corollary 3.10 we have $D_{H',n} \subset M^{G'\text {-la}}$ and we know M descends to $D_{H',n}$ , so the $\Lambda _{H',n}$ -basis of $D_{H',n}$ gives a $\widetilde {\Lambda }$ -basis of M consisting of $G'$ -locally analytic elements. By Proposition 2.16 we have $M^{G' \text {-la}} = \widetilde {\Lambda }^{G'\text {-la}} \otimes _{\Lambda _{H',n}} D_{H',n}$ and so the natural isomorphism $\widetilde {\Lambda } \otimes _{\Lambda _{H',n}} D_{H',n} \xrightarrow {\cong } M$ is identified with the map $\widetilde {\Lambda } \otimes _{{\Lambda }^{G'\text {-la}}} M^{G'\text {-la}} \rightarrow M$ , which concludes the proof.

Warning 3.12. The functoriality of c-smallness does not coincide with that of $(G_0, \lambda ) \mapsto \mathcal {C}^{\lambda \text {-an}}(G_0,M)$ . Namely, if the pair $(G_0, \lambda )$ maps to the pair $(G_0', \lambda ')$ in the direct limit defining the $\mathrm {R}^{i}_{\mathrm {la}}$ then it is not true in general that $(G_0', \lambda ')$ is still c-small. This is because $(G_0, \lambda )$ maps to $(G_0', \lambda ')$ when $G_0' \subset G_0$ and $\lambda ' \leq \lambda $ , but the c-smallness of $(G_0', \lambda ')$ is only guaranteed when $\lambda ' \geq \lambda $ .

Proof. Each basis element of each $M_{\underline {n},\lambda }$ is of the form $m_i\varpi ^{\lfloor p^{\lambda }\underline {k} \rfloor }\binom {\underline {x}}{\underline {k}}$ for some $m_i$ for $1 \leq i \leq \mathrm {rank}(M)$ and $\underline {k} \leq {\underline {n}}.$ Using the identity

$$ \begin{align*}(g-1)(abc) = (g-1)(a)bc + a(g-1)(b)g(c) + g(a)g(b)(g-1)(c),\end{align*} $$

one reduces to showing each of the valuations of $(g-1)(m_i)$ , $(g-1)(\varpi ^{\lfloor p^{\lambda }\underline {k} \rfloor })$ and $(g-1)(\binom {\underline {x}}{\underline {k}})$ are $> c$ .

First, the valuation of the $(g-1)(m_i)$ is $> c$ by the c-smallness of $G_0$ . Second, the valuation of the term $(g-1)(\varpi ^{\lfloor p^{\lambda }\underline {k} \rfloor })$ is $\geq \mathrm {val}((g-1)(\varpi ))> c$ by virtue of the identity $(g-1)(\varpi ^n) = (g-1)(\varpi ^{n-1})g(\varpi )+(g-1)(\varpi )\varpi .$

Finally, we need to estimate $(g-1)(\binom {\underline {x}}{\underline {k}})$ . First, by using the identity

$$ \begin{align*}(gh-1)(a) = g((h-1)(a)) + (g-1)(a),\end{align*} $$

we may reduce to the case where g is a basis element, and hence its action is given by $\mathrm {sh}_{p^l1_i}$ for some $1 \leq i \leq d$ . We then have the Vandermonde identity

$$ \begin{align*}(\mathrm{sh}_{p^l 1_i}-1)(\binom{\underline{x}}{\underline{k}}) = \sum_{j=1}^{k_i}\binom{\underline{x}}{\underline{k}-j1_i}\binom{p^l}{j}.\end{align*} $$

We have $\mathrm {val}_{ \lambda }(\binom {\underline {x}}{\underline {k}}) = -\lfloor p^\lambda \underline {k} \rfloor $ and $\mathrm {val}_{ \lambda }(\binom {\underline {x}}{\underline {k}-j1_i} = -\lfloor p^\lambda (\underline {k}-j1_i) \rfloor $ , so it suffices to show

$$ \begin{align*}-\lfloor p^\lambda(\underline{k}-j1_i) \rfloor + \mathrm{val}(\binom{p^l}{j})> -\lfloor p^\lambda\underline{k} \rfloor+c\end{align*} $$

for every $j \leq k_i$ . This inequality follows in turn from the simpler inequality

$$ \begin{align*}\lfloor p^\lambda j \rfloor + \mathrm{val}(\binom{p^l}{j})> c.\end{align*} $$

If $(G_0,\lambda )$ is c-small then $p^\lambda> c + 1$ and the inequality holds for all $j \geq 1$ and $l=0$ . Otherwise, it is clear one can choose l large enough so that the inequality holds for all $j \geq 1$ .

Proof. Given $\underline {k} \in \mathbf {Z}^d_{\geq 0}$ let ${\underline {k}}_{\Gamma _0}$ denote the first coordinate of $\underline {k}$ (corresponding to $\Gamma _0$ ) and let ${\underline {k}}_{H_0}$ denote $\underline {k}$ without the first coordinate (corresponding to $H_0$ ). The description of the $\widetilde {\Lambda }^+$ -basis of $M^+_{\underline {k},\lambda }$ shows that there is an isomorphism

$$ \begin{align*}M^+_{\underline{k},\lambda} \cong M^+ \otimes_{\widetilde{\Lambda}^+} \mathcal{C}^{\lambda\text{-an}}(H_0,\widetilde{\Lambda}^+)^{\Delta^{\underline{k}_{H_0}}=0}\otimes_{\widetilde{\Lambda}^+} \mathcal{C}^{\lambda\text{-an}}( \Gamma_0,\widetilde{\Lambda}^+)^{\Delta^{\underline{k}_{\Gamma_0}}=0},\end{align*} $$

which can be rewritten as

$$ \begin{align*}M^+_{\underline{k},\lambda} \cong\mathcal{C}^{\lambda\text{-an}}(H_0,M^+)^{\Delta^{\underline{k}_{H_0}}=0}\otimes_{\widetilde{\Lambda}^+} \mathcal{C}^{\lambda\text{-an}}( \Gamma_0,\widetilde{\Lambda}^+)^{\Delta^{\underline{k}_{\Gamma_0}}=0}.\end{align*} $$

Now one observes that $\mathcal {C}^{\lambda \text {-an}}( \Gamma _0,\widetilde {\Lambda }^+)^{\Delta ^{\underline {k}_{\Gamma _0}}=0}$ has a basis given by $\varpi ^{\lfloor p^\lambda m \rfloor }\binom {x}{m}$ for $0 \leq m < \underline {k}_{\Gamma _0}.$ It is c-fixed for the $\Gamma _0$ -action because $(G_0,\lambda )$ is c-small (this is the same argument appearing in Lemma 3.13). Hence by Proposition 4.10 of [Reference PoratPor22b] there is a natural isomorphism

$$ \begin{align*}D_{H_0,n}^+(\mathcal{C}^{\lambda\text{-an}}( \Gamma_0,\widetilde{\Lambda}^+)^{\Delta^{\underline{k}_{\Gamma_0}}=0}) \cong \mathcal{C}^{\lambda\text{-an}}( \Gamma_0,\Lambda_{H_0,n}^+)^{\Delta^{\underline{k}_{\Gamma_0}}=0}.\end{align*} $$

We then conclude by applying Lemma 3.8, taking the limit over $\underline {k}$ and taking the completion.

Proof. Inverting $\varpi $ in the isomorphism (5), we have natural isomorphisms

$$ \begin{align*}\widetilde{\Lambda} \otimes_{\Lambda_{H_0,n}} D^{\lambda}_{H_0,n}(M) \cong \mathcal{C}^{\lambda\text{-an}}(G_0,M).\end{align*} $$

By Proposition 5.8 of [Reference PoratPor24], the cohomology $\mathrm {H}^i(H_0,\mathcal {C}^{\lambda \text {-an}}(G_0,M))$ is zero for $i \ge 1$ . Note that in that setting one assumes that the valuation is p-adic, but the same proof works for a $\varpi $ -adic valuation defined by an arbitrary topologically nilpotent unit $\varpi $ . We conclude the proof by applying the Hochschild-Serre spectral sequence (see [Reference KedlayaKed16, Lemma 3.3] for a version which applies in our setting).

Proof. In the above decomposition of the $\Gamma _0$ -cohomology of

$$ \begin{align*}\widetilde{\Lambda}^{H_0} \otimes_{\Lambda_{H_0,n}} D^{\lambda}_{H_0,n}(M) \cong \mathcal{C}^{\lambda\text{-an}}(G_0,M)^{H_0}\end{align*} $$

the left-hand side does not depend on n. Hence, the same holds for the right-hand side. Therefore, it suffices to show that

$$ \begin{align*}\varinjlim_{n}\mathrm{H}^1(\Gamma_0,X_{H_0,n}\otimes_{\Lambda_{H_0,n}} D^{\lambda}_{H_0,n}) = 0.\end{align*} $$

Write $\gamma $ for a generator of $\Gamma _0$ . It suffices to show that every

$$ \begin{align*}a\otimes b \in X_{H_0,n} \otimes_{\Lambda_{H_0,n}} D_{H_0,n}^{\lambda}(M)\end{align*} $$

is in the image of $\gamma -1$ for some $X_{H_0,n'} \otimes _{\Lambda _{H_0,n'}} D_{H_0,n'}^{\lambda }(M)$ , with $n'$ depending only on n.

By Proposition 3.9, we know that there exists some $s> 0$ such that

$$ \begin{align*}(\gamma-1)(D_{H_0,n}^{\lambda,+}(M^+)) \subset \varpi^s\cdot D_{H_0,n}^{\lambda,+}(M^+).\end{align*} $$

For some power $\gamma '$ of $\gamma $ we therefore have

$$ \begin{align*}(\gamma'-1)(D_{H_0,n}^{\lambda,+}(M^+)) \subset \varpi^{2c_2}\cdot D_{H_0,n}^{\lambda,+}(M^+).\end{align*} $$

Take $n' \geq n$ so that $n(\gamma ') \leq n'$ . For such $n'$ , we know that $\mathrm {val}{(\gamma '-1)(x)}\geq \mathrm {val}(x)-c_2$ for $x \in X_{H_0,n'}$ .

Let $\alpha $ be such that $(\gamma ^{\prime -1}-1)(\alpha ) = a$ . Consider the series

$$ \begin{align*}y = \sum_{i=0}^\infty (\gamma^{\prime-1}-1)^{-i}(\alpha) \otimes (\gamma'-1)^{i}(b).\end{align*} $$

The series converges because $\mathrm {val}(\gamma ^{\prime -1}-1)^{-1}(x) \geq \mathrm {val}(x) -c_3$ while $\mathrm {val}((\gamma '-1)(x))\geq \mathrm {val}(x) +2c_3$ by our choice of $n'$ . A direct computation shows that $(\gamma '-1)(y) = a \otimes b$ . As $\gamma '-1$ is divisible by $\gamma -1$ , this proves $a \otimes b$ is in the image of $\gamma -1$ , as required.

Proof. As always, we fix some $c> c_1 +2c_2 + 2c_3$ . We claim that c-small pairs $(G_0,\lambda )$ are cofinal in the above direct limit. This is a consequence of Corollary 2.7 together with the observation that if $G_0$ is c-small, so is $G_l$ . In particular, we have seen that $\mathrm {H}^i(G_0,\mathcal {C}^{\lambda \text {-an}}(G_0,M) = 0$ for $i\geq 2$ and c-small pairs $(G_0,\lambda )$ , so it automatically follows that $\mathrm {R}^i_{\mathrm {la}}(M) = 0$ for $i \geq 2$ . The nontrivial part is showing the vanishing of $\mathrm {R}^1_{\mathrm {la}}(M).$ Using the cofinality of c-small pairs, this is a consequence of the following proposition.

Proof. Recall that in 3.17 we have shown an isomorphism for $n \geq n(M,\lambda )$

$$ \begin{align*}\mathrm{H}^1(G_0,\mathcal{C}^{\lambda \text{-an}}(G_0,M)) \cong \mathrm{H}^1(\Gamma_0,D_{H_0,n}(M)),\end{align*} $$

and so $f \in \mathrm {H}^1(\Gamma _0,D_{H_0,n}(M)).$ Using $\mathrm {H}^1(\Gamma _0,D_{H_0,n}(M)) \cong D_{H_0,n}(M)/(\gamma -1),$ we may think of f as an element of $D_{H_0,n}(M)$ . Let $s = s(G_0,n)$ be the constant appearing in Proposition 3.9, and choose $\lambda '$ small enough that $p^{\lambda '} < s$ . Choose $l \geq 0$ large enough so that 1 of Lemma 3.13 applies. We are going to show that f is mapped to $0$ in $\mathrm {H}^1(G_l,\mathcal {C}^{\lambda ' \text {-an}}(G_l,M)).$

By Proposition 3.14 we may write f as function (in the variable of $\Gamma _0$ )

$$ \begin{align*}f(x) = \sum_{n \geq 0}{m_n}\binom{x}{n}\end{align*} $$

where $m_n \in D^{\lambda }_{H_0,n}(\mathcal {C}^{\lambda \text {-an}}(H_0,M))$ and $\mathrm {val}_\lambda (m_n) - \lfloor p^\lambda n \rfloor \rightarrow \infty .$

Consider now the function given by

$$ \begin{align*}F(x) = \sum_{k \geq 0} (-1)^k \sum_{n \geq 0} (\gamma-1)^k(\gamma^{-k}(m_n))\binom{x}{n+k+1}.\end{align*} $$

Recalling that $\gamma $ acts on $\binom {x}{n+k+1}$ by shifts, one easily checks that $(\gamma -1)(F) = f.$ The problem is that this sum is not guaranteed to converge in $D^{\lambda }_{H_0,n}(M)$ . Indeed, what we have is

$$ \begin{align*}\mathrm{val}_\lambda((\gamma-1)^k(m_n)) \geq sk + \mathrm{val}_{\lambda}(m_n)\end{align*} $$

and

$$ \begin{align*}\mathrm{val}_\lambda(\binom{x}{n+k+1}) = -\lfloor p^\lambda (n+k+1)\rfloor\end{align*} $$

so we get the estimate

$$ \begin{align*}\mathrm{val}_\lambda((\gamma-1)^k(\gamma^{-k}(m_n))\binom{x}{n+k+1}) \geq k(s-p^\lambda) + (\mathrm{val}_{\lambda}(m_n) - \lfloor p^\lambda n \rfloor) + O(1).\end{align*} $$

This estimate would be good enough (i.e., tend to $\infty $ as $k, n \rightarrow \infty $ ) if we had $s> p^\lambda .$ Unfortunately, there is no reason why this would be the case.

There is a trick which fixes this problem: the estimate is good enough when $\lambda $ is replaced with $\lambda ',$ because $s> p^{\lambda '}$ . In other words, by the same estimate we see that under the natural map

$$ \begin{align*}D^\lambda_{H_0,n}(M) \rightarrow D^{\lambda'}_{H_l,n}(M)\end{align*} $$

the sum does converge (using $\mathrm {val}_{\lambda '} \geq \mathrm {val}_{\lambda }$ ). This means that the image of f in $D^{\lambda '}_{H_l,n}(M)$ lies in the image of $\gamma -1$ . Thus, decomposing as in 3.15 and the subsequent paragraph to get an isomorphism of $\mathrm {H}^1(G_0,\mathcal {C}^{\lambda ' \text {-an}}(G_0,M))$ with

$$ \begin{align*}\mathrm{H}^1(G_0/H_l,X_{H_l,n}\otimes_{\Lambda_{H_l,n}} D^{\lambda'}_{H_l,n}(M)) \oplus \mathrm{H}^1(G_0/H_l,D^{\lambda'}_{H_l,n}(M)),\end{align*} $$

we obtain that under the natural map

$$ \begin{align*}\mathrm{H}^1(G_0,\mathcal{C}^{\lambda \text{-an}}(G_0,M))\rightarrow \mathrm{H}^1(G_0,\mathcal{C}^{\lambda' \text{-an}}(G_0,M))\end{align*} $$

the element f lands in

$$ \begin{align*}\ker(\mathrm{H}^1(G_0/H_l,D^{\lambda'}_{H_l,n}(M))\rightarrow \mathrm{H}^1(\Gamma_0,D^{\lambda'}_{H_l,n}(M))).\end{align*} $$

By the inflation restriction sequence, this is the same as $\mathrm {H}^1(H_0/H_l,D^{\lambda '}_{H_l,n}(M)^{\Gamma _0})$ . Thus composing with the restriction map to $\mathrm {H}^1(G_l,\mathcal {C}^{\lambda ' \text {-an}}(G_0,M))$ (and hence also with the further composition to $\mathrm {H}^1(G_l,\mathcal {C}^{\lambda ' \text {-an}}(G_l,M))$ ) we see that f is mapped to zero, as required!

Proof. The inclusion of the right-hand side in the left-hand side is clear. Conversely, suppose x lies on the right-hand side. By applying $\varphi ^n$ , we may reduce to the case $n = 0$ . So we have $x \in \widetilde {\mathbf {A}}^{(0,r]}_{\mathbf {Q}_p} \cap \mathbf {B}^{[s,r]}_{\mathbf {Q}_p},$ whence x lies in the larger ring $\widetilde {\mathbf {B}}^{(0,r]}_{\mathbf {Q}_p} \cap \mathbf {B}^{[s,r]}_{\mathbf {Q}_p} = \mathbf {B}^{(0,r]}_{\mathbf {Q}_p}.$ Thus we reduce to showing $\mathbf {B}^{(0,r]}_{\mathbf {Q}_p}/\mathbf {A}^{(0,r]}_{\mathbf {Q}_p} \rightarrow \widetilde {\mathbf {B}}^{(0,r]}_{\mathbf {Q}_p}/\widetilde {\mathbf {A}}^{(0,r]}_{\mathbf {Q}_p}$ is injective. This reduces further to showing $\widetilde {\mathbf {A}}^{(0,r]}/\mathbf {A}^{(0,r]}_{\mathbf {Q}_p}$ is p-torsionfree. In fact it suffices to show $\widetilde {\mathbf {A}}^{(0,r],\circ }/\mathbf {A}^{(0,r],\circ }_{\mathbf {Q}_p}$ is p-torsionfree, because $\widetilde {\mathbf {A}}^{(0,r]}/\mathbf {A}^{(0,r]}_{\mathbf {Q}_p}$ is its $[\varpi ]$ -adic localization.

Next, recall that $\mathbf {A}^{(0,r],\circ }_{\mathbf {Q}_p} = \widetilde {\mathbf {A}}^{(0,r],\circ }\cap \widetilde {\mathbf {A}}_{\mathbf {Q}_p}$ , so $\widetilde {\mathbf {A}}^{(0,r]}/\mathbf {A}^{(0,r]}_{\mathbf {Q}_p}$ injects into $\widetilde {\mathbf {A}}/\mathbf {A}_{\mathbf {Q}_p}$ . This will be p-torsionfree provided that $\mathbf {A}_{\mathbf {Q}_p}\otimes {\mathbf {F}_p} \rightarrow \widetilde {\mathbf {A}} \otimes {\mathbf {F}_p}$ is injective. But this is easy: it is simply the map $\mathbf {F}_p((X)) \rightarrow \mathbf {C}^\flat _p$ .

Proof. Indeed, choosing some arbitrary s, we have

$$ \begin{align*}\widetilde{\mathbf{A}}^{(0,r],\mathrm{la}}_{\mathbf{Q}_p} \subset \widetilde{\mathbf{A}}^{(0,r]}_{\mathbf{Q}_p} \cap \widetilde{\mathbf{B}}^{[s,r],\mathrm{la}}_{\mathbf{Q}_p} = \widetilde{\mathbf{A}}^{(0,r]}_{\mathbf{Q}_p} \cap (\cup_n \varphi^{-n}(\mathbf{B}^{[p^{-n}s,p^{-n}r]}_{\mathbf{Q}_p})).\end{align*} $$

So by Lemma 4.3, the ring $ \widetilde {\mathbf {A}}^{(0,r],\mathrm {la}}_{\mathbf {Q}_p}$ is contained in $\cup _n \varphi ^{-n}(\mathbf {A}^{(0,p^{-n}r]}_{\mathbf {Q}_p})$ . Conversely, any element in $\cup _n \varphi ^{-n}(\mathbf {A}^{(0,p^{-n}r]}_{\mathbf {Q}_p})$ is locally analytic by Corollary 3.10.

Proof. When $K^{\mathrm {cyc}} \subset K^{\mathrm {LT}}$ this follows from part 2 of Proposition 4.1 and part 1 of Theorem 3.1. In general, one can descend along an unramified twist (see § $8$ [Reference BergerBer16] for a similar argument).

Proof. We have

$$ \begin{align*}\widetilde{\mathbf{{A}}}_{\mathbf{Q}_p}^{(0,r]}/p^n = \widetilde{\mathbf{{A}}}_{\mathbf{Q}_p}/p^n\end{align*} $$

by devissage to the case $n=1$ (where both are equal to $\widetilde {\mathbf {E}}_{\mathbf {Q}_p}$ ). By Theorem 3.1 we have $\mathrm {R}^1_{\mathrm {la}}(p^{n-1}\widetilde {\mathbf {{A}}}^{(0,r]}_{\mathbf {Q}_p}/p^n) = 0$ . Hence by devissage we deduce

$$ \begin{align*}(\widetilde{\mathbf{{A}}}_{\mathbf{Q}_p}^{(0,r]}/p^n)^{\mathrm{la}} = (\cup_m\varphi^{-m}(\mathbf{{A}}_{\mathbf{Q}_p}^{(0,p^{-m}r]}))/p^n\end{align*} $$

(the case $n=1$ being Theorem 4.2). Finally, we need to explain why

$$ \begin{align*}(\cup_m\varphi^{-m}(\mathbf{{A}}_{\mathbf{Q}_p}^{(0,p^{-m}r]}))/p^n =\varphi^{-\infty}(\mathbf{{A}}_{\mathbf{Q}_p})/p^n,\end{align*} $$

but this is once again true by devissage.

Proof. Part 1 follows from Theorem 4.4 of [Reference BergerBer16] and Lemma 4.3. Part 2 follows from Corollary 3.10.

Proof. We start by proving part 2. There is a surjection

$$ \begin{align*}\varphi^{-m}(\mathbf{{A}}_{\mathbf{Q}_p}^{(0,rp^{-m}]}) \twoheadrightarrow \varphi^{-m}(\mathbf{{A}}_{\mathbf{Q}_p}^{(0,rp^{-m}]})/p^n.\end{align*} $$

Hence by part 2 of the previous lemma and functoriality, there exists some pair $\Gamma , \lambda $ , independent of n, such that every element of $\varphi ^{-m}(\mathbf {{A}}_{\mathbf {Q}_p}^{(0,rp^{-m}]})/p^n$ lies in $(\widetilde {\mathbf {{A}}}_{\mathbf {Q}_p}^{(0,r]}/p^n)^{\Gamma ,\lambda \text {-an}}$ . The proof of part 1 is similar but we need another trick, because we do not know that $(\widetilde {\mathbf {{A}}}_{\mathbf {Q}_p}^{(0,r]})^{\Gamma ,\lambda \text {-an}}$ surjects onto $(\widetilde {\mathbf {{A}}}_{\mathbf {Q}_p}^{(0,r]}/p^n)^{\Gamma ,\lambda \text {-an}}$ . Rather, we note that by Remark 3.19, the entirety of $\mathrm {R}^1_{\Gamma ,\lambda }(\widetilde {\mathbf {{A}}}_{\mathbf {Q}_p}^{(0,r]})$ maps to zero in some $\mathrm {R}^1_{\Gamma ',\lambda '}(\widetilde {\mathbf {{A}}}_{\mathbf {Q}_p}^{(0,r]})$ . This implies that given a pair $(\Gamma ,\lambda )$ there exists some other pair $(\Gamma ',\lambda ')$ so that for all $n\geq 1$ we have that elements of $(\widetilde {\mathbf {A}}_{\mathbf {Q}_p}^{(0,r]}/p^n)^{\Gamma ,\lambda \text {-an}}$ lift to $(\widetilde {\mathbf {A}}_{\mathbf {Q}_p}^{(0,r]})^{\Gamma ',\lambda ' \text {-an}}$ . We can now argue as before by applying part 1 of the previous lemma (for the pair $\Gamma ',\lambda '$ ).

Proof. The proposition shows that the direct systems

$$ \begin{align*}\{\varprojlim_{n}(\mathbf{{A}}_{\mathbf{Q}_p}^{(0,r]}/p^n)^{\Gamma,\lambda \text{-an}}\}_{\Gamma,\lambda}\end{align*} $$

and

$$ \begin{align*}\{\varprojlim_{n}\varphi^{-m}(\mathbf{{A}}_{\mathbf{Q}_p}^{(0,rp^{-m}]})/p^n \}_m\end{align*} $$

are cofinal. So the corresponding direct limits ranging over $\Gamma ,\lambda $ and over m are naturally isomorphic. These give $\mathbf {{A}}_{\mathbf {Q}_p}^{\mathrm {la}}$ and $\varphi ^{-\infty }(\mathbf {{A}}_{\mathbf {Q}_p})$ , respectively, and hence we establish the desired equality.