Continuous functions on Berkovich spaces

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Continuous functions on Berkovich spaces

Junyi Xie

To cite this version:

Junyi Xie. Continuous functions on Berkovich spaces. 2020. �hal-03039682�

JUNYI XIE

Abstract. Let k be a perfect complete valued field with a nontrivial non- archimedean norm | · | andω ∈k with 0<|ω|<1. Let X be a reduced and normalk-analytic space. ThenO^◦'lim

← O^◦/ωⁿ.

1. Introduction

1.1. A motivation. Let X be a complex manifold. Cauchy’s convergence the- orem says that a Cauchy sequence of continuous (resp. holomorphic) functions for the L^∞-norm convergences to a continuous (resp. holomorphic) on X.

Let us consider a sheaf version of this theorem. Denote by C (resp. O) the sheaf ofC-valued continuous (resp. holomorphic) functions. SetF =C orO.For every open subset U ⊆X, denote byρ_U :F(U)→[0,+∞] theL^∞-norm. Define a piecewise F-function on U to be a collection {(U_i, F_i), i ∈ I} where U_i, i ∈ I is an open covering of U and F_i ∈ F(U_i). Denote by P(U) the set of piecewise F-function on U.

For F ={(U_i, F_i), i∈I}, G={(V_j, G_j), j ∈J} ∈ P(U), define d(F, G) := sup

i∈I,j∈J

ρ((F_i|_Ui∩U_j−G_j|_Ui∩U_j)).

The functiond is similar to a semi-metric, except that d(F, F) may take positive value. We may define the Cauchy sequences onP(U) as usual, to be the sequences F_n, n≥0, for which, for every >0, there isN ≥0 such that for everym, n≥N, d(F_n, F_m)≤.We define the equivalence relation of the Cauchy sequences in the usual way. Define ˆF to be the presheaf onX sendingU to the equivalence classes of Cauchy sequences in P(U). Easy to see that F is a sub-presheaf of ˆF.

When F = C, we have ˆF = ˆC = C. This fact can be viewed as a sheaf version of Cauchy’s convergence theorem for C. But this sheaf version is not true for O when dimX ≥ 1. Observe that every bounded continuous function can be uniformly approximated by piecewise locally constant functions (hence by piecewise holomorphic functions). So ˆO(X) contains all bounded continuous functions on X. In fact, it is easy to see thatC is exactly the sheafification of ˆF. This gives us a way to define the sheaf of continuous functions from the sheaf of holomorphic functions.

In non-archimedean geometry (in the sence of Berkovich), we a priori have a structure sheaf which is the analogy of the sheaf of holomorphic functions. Apply- ing the above construction, we get an analogy of the sheaf of continuous functions.

Date: December 4, 2020.

The author is partially supported by project “Fatou” ANR-17-CE40-0002-01.

1

One may ask what does it look like? In particular, are there ”continuous func- tions” which are not analytic? In this paper we show that under mild conditions, the sheaf version of Cauchy’s convergence theorem holds for the structure sheaf.

In other words, all ”continuous functions” in this sense are analytic. We will also study the analytic and continuous functions on subsets of a Berkovich space. In particular, there are continuous functions on some closed subsets which are not analytic.

In the sequels to the papers, we will apply the results in this paper to study non-archimedean dynamics. In particular, Theorem 1.1 will be used to generate [7, Appendix B] to the global setting.

1.2. The statement. Denote byka complete valued field with a nontrivial non- archimedean norm | · |. Denote by k^◦ :={f ∈k| |f| ≤ 1} the valuation ring and k^◦◦ := {f ∈ k| |f| < 1} its maximal ideal. Denote by ek := k^◦/k^◦◦ the residue field. We denote by ρ_X the spectral norm on X. Some times, we write ρ for ρ_X for the simplicity.

LetXbe a reducedk-analytic space. LetO(orO_X when we want to emphasize the space X) be the structure sheaf on X w.r.t the G-topology. Let O^◦(X) the subsheaf of O satisfying O^◦(U) =O(U)^◦ for every analytic subdomain U of X.

Let ω be any element in k satisfying |ω| ∈ (0,1). Denote O^◦/ω be the cokernal of the morphism ω× ·:O^◦ → O^◦. We prove the following result.

Theorem 1.1. Assume that kif perfect and X is normal. Let ω be any element in k satisfying |ω| ∈ (0,1). Denote by α_n :O^◦ → O^◦/ωⁿ the quotient morphism.

Then the natural morphism α := lim

← α_n:O^◦ →lim

← O^◦/ωⁿ is an isomorphism.

Remark 1.2. It is easy to see that the morphismO^◦ →lim

← O^◦/ωⁿ is an isomor- phism if and only if the natural morphism O →lim

← O/ωⁿO^◦ is an isomorphism.

Remark 1.3. As Ruochuan Liu pointed out, when kis of mixed characteristic, there is an alternative proof of Theorem 1.1 using the almost vanishing of the higher cohomologies ofO⁺ on affinoid perfectoid spaces [6, Theorem 6.3] and the Ax-Sen-Tate for rigid analytic spaces [5, Proposition 8.2.1]. He also pointed out that when k is of mixed characteristic, Theorem 1.1 is implied by [4, Theorem 10.3].

Remark 1.4. If X is not normal, α can be non-surjective. An example is as follows: Let X := M(k{T₁, T₂}/(T₂² − T₁³)). For n ≥ 0, let U_n := {x ∈ X| |T1(x)| ≤ |ω²ⁿ|} and Vn := {x ∈ X| |T1(x)| ≥ |ω²ⁿ|}. We note that for x ∈V_n, T₂/T₁ ∈ O(V₁) and |T₂/T₁(x)| =|T₁(x)|^1/2. So we have T₂/T₁ ∈ O(V_n)^◦. Set v_n := α_n(T₂/T₁) ∈ O^◦/ωⁿ(V_n). Since for every x ∈ U_n ∩V_n, |T₂/T₁(x)| =

|T₁(x)|^1/2 = |ωⁿ|, we have v_n|_Un∩Vn = 0 ∈ O^◦/ωⁿ(V₁ ∩V₂). So there is a unique section t_n ∈ O^◦/ωⁿ(X) such that t_n|_Vn = v_n and t_n|_Un = 0. It is clear that for m ≥n, the image of t_m inO^◦/ωⁿ(X) ist_n.Set T := lim

← t_n ∈lim

← O^◦/ωⁿ.Leto be the k-point in X defined by the maximal ideal (T1, T2) in k{T1, T2}/(T₂² −T₁³).

We note that on Y := X\ {o}, we have T|_Y = α(T₂/T₁). But T₂/T₁ does not extend to a regular function on X. So we haveT 6∈α(O^◦(X)).

The quotient morphism α_ω :O^◦ → O^◦/ω induces a morphism α_ω(X) :O^◦(X)→ O^◦/ω(X).

Even when X is affinoid, α_ω(X) is not surjective in general.

Example 1.5. Let k = C((T)), ω := T and X := M(k{x, y}/(y² −x³ −T⁶)).

Observe that O^◦(X) = C[[T]][x, y]/(y²−x³−T⁶) and O^◦(X)/ω =C[x, y]/(y²− x³). Set U₁ := {z ∈ X| |x³| ≤ |y²|} and U₂ := {z ∈ X| |y²| ≤ |x³|}. Both U₁ and U₂ are rational subdomain of X.DefineF₁ :=x²/y ∈O(U₁) andF₂ =y/x∈ O(U₂).

We claim ρ_U1(F₁) ≤ 1 and ρ_U2(F₂) ≤ 1. For every z ∈ X, if R := |x³(z)| =

|y²(z)|, then 0< R≤1, z ∈U₁∩U₂ and

|F₁(z)|=|(x²/y)(z)|=|(y/x)(z)|=|F₂(z)|=R^1/6 ≤1.

For every z ∈X, if|x³(z)|<|y²(z)|, thenz ∈U₁\U₂, |x(z)|<|T²|and |y(z)|=

|T³|. So |F₁(z)| = |(x²/y)(z)| < |T| < 1. For every z ∈ X, if |x³(z)| > |y²(z)|, then z ∈U₂\U₁, |x(z)|=|T²| and |y(z)|<|T³|. So |F₂(z)|=|(y/x)(z)|<|T|<

1. Then we proved our claim.

For z ∈ U₁ ∩U₂, we have |x³(z)| = |y²(z)| ≥ |T⁶|. Then we have |xy(z)| ≥

|T⁵|. We have F₁ −F₂ = x²/y −y/x = (x³ −y²)/xy = −T⁶/xy. So we have

|(F₁ −F₂)(z)| = |(T⁶/xy)(z)| ≤ |T⁶|/|T⁵| ≤ |T|. It shows that α_ω(F₁)|_U1∩U2 = αω(F2)|U1∩U2. So we may glue αω(F1) and αω(F1) to a section F ∈ O^◦/ω(X).

We claim that F is not contained in the image of αω(X). Otherwise F = α_ω(X)(G) whereG∈C[[T]][x, y]/(y²−x³−T⁶).DefineU :={z ∈X| |x(z)|= 1}.

We have U ⊆U₁∩U₂.Then on U, we get

y/x=F|_U =G mod ω∈C[[T]][x, y]/(y²−x³−T⁶)/(T) = C[x, y]/(y²−x³) which is a contradiction.

The proof of Theorem 1.1 relies on the following lemma.

Lemma 1.6. Letωbe any element inksatisfying|ω| ∈(0,1).Then the morphism α_ω(D^r(R₁, . . . , R_r)) :O^◦(D^r(R₁, . . . , R_r))→ O^◦/ω(D^r(R₁, . . . , R_r))

is surjective for every r≥0.

The following is a simple application of Lemma 1.6.

Corollary 1.7. Let ω be any element in k satisfying |ω| ∈ (0,1). Then we have O^◦/ω(A^r,an) = k^◦/ω for r ≥0.

Acknowledgement. I would like to thank Bernard Le Stum, Xinyi Yuan, J´erˆome Poineau for useful discussions. I thank Antoine Ducros for telling me that we need the condition ”generically quasi-smooth” in Lemma 2.2. I thank Ruochuan Liu for sharing me his idea of an alternative proof of Theorem 1.1 in mixed charac- teristic using relative p-adic Hodge theory.

2. A Noether normalization lemma

Assume that the norm onkis non-trivial. LetX =M(A) be a strictk-affinoid space. We say thatX isgenerically quasi-smooth if there is a Zariski dense open subset U of X such thatU is quasi-smooth.

Remark 2.1. Ifkis perfect andXis reduced, thenXis generically quasi-smooth.

We need the following version of Noether normalization lemma for the proof of Theorem 1.1.

Lemma 2.2. Assume that X is irreducible and generically quasi-smooth of di- mension r. Then there exists a finite morphism

π:X →D^r =M(k{T1, . . . , Tr}),

such thatπ^∗ :O(D^r) = k{T₁, . . . , T_r},→ O(X) is injective and the field extension π^∗ : FracO(D^r),→Frac (O(X)) to be separable.

Proof of Lemma 2.2. By [3, 6.1.2. Theorem 1.(i)], there exists an admissible surjection F : k{U₁, . . . , U_n} A whose restriction G := F|_k{U1,...,Ur} is fi- nite and injective. Since X is generically quasi-smooth, there exists a rigid point x ∈ X such that X is quasi-smooth at x. Then there exists an affi- noid subdomain W of Dⁿ = M(k{U₁, . . . , U_n}) containing x such that X ∩W is defined by the ideal (g₁, . . . , gn−r) and moreover the Jacobian matrix at x J(x) := (∂_Ujg_i(x))1≤i≤n−r,1≤j≤n ∈ Mn×(n−r)(H(x)) is of full rank. For j = 1, . . . , n, set J_j := (∂_Ujg₁(x), . . . , ∂_Ujgn−r(x)). Then there are 1 ≤ j₁ < · · · <

jn−r ≤ n such that the vectors J_j1, . . . , J_jn−r are linearly independent. Write {1, . . . , n} \ {j₁, . . . , jn−r} = {s₁, . . . , s_r} where s₁ < · · · < s_r. For a ∈ k, denote by Φ_a : Dⁿ → D^r sending (U₁, . . . , U_n) to (U₁ + aU_s1, . . . , U_r +aU_sr) and set π_a := Φ_a|_X. Then there exists R > 0 such that when 0 < |a| < R, (dπ_a)(x) is invertible and the reduction πe_a^∗ : k{T^₁, . . . , T_r} → Ae equal to the reduction Ge : k{T^₁, . . . , T_r} → A. Definee π := π_a for some a ∈ k satisfying 0 < |a| < R. Since G is finite, Ge = πe^∗ is finite [3, 6.3.5. Theorem 1]. Then π is finite [3, 6.3.5. Theorem 1]. Since (dπ)(x) is invertible, the field extension π^∗ : FracO(D^r),→Frac (O(X)) is separable, which concludes the proof.

3. The piecewise analytic functions

Let X be a reduced k-analytic space. To avoid some set theoretical problem, we fix a cardinal number κ≥max{ℵ0,2^#X}.

Apiecewise analytic functions onXis defined to be a collection{(U_i, f_i), i∈I}

indexed by an ordinal number I of cardinal at most κ, where U_i, i ∈ I is a G- covering of X (by analytic domains) and fi ∈ O(Ui).Denote by P(X) to be the set of piecewise analytic functions on X. In particular, for F = {(Ui, fi), i ∈ I}, G = {(V_i, g_i), i ∈ J}, we say that F = G if I = J and for every i ∈ I = J, U_i =V_i and f_i =g_i.

For F ={(U_i, f_i), i ∈ I}, denote by U_F :={U_i, i ∈ I} the covering associated to F.For F ={(U_i, f_i), i∈I}, G={(V_i, g_i), j ∈J}, we define

F +G={(U_i∩V_j, f_i+g_j),(i, j)∈I×J} and

F ×G={(U_i ∩V_j, f_ig_j),(i, j)∈I×J}.

Here I ×J is the multiplication of ordinal numbers i.e. the order type of I×J with the lexicographical order.

Remark 3.1. When #I = 1 or #J = 1, we have F ×G = G×F. In general, F ×G6=G×F. Because usually the ordering on I×J and J ×I are different.

For F, G, H ∈ P(X), we have

F +G=G+F,(F +G) +H =F + (G+H) and (F ×G)×H =F ×(G×H).

In general F ×(G+H)6=F ×G+F ×H.

We have an embeddingO(X),→ P(X), sendinghto{(X, h)}.This embedding preserves the addition and the multiplication. For every F ∈ O(X), G, H ∈ P(X), we have

F ×(G+H) =F ×G+F ×H.

We note that, for F ={(U_i, f_i), i∈I}, we have

F −F ={(U_i∩U_j, f_i−f_j),(i, j)∈I×J}

and

0×F ={(Ui,0), i∈I}

are not necessarily equal to 0.

For F ={(U_i, f_i), i∈I} ∈ P(X),define ρ(F) = sup

i∈I

ρ(f_i)∈[0,+∞]

and

∆(F) :=ρ(F −F),

where ρ is the spectral norm. We may view F as a multi-valued function on X.

For everyx∈X, the value ofF atxis denoted by F(x) := {f_i ∈ H(x)| x∈U_i}.

Then we have

ρ(F) = sup

x∈X

sup{|f_i(x)||x∈U_i} and

∆(F) = sup

x∈X

sup{|(f_i−f_j)(x)|| x∈U_i∩U_j}.

For F, G∈ P(X), a∈k, we have

ρ(F +G)≤max{ρ(f), ρ(g)}, ρ(F G)≤ρ(F)ρ(G);

∆(F +G)≤max{∆(F),∆(G)},∆(F G)≤max{ρ(F)∆(G), ρ(G)∆(F)}.

and

ρ(aF) = |a|ρ(F),∆(aF) =|a|∆(F).

Fors∈[0,∞], denote byP^s(X) (resp. P_s(X) ) the set ofF ∈ P(X) satisfying ρ(F)≤s(resp. ∆(F)≤s.) Set P_t^s(X) :=P^s(X)∩ P_t(X).They are stable under addition. It is easy to see the following properties

(i) P(X) =P^∞(X);

(ii) P(X)^s_t is increasing for both s and t;

(iii) P_t^s(X) = P^s(X) if s≤t;

(iv) P^s(X)⊆ P_s(X);

(v) for every s₁, s₂, t₁, t₂ ∈[0,∞],P_t^s₁¹(X)× P_t^s₂²(X)⊆ P_max{s^s1s2

1t2,s2t1}(X).

For F ∈ P(X), ∆(F) = 0 if and only if F takes form {(U_i, f|_Ui), i ∈ I} where f ∈ O(X); if and only if there exists f ∈ O(X) such that ρ(F −f) = 0.

Let Y be a reduced k-analytic space satisfying 2^#Y ≤κ. Let f :Y →X be a morphism. Then we have the pull back map

f^∗ :P(X)→ P(Y)

sending F := {(U_i, f_i), i ∈ I} to f^∗F := {(f⁻¹(U_i), f^∗(f_i)), i ∈ I}. This map preserves the addition and the multiplication. The restriction of f^∗ on O(X) is exactly the usual pull back of regular functions. We have ρ(f^∗(F)) ≤ ρ(F) and

∆(f^∗F)≤∆(F).

In particular, for any analytic subdomainU ofX, we may define the restriction map P(X) → P(U) by applying the pull back map for the inclusion. Observe that, if V_i, i ∈ I is a covering of analytic subdomains of X and F ∈ P(X), we have

ρ(F) = max{ρ(F|_Vi), i= 1, . . . , m}.

Let I be an ordinal number with cardinal at most κ and V_i, i ∈ I be a G- covering of X by analytic subdomains and F_i = {(V_jⁱ, f_jⁱ), j ∈ J_i} ∈ P(V_i), we define

∨i∈IF_i :={(V_jⁱ, f_jⁱ),(i, j)∈ ts∈IJ_s} ∈ P(X).

Here we write (i, j)∈ t_s∈IJ_s for i∈ I and j ∈J_i; and we think t_i∈IJ_i as a well ordinal set with the following ordering: (a, b)≤(c, d) ifa < c ora=c, b≤dand identify it with its order type.

We have

ρ(∨i∈IF_i) = sup

i∈I

ρ(F_i) and

∆(∨i∈IF_i)≤sup{ρ(F_i|_Vi∩V_j −F_j|_Vi∩V_j), i, j ∈I}.

LetU ={U_i, i∈I}be a G-covering ofXby analytic subdomains. A refinement of U is a G-covering V ={V_j, j ∈J}with a map σ :I →J such that V_i ⊆U_σ(i). For F = {(U_i, f_i), i∈ I} ∈ P(X), if J is an ordinal number of cardinal at most κ, α:= (V ={V_j, j ∈J}, σ) is a refinement of U_F. We defineF_α∈ P(X) by

F_α :={(V_i, f_σ(i)|_Vi), j ∈J}.

We have

ρ(F_α)≤ρ(F),∆(F_α)≤∆(F) and ρ(F_α−F)≤∆(F).

Let F_n, n ≥ 0 be a sequence in P(X). We say that F_n, n ≥ 0 is a Cauchy sequence if for every > 0, there exists N ≥ 0 such that for every n, m ≥ N, we have ρ(F_n−F_m) ≤ . In fact, F_n, n ≥ 0 in P(X) is Cauchy if and only if

n→∞lim ∆(Fn) = 0 and lim

n→∞ρ(Fn−Fn+1) = 0. For f ∈ O(X)⊆ P(X), we say that

n→∞lim F_n=f if ρ(F_n−f)→0 as n→ ∞. It is clear that if lim

n→∞F_n =f, then we have

(3.1) ρ(F_n−f) = lim

m→∞ρ(F_n−F_m).

Two Cauchy sequences F_n, n ≥ 0 and G_n, n ≥ 0 are defined to be equivalent if the lim

n→∞(Fn−Gn) = 0.Denote by ˆO(X) the space of equivalent classes of Cauchy sequences in P(X).For every Cauchy sequencesF_n, n≥0 inP(X), we denote by [F_n] its image in ˆO(X).Then we have a natural inclusionO(X)⊆O(X) sendingˆ f ∈ O(X) to the image of the constant sequence F_n := f, n ≥ 0. We identify O(X) with its image in ˆO(X).

It is easy to see that the addition and multiplication in P(X) induce addition and multiplication on ˆO(X) such that for every Cauchy sequences F_n, G_n, n≥0 in P(X), we have

[F_n] + [G_n] = [F_n+G_n] and [F_n][G_n] = [F_nG_n].

The norm ρ induces a norm on P(X) (which is still denoted by ρ), satisfying ρ([F_n]) = lim

n→∞ρ(F_n).

Observe that ρ([F_n]) = 0 if and only if [F_n] = 0.

We may view [F_n] as a function on X. Indeed, for every point x ∈ X, there exists a unique limit point [F_n](x) for the sequences of subsetF_n(x) inH(x). This point does not depend on the choice of F_n, n ≥0 in its equivalent class. We say that [F_n](x) is the value of [F_n] at x.Then we have

(3.2) ρ([F_n]) = sup

x∈X

|[F_n](x)|.

Since ρ([Fn]−[Gn]) = sup_x∈X|[Fn−Gn](x)|= sup_x∈X |[Fn](x)−[Gn](x)|, [Fn] = [Gn] if and only if they takes the same value at every point in X.

The restriction of ρ on O(X) is the spectral norm on O(X). When X is compact, ρ takes finite values on ˆO(X) and it makes ˆO(X) to be a Banach k-algebra.

The spaces P(X),P^s(X),Pt(X),P_t^s(X) depends on the choice of the cardinal number κ. However, we have the following result.

Proposition 3.2. The space O(X)ˆ does not depend on the choice of κ.

Proof of Proposition 3.2. Letβbe a cardinal number at leastκ.Denote byP(X)⁰ and ˆO(X)⁰ the spaces constructed in the same way as P(X) and ˆO(X) except replacingκbyβ.We have a natural inclusionP(X)⊆ P(X)⁰. It induces a natural morphism ˆO(X) → O(X)ˆ ⁰. Easy to see that this morphism is well defined.

Moreover, it is a morphism of k-algebra and it preserves the spectral normρ. In

particular, this morphism is injective. Then we may view ˆO(X) as a subspace of O(X)ˆ ⁰.

Lemma 3.3. For everyF⁰ ∈ P(X)⁰, there existsF ∈ P(X)such thatρ(F−F⁰)≤

∆(F⁰).

For every [F_n⁰]∈O(X)ˆ ⁰, pickFn ∈ P(X), n ≥0 such thatρ(Fn−F_n⁰)≤∆(F_n⁰).

Since ∆(F_n⁰)→ 0 as n → ∞, Fn is a Cauchy sequence and [F_n⁰] = [Fn] ∈O(X).ˆ

This concludes the proof.

Proof of Lemma 3.3. Write F⁰ = {(U_i, F_i), i ∈ I}. Since the cardinal of the set of analytic subdomains of X is at most 2^#X ≤ κ, there exists a subset J of I of cardinal at mostκsuch that for every i∈I, there existsj ∈J such thatU_i =U_j. We note that J is a well-order set. We identify it with its order type. Define F :={(U_j, F_j), i∈ J} ∈ P(X). We get ρ(F −F⁰)≤∆(F⁰), which concludes the

proof.

For everys >0, denote by ˆO^s(X) the set ofF ∈O(X) satisfyingˆ ρ(F)≤s.For every F ∈Oˆ^s(X), we may write is as F = [F_n] whereF_n is a Cauchy sequence in P^s(X).Indeed, we may writeF = [F_n⁰] for some Cauchy sequence inP(X).Then there exists N ≥ 0, such that ρ(F_n⁰) ≤ s for n ≥ N. Then we have F = [F_N⁰ _+n] and F_N+n⁰ ∈ P^s(X) forn ≥0.

Remark 3.4. If we defineOc^s(X) to be the equivalent classes of Cauchy sequences of in P^s(X), the above argument just shows that Oc^s(X) = ˆO^s(X).

3.1. Sheaf properties.

Proposition 3.5. The following presheaves

P (resp. P^s,P_t,P_t^s,O,ˆ Oˆ^s) :U 7→ P(U) (resp. P^s(U),P_t(U),P_t^s(U),O(U),ˆ Oˆ^s(U)) are sheaves on X.

Remark 3.6. When X is a complex manifold and O is the sheaf of homeomor- phisms, we may define the presheaf ˆOin a same way. But usually it is not a sheaf and its sheafification is the sheaf of continuous functions.

Proof of Proposition 3.5. We first prove it for P. We need to show that for any analytic subdomain U of X, every G-covering V_j, j ∈ J of U, the first arrow in the following diagram is an equalizer:

P(U)→Y

i∈J

P(V_i)^−→_−→ Y

i,j∈J

P(V_i∩V_j).

We first show that the first arrow is injective. Let F = {(U_i, F_i), i ∈ I₁)}, G = {(W_i, G_i), i∈I₂)}be two sections ofP(U),such that for everyj ∈J,F|_Vj =G|_Vj. Then we get I₁ =I₂. Moreover, for every i∈I₁ =I₂, we have U_i∩V_j =W_i∩V_j and F_i|_Ui∩Vj = G_i|_Wi∩Vj. It follows that U_i =W_i and F_i = G_i for all i∈ I₁ =I₂. Then we getF =G.Now assume that (F_j, j ∈J)∈Q

jP(V_j) is contained in the equalizer. We only need to show that it is the image of some section of P(U).

WriteF_j ={(U_i^j, F_i^j), i∈I_j}.Forj, k ∈J, we haveF_j|_Vj∩Vk =F_k|_Vj∩Vk.It follows that I := I_j are the same for all j ∈ J. For every i ∈ I, define U_i := ∪_j∈JU_i^j. Since for every i ∈ I, j, k ∈ J, we have U_i^j ∩V_j ∩V_k = U_i^k ∩V_j ∩V_k, then for every j ∈ J, U_i^j = U_i ∩V_j. Since for every j, k ∈ J F_i^j|_Ui∩V_j∩V_k = F_i^k|_Ui∩V_j∩V_k, there exists F_i ∈ O(U_i) such thatF_i^j =F_i|_Ui∩Vj. It follows that (F_j, j ∈J) is the image of {(F_i, U_i), i ∈ I} ∈ P(U). We conclude the proof for P. The proofs for P^s,Pt,P_t^s are the same as the proof for P.

We only need to prove it for ˆO. We need to show that for any analytic sub- domain U of X, every G-coveringV_j, j ∈J of U, the first arrow in the following diagram is an equalizer:

O(U)ˆ →Y

i∈J

O(Vˆ _i)^−→_−→ Y

i,j∈J

O(Vˆ _i ∩V_j).

By Equality 3.2, the first arrow is injective. By Proposition 3.2, we may assume that κ ≥ #J. We may endow J a well-order and assume it to be an ordinal number. Assume that ([F_n^j], j ∈J) is an element in the equalizer in Q

i∈JO(Vˆ _i).

Define F_n := ∨j∈JF_n^j ∈ P(X). It is easy to check that F_n, n ≥ 0 is a Cauchy sequence and [F_n]|_Vj = [F_n^j]|_Vj. We conclude the proof.

Let ω be an element in k satisfying 0 < |ω| < 1. For every element F = {(U_i, F_i), i ∈ I} ∈ P_|ω|¹ (X), we denote by Fe the section of O_X^◦ /ω(X) which is glued by Fe_i ∈ O_X^◦/ω(U_i) where Fe_i is the image of F_i in O^◦_X(V_i)/ω. Denote by ψ_X : P_|ω|¹ (X) → O^◦_X/ω(X) the map defined by F 7→ F .e Sometimes we write ψ^ω_X for ψ_X to emphasize ω. We note that for F₁, F₂ ∈ P_|ω|¹ (X), ψ_X(F₁) =ψ_X(F₂) if and only if ρ(F₁−F₂)≤ |ω|. We have the following result.

Proposition 3.7. The mapψ_X :P_|ω|¹ (X)→ O^◦/ω(X)(resp. P|ω|(X)→ O/ω(X)) is surjective.

Proof of Proposition 3.7. The P_|ω|¹ part of this proposition implies the P|ω| part.

So we only need to proof the P_|ω|¹ part.

Let Fe be a section in O/ω(X). Since the quotient morphism φ : O^◦ → O^◦/ω(X) is surjective, there exists a G-covering V_j, j ∈ J of X and F_j ∈ O^◦(V_j), j ∈ J such that φ|_Vj(F_j) = Fe|_Vj. Denote by β a cardinal number at least max{#J, κ}. Denote by P(X)⁰,P(X)⁰¹_ω the spaces constructed in the same way as P(X),P(X)¹_ω except replacing κ by β. View P(X),P(X)¹_ω as subspaces of P(X)⁰,P(X)⁰¹_ω. Define the morphism ψ⁰_X : P_|ω|⁰¹(X) → O^◦/ω(X) in the same way as ψ_X. We haveψ⁰_X|_P(X)¹_ω =ψ_X.

After giving a well-order onJ and identifying it with its order type, we may as- sume thatJ is an ordinal number of cardinal at most β.ThenF⁰ :={(V_j, F_j), j ∈ J} defines an element in P_|ω|⁰¹ (X) satisfying ψ⁰_X(F⁰) = F .e By Lemma 3.3, there exists F ∈ P_ω¹(X) such that ρ(F −F⁰) ≤ ∆(F⁰) ≤ |ω|. Then we have ψ_X(F) = ψ⁰_X(F) = ψ_X⁰ (F⁰) = Fe, which concludes the proof.

For every analytic subdomain U of X, [F_n] ∈Oˆ¹(U), there exists N ≥0 such that F_n ∈ P_|ω|¹ (U) for n ≥ N and ψ_X^ω(F_n) = ψ_X^ω(F_N). The value ψ_X^ω(F) :=

ψ^ω_X(F_n) does not depend on the choice of N and the sequence F_n. Denote by β_Uⁿ : ˆO¹(U) → O^◦/ωⁿ(U) the morphism sending F to ψ_X^ωn(F). They define a morphism of sheaves βⁿ : ˆO¹ → O^◦/ωⁿ. Moreover β_n induces a morphism of sheaves β := lim

← β_n : ˆO¹ →lim

← O^◦/ωⁿ.

Proposition 3.8. The morphism β : ˆO¹ →lim

← O^◦/ωⁿ (resp. O →ˆ lim

← O/ωⁿO^◦) is an isomorphism.

Proof of Proposition 3.8. The ˆO¹ part of this proposition implies the ˆO part. So we only need to proof the ˆO¹ part.

We only need to show that for affinoid subdomainU,β_U : ˆO¹(U)→lim

← O^◦/ωⁿ(U) is an isomorphism.

For F = [Fn] ∈ Oˆ¹(U), we may assume that ρ(Fn) ≤ 1. If F 6= 0, there exists l ≥ 0 such that ρ(F) > |ω^l|. There exists N ≥ 0, such that for n ≥ N, ψ^ω_X^l(F_n) = ψ^ω_X^l(F_N). Denote by q_l : lim

← O^◦/ωⁿ(U) → O^◦/ω^l(U) the quotient morphism, we have

q_l(β(F)) =β_n(F) =ψ^ω_X^l(F_N)6= 0.

This shows that β is injective.

For every (g_n)n≥0 ∈ lim

← (O^◦/ωⁿ(U)) = lim

← O^◦/ωⁿ(U), by Proposition 3.7, we may writeg_n =ψ_U^ωn(G_n) for someG_n∈ P_|ω¹n|(U).We may check thatG_n, n≥0 is a Cauchy sequence in P¹(U) and β_U([G_n]) = (g_n)n≥0.Thenβ is surjective, which

concludes the proof.

3.2. An induced continuousR-valued function. Forb >0,F ={(U_i, F_i), i∈ I} ∈ P_b(X), we define a functionφ^b_F :X →R onX as follows: for every x∈X,

φ^b_F(x) := max{log(|F_i(x)|),log(b)},

wherex∈U_i.It is well defined, because fori, j ∈Isatisfyingx∈U_i∩U_j, we have

|(F_i−F_j)(x)| ≤b, then max{log(|F_i(x)|),log|b|}= max{log(|F_j(x)|),log|b|}. We note that for F⁰ ∈ P_b(X), ifρ(F −F⁰)≤b, then we have φ^b_F =φ^b_F0.

Proposition 3.9. The function φ^b_F is continuous on X.

Proof of Proposition 3.9. For every i ∈ I, φ^b_F|_Ui = max{log(|F_i|_Ui|),log(b)} is continuous. We conclude the proof by the following lemma.

Lemma 3.10. Let U_i, i ∈ I be an affinoid G-covering of X. Then a function φ : X → R is continuous if and only if for every i ∈ I, φ|_Ui : U_i → R is continuous.

Proof of Lemma 3.10. It is clear that if φ is continuous, φ|_Ui is continuous for every i∈I.

Now assume that φ|_Ui is continuous for every i∈ I. Let K be a closed subset of R, we only need to show that φ⁻¹(K) is closed in X. Let x be a point in X \φ⁻¹(K), since U_i, i ∈ I is a G-covering of X, there is a finite subset I_x of I such that

x∈(∩_i∈IxU_i)∪((∪_i∈IxU_i)^◦).

Then (∪i∈IxU_i)∩φ⁻¹(K) = ∪i∈Ix(φ|_Ui)⁻¹(K) is closed in X. So φ⁻¹(K) contains the open neighborhood ((∪i∈IxU_i)^◦)\φ⁻¹(K) ofx, which concludes the proof.

4. Proof of Lemma 1.6 and Corollary 1.7

Proof of Lemma 1.6. Denote byX :=D^r(R₁, . . . , R_r) =M(k{R⁻¹₁ T₁, . . . , R⁻¹_r T_r}).

Set b :=|ω|. When r= 0, there is nothing to prove. Now we assume that r≥1.

By Proposition 3.7, we only need to show that for every elementF ={(U_i, F_i), i∈ I} ∈ P_b¹(X), there exists a section G ∈ O^◦(X) such that for every i ∈ I, ρ(G|Ui−Fi)≤b.

4.1. Reduce to the finite covering by rational subdomains. Since for every i∈I,Ui has a G-covering of (at mostκ) rational subdomains, after a refinement, we may assume that U_i, i∈I are rational subdomains. In particular, allU_i, i∈I are compact.

For every x∈X, there exists a finite subsetI_x ⊆I such that x∈(∪i∈I_xU_i)^◦∩(∩i∈I_xU_i).

SinceXis compact, there exists a finite setS⊆X, such thatX ⊆ ∪_x∈S(∪_i∈IxU_i)^◦. Lemma 4.1. The collection U_i, i∈ ∪x∈SI_x is a G-covering of X.

Proof of Lemma 4.1. For everyx∈X, there existsy∈Ssuch thatx∈(∪i∈IyU_i)^◦. Let I₁ be the set of i∈I_y such that x∈U_i. Then we get x∈ ∩i∈I₁U_i and

x∈(∪i∈IyU_i)^◦\(∪j∈I_y\I₁U_j)⊆ ∪i∈I₁U_i.

Since (∪i∈I_yU_i)^◦\(∪j∈I_y\I₁U_j) is open, we get x∈(∪i∈I₁U_i)^◦. This concludes the

proof.

After replacing I by ∪x∈SI_x, we may assume that I is finite.

4.2. Reduce to the strict case.

Lemma 4.2. Assume thatR∈R>0\|k^∗|.LetKbe the fieldk{R⁻¹S₁, RS₂}/(S₁S₂− 1). If Lemma 1.6 holds for the K analytic space X_K :=X⊗b_kK, then it holds for X.

Proof of Lemma 4.2. There exists G_K ∈ O^◦(X_K) such that for every i ∈ I, ρ(G_K|_U

i⊗b_kK −F_i) ≤ b. Write G_K = P

i∈ZG_iS₁ⁱ where G_i ∈ O(X). Then for every i ∈ I, we have ρ(P

j∈Z^∗Gj|UiS₁^j + (G0−Fi)) ≤ b. By [1, Section 2.1], we have ρ(P

j∈Z^∗G_j|_UiS₁^j+ (G₀−F_i)) = max{ρ(G₀|_Ui −F_i),maxj6=0ρ(G_j|_Ui)R^j}. It follows that ρ(G0|Ui −Fi)≤b, i∈I, which concludes the proof.

Applying Lemma 4.2, we may assume that R_j ∈ |k^∗|, j = 1, . . . , r. Then we may assume that X :=D^r =M(k{T₁, . . . , T_r}).

4.3. The case r = 1. In this case X = M(k{T}) is a rooted R-tree. Denote by ≤ the natural partial ordering on X by declaring that x ≤ y if and only if

|f(x)| ≤ |f(y)| for all f ∈ k{T}. The Gauss point ξ : f ∈ k{T} 7→ ρ(f) is the unique maximal element.

For every element H = {(U_i⁰, Hi), i ∈ I⁰} ∈ Pb(X) and c ≥ b, we define a function φ^c_H :X →R onX as follows: for every x∈X,

φ^c_H(x) := max{log(|H_i(x)|),log(c)},

wherex∈U_i⁰.It is well defined. Because fori, j ∈I⁰satisfyingx∈U_i⁰∩U_j⁰, we have

|(Hi−Hj)(x)| ≤b ≤c, then max{log(|Hi(x)|),log|c|}= max{log(|Hj(x)|),log|c|}.

We note that for H⁰ ∈ Pb(X), ρ(H−H⁰)≤c, we have φ^c_H =φ^c_H0. Lemma 4.3. For every c≥b, the function φ^c_H is ξ-subharmonic on X.

Proof of Lemma 4.3. We first treat the case where c > b.

By the argument in Section 4.1, we may assume that I⁰ is finite and U_i⁰, i∈ I are rational subdomains. Since U_i⁰, i ∈ I⁰ are rational, k(T)∩ O(U_i⁰) is dense in O(U_i⁰).So we may assume thatH_i ∈k(T)∩O(U_i⁰).SetI_i :={j ∈I⁰| U_i⁰∩U_j⁰ 6=∅}.

Since U_i⁰, i∈I⁰ are compact, there are open sets Vi, i∈I⁰ such that (i) U_i⁰ ⊆V_i;

(ii) V_i∩V_j =∅if and only if U_i⁰∩U_j⁰ =∅.

For i ∈ I⁰, define an open set W_i := ∩j∈I_i{x ∈ V_i| |H_i −H_j| < c}. Since U_i⁰ ⊆ W_i, W_i, i ∈ I forms an open covering of X. For every x ∈ W_i, we have φ^c_H = max{log(|H_i|),log(c)},which is ξ-subharmonic. Then φ^c_H is ξ-subharmonic.

Now we only need to treat the case b = c. Since φ^b_H is the uniform limit of the sequence of ξ-subharmonic functions φ^b+(1−b)/n_H , n ≥0, φ^b_H is ξ-subharmonic.

This concludes the proof.

Since U_i, i∈I are rational,k(T)∩ O(U_i) is dense in O(U_i).So we may assume that F_i ∈k(T)∩ O(U_i). We may assume that ξ∈U₀.

Lemma 4.4. For every > 0, there exists a rational function P/Q ∈ k(T)∩ O(U₀), P, Q ∈ k[T] such that ρ₀(F₀ −P/Q) ≤ and the norms of all roots of Q= 0 are at most 1.

Proof of Lemma 4.4. Write F₀ = P/Q where P, Q ∈ k[T] and Q 6= 0. We may assume that Q is monic. We note that for every monic irreducible factor Q⁰ of Q, the norm of all roots of Q⁰ = 0 are the same. So we may write Q = Q1Q2

where Q_i, i = 1,2 are monic polynomials such that the norms of all roots of Q₁ = 0 are at most 1 and all roots of Q₂ = 0 are strictly larger than 1. Since 1/Q₂ ∈ O(D) = k{T}, after replacing Q⁻¹₂ by some polynomial in k[T] which is close enough to 1/Q₂ ∈ k{T}, we may assume that Q=Q₁. In other words, we may assume that the norms of all roots of Q = 0 are at most 1. This concludes

the proof.

By Euclidean division, we may write P = GQ+R where G, R ∈ k[T] and degR < degQ. Then we have F₀ = G+R/Q. Set L := F −G ∈ P_b(X). By

Lemma 4.3, φ^b_L is a ξ-subharmonic onX. In particular, we have X

v∈Tan_ξX

D_vφ^c_L≤0.

Lemma 4.5. We have |R/Q(ξ)| ≤b.

Proof of Lemma 4.3. Assume that |R/Q(ξ)|> b.

For every i ∈ I such that ξ ∈ U_i, L_i = F_i −G takes form L_i = R/Q+ωS where S ∈k(T) and|S(ξ)| ≤1. Then there exists an open neighborhoodV ofξ, such that for every i∈I, x∈V ∩U_i, we have|L_i(x)|=|R/Q(x)|> b. Then, for v ∈Tan_ξX, we have

Dvφ^b_L =Dv(log|R/Q|) =Dv(log|R|)−Dv(log|Q|).

We have

X

v∈Tan_ξX

D_v(log|R|)≥ −degR.

Since the norms of all roots of Q= 0 are at most 1, we have X

v∈Tan_ξX

D_v(log|Q|) = −degQ.

It follows that X

v∈Tan_ξX

D_vφ^b_L= X

v∈Tan_ξX

D_v(log|R/Q|)≥degQ−degR >0,

which is a contradiction.

By Lemma 4.3, we have

φ^b_L(ξ) = max{log|L₀(ξ)|,log(b)}= max{log|R/Q(ξ)|,log(b)}= log(b).

Since φ^b_L is decreasing on X, we get φ^b_L= logb onX. It follows that ρ(F −G) = ρ(L)≤b <1.

It follows that G ∈ O_X^◦(X) and ρ(G|_Ui −F_i) ≤ b for i ∈ I. This concludes the proof when r= 1.

4.4. The general case. Now we prove Lemma 1.6 by induction on r ≥ 1. We may assume that r≥2.

Since U_i, i∈I is rational,k(T₁, T₂, . . . , T_r)∩ O(U_i) is dense in O(U_i). We may assume that F_i ∈k(T₁, T₂, . . . , T_r) for all i∈I.

Consider the morphism

π :X =M(k{T₁, . . . , T_r})→D:=M(k{T₁}) defined by

π^∗ :T1 7→T1. For every x∈D, denote by

X_x :=M(H(x){T₂, . . . , T_r})'D^r−1H(x)

the fiber of π above x. We have F_x := F|_Xx = {(U_i ∩X_x, F_i|_Ui∩Xx), i ∈ I} ∈ P_b¹(X_x). For every c≥b, define a function ψ_F^c :D→R by

x7→max{ρ(F|Xx),log(c)}.

Let ξ be the Gauss point in D.

Lemma 4.6. For every c≥b, the function ψ_F^c is ξ-subharmonic on D. Proof of Lemma 4.6. We first treat the case where c > b.

By the induction hypotheses, for every x ∈ D, there exists H_x ∈ O_X^◦

x(X_x) = H(x){T₂, . . . , T_r} such that ρ(H_x−F_x) ≤b. There existsg_x ∈ k[T₁]\ {0}, G_x ∈ k[T₁]_(gx)[T₁, . . . , T_r] such that g_x(x) 6= 0 and ρ(H_x−G_x|_Xx) ≤ b. There exists a rational neighbourhood W_x of x inV_x :={y∈D| g_x(y)6= 0}, such that

(i) π⁻¹(Wx)∩Ui =∅ if and only if Xx∩Ui =∅;

(ii) for every i∈I, ρ(Gx|_π⁻¹_(Wx)∩U_i−Fi|_π⁻¹_(Wx)∩U_i)< c.

Then for every z ∈π⁻¹(Wx), we have

max{log|G_x(z)|,log(c)}=ψ_F^c(z).

Since D is compact, there exists a finite set {x₀, . . . x_m} ⊆D such that

∪^m_j=0W_xj =D.

Set H :={(π⁻¹(W_xi), H_i), i ∈ {0, . . . , m}} ∈ P_c¹(X), where H_i :=G_xi, we have ψ^c(H) = ψ^c(F). Set W_i := W_xi, i = 0, . . . , m. There exists l ≥ 1, such that, for every i= 0, . . . , m,

Hi = X

|J|≤l

A^J_i(T1)T^J

whereJ = (j₂, . . . , j_r)∈Z^r−1≥0 is a multi-index,|J|=j₂+· · ·+j_r,T^J :=T₂^j2. . . T_r^jr and A^J_i(T₁)∈k(T₁)∩ O_D(W_i). For|J| ≤l, set

H^J :={(Wi, A^J_i), i∈ {0, . . . , m}} ∈ Pc(D).

Define φ^c_HJ as in Section 4.3. By Lemma 4.3, φ^c_HJ isξ-subharmonic.

For every x∈W_i, we have

max{log(ρ(H|_Xx)),log(c)}= max{log(ρ(H_i|_Xx)),log((c)}

= max

|J|≤lmax{log|A^J_i(x)|,log((c)}= max

|J|≤lφ^c_HJ. It follows that

ψ^c(F) = ψ^c(H) = max

|J|≤lφ^c_HJ

is ξ-subharmonic.

Now we treat the case c= b. Since ψ^b_F is the uniform limit of the sequence of ξ-subharmonic functions φ^b+(1−b)/n_F , n ≥ 0, φ^b_F is ξ-subharmonic. This concludes

the proof.

By Lemma 4.6, we have

ψ_F^b(ξ) = sup

x∈D

ψ_F^b(x).

Let η be the Gauss point of X = D^r. It can be also viewed as the Gauss point of Xξ 'D^r−1H(η). By the induction hypotheses, there exists H ∈ O_X^◦_ξ(Xξ) = H(ξ){T₂, . . . , T_r}such that ρ(H−F_ξ)≤b. It shows that

sup

x∈D

ψ^b_F(x) =ψ^b_F(ξ) = max{logρ(F|_ξ),log(b)}

= max{logρ(H),log(b)}= max{log|H(η)|,log(b)}

= max{log|F(η)|,log(b)}, (4.1)

whereF(η) := sup{|Fi(η)|| η ∈Ui}. This equation can be viewed as the maximal principle for elements in Pb(D^r).

Sincek(T₁) is dense inH(ξ), we may assume that H ∈k(T₁)[T₂, . . . , T_r].Write H =G(T1, . . . , Tr) + X

|J|≤l

AJ(T1)/B(T1)T^J

where J = (j₂, . . . , j_r) ∈ Z^r−1≥0 is a multi-index, |J| = j₂ + · · ·+ J_m, T^J :=

T₂^j2. . . T_r^jr, G(T₁, . . . , T_r)∈ k[T₁, . . . , T_n], A_J(T₁), B(T₁)∈ k[T₁]. By Lemma 4.4, we may assume that the norms of all roots of B(T₁) = 0 are at most one. By euclidean division, we may assume that degAJ <degB.SetL:=F−G∈ Pb(X).

We claim that max|J|≤l{ρ(AJ(T1)/B(T1))} ≤b. Otherwise there exists J, with

|K| ≤l and ρ(A_K(T₁)/B(T₁))> c > b. There exists a rational neighbourhood W of ξ, such that

(i) π⁻¹(W)∩U_i =∅if and only if X_ξ∩U_i =∅;

(ii) for every i∈I, ρ(H|_π⁻¹_(W)∩U_i−F_i|_π⁻¹_(W)∩U_i)< c;

(iii) for every x∈W, |A_K(x)/B(x)|> c.

It follows that R :=L∨ {(π⁻¹(W), H −G)} ∈ P_c(X). Apply Lemma 4.6 for R, the function ψ^c_R onD is ξ-subharmonic. For x∈W, we have

ψ_R^c(x) = max{max

|J|≤llog|AJ(x)/B(x)|,log(c)} ≥log|AK(x)/B(x)|.

Then we have

0≥ X

v∈Tan_ξD

D_vψ_R^c ≥ X

v∈Tan_ξX

D_v(log|A_K/B|)≥degA_K−degB >0,

which is a contradiction. This concludes the claim.

Then we have

ρ((F −G)|_Xξ)≤max{ρ((H−G)|_Xξ), ρ(F|_Xξ −H)} ≤b.

We note that F −G∈ P_b(X). Apply Equality4.1 for it, we get ρ(F −G)≤ρ((F −G)|_Xξ)≤b.

This concludes the proof of Lemma 1.6.