4Some padic analysis
III Local Fields
4 Some padic analysis
We are now going to do some fun things that is not really related to the course.
In “normal” analysis, the applied mathematicians hold the belief that every
function can be written as a power series
f(x) =
∞
X
n=0
a
n
x
n
.
When we move on to
p
adic numbers, we do not get such a power series expansion.
However, we obtain an analogous result using binomial coefficients.
Before that, we have a quick look at our familiar functions
exp
and
log
, which
we shall continue to define as a power series:
exp(x) =
∞
X
n=0
x
n
n!
, log(1 + x) =
∞
X
n=1
(−1)
n−1
x
n
n
The domain will no longer be all of the field. Instead, we have the following
result:
Proposition.
Let
K
be a complete valued field with an absolute value
 · 
and
assume that
K ⊇ Q
p
and
 · 
restricts to the usual
p
adic norm on
Q
p
. Then
exp
(
x
) converges for
x < p
−1/(p−1)
and
log
(1 +
x
) converges for
x <
1, and
then define continuous maps
exp : {x ∈ K : x < p
−1/(p−1)
} → O
K
log : {1 + x ∈ K : x < 1} → K.
Proof.
We let
v
=
−log
p
 · 
be a valuation extending
v
p
. Then we have the
dumb estimate
v(n) ≤ log
p
n.
Then we have
v
x
n
n
≥ n · v(x) − log
p
n → ∞
if v(x) > 0. So log converges.
For exp, we have
v(n!) =
n − s
p
(n)
p − 1
,
where s
p
(n) is the sum of the padic digits of n. Then we have
v
x
n
n!
≥ n · v(x) −
n
p − 1
= n ·
v(x) −
1
p − 1
→ ∞
if v(x) > 1/(p − 1). Since v
x
n
n!
≥ 0, this lands in O
K
.
For the continuity, we just use uniform convergence as in the real case.
What we really want to talk about is binomial coefficients. Let
n ≥
1. Then
we know that
x
n
=
x(x − 1) ···(x − n + 1)
n!
is a polynomial in
x
, and so defines a continuous function
Z
p
→ Q
p
by
x 7→
x
n
.
When n = 0, we set
x
n
= 1 for all x ∈ Z
p
.
We know
x
n
∈ Z
if
x ∈ Z
≥0
. So by density of
Z
≥0
⊆ Z
p
, we must have
x
n
∈ Z
p
for all x ∈ Z
p
.
We will eventually want to prove the following result:
Theorem
(Mahler’s theorem)
.
Let
f
:
Z
p
→ Q
p
be any continuous function.
Then there is a unique sequence (a
n
)
n≥0
with a
n
∈ Q
p
and a
n
→ 0 such that
f(x) =
∞
X
n=0
a
n
x
n
,
and moreover
sup
x∈Z
p
f(x) = max
k∈N
a
k
.
We write
C
(
Z
p
, Q
p
) for the set of continuous functions
Z
p
→ Q
p
as usual.
This is a Q
p
vector space as usual, with
(λf + µg)(x) = λf(x) + µg(x)
for all λ, µ ∈ Q
p
and f, g ∈ C(Z
p
, Q
p
) and x ∈ Z
p
.
If f ∈ C(Z
p
, Q
p
), we set
kfk = sup
x∈Z
p
f(x)
p
.
Since
Z
p
is compact, we know that
f
is bounded. So the supremum exists and
is attained.
Proposition.
The norm
k · k
defined above is in fact a (nonarchimedean)
norm, and that C(Z
p
, Q
p
) is complete under this norm.
Let
c
0
denote the set of sequences (
a
n
)
∞
n=0
in
Q
p
such that
a
n
→
0. This is
a Q
p
vector space with a norm
k(a
n
)k = max
n∈N
a
n

p
,
and
c
0
is complete. So what Mahler’s theorem gives us is an isometric isomor
phism between c
0
and C(Z
p
, Q
p
).
We define
∆ : C(Z
p
, Q
p
) → C(Z
p
, Q
p
)
by
∆f(x) = f(x + 1) − f(x).
By induction, we have
∆
n
f(x) =
n
X
i=0
(−1)
i
n
i
f(x + n − i).
Note that ∆ is a linear operator on C(Z
p
, Q
p
), and moreover
∆f(x)
p
= f(x + 1) − f(x)
p
≤ kfk.
So we have
k∆fk ≤ kfk.
In other words, we have
k∆k ≤ 1.
Definition
(Mahler coefficient)
.
Let
f ∈ C
(
Z
p
, Q
p
). Then
n
thMahler coeffi
cient a
n
(f) ∈ Q
p
is defined by the formula
a
n
(f) = ∆
n
(f)(0) =
n
X
i=0
(−1)
i
n
i
f(n − i).
We will eventually show that these are the
a
n
’s that appear in Mahler’s
theorem. The first thing to prove is that these coefficients do tend to 0. We
already know that they don’t go up, so we just have to show that they always
eventually go down.
Lemma. Let f ∈ C(Z
p
, Q
p
). Then there exists some k ≥ 1 such that
k∆
p
k
fk ≤
1
p
kfk.
Proof.
If
f
= 0, there is nothing to prove. So we will wlog
kfk
= 1 by scaling
(this is possible since the norm is attained at some
x
0
, so we can just divide by
f(x
0
)). We want to find some k such that
∆
p
k
f(x) ≡ 0 mod p
for all x. To do so, we use the explicit formula
∆
p
k
f(x) =
p
k
X
i=0
(−1)
i
p
k
i
f(x + p
k
− i) ≡ f(x + p
k
) − f(x) (mod p)
because the binomial coefficients
p
k
i
are divisible by
p
for
i 6
= 0
, p
k
. Note that
we do have a negative sign in front of
f
(
x
) because (
−
1)
p
k
is
−
1 as long as
p
is
odd, and 1 = −1 if p = 2.
Now
Z
p
is compact. So
f
is uniformly continuous. So there is some
k
such
that
x − y
p
≤ p
−k
implies
f
(
x
)
− f
(
y
)

p
≤ p
−1
for all
x, y ∈ Z
p
. So take this
k, and we’re done.
We can now prove that the Mahler’s coefficients tend to 0.
Proposition.
The map
f 7→
(
a
n
(
f
))
∞
n=0
defines an injective normdecreasing
linear map C(Z
p
, Q
p
) → c
0
.
Proof. First we prove that a
n
(f) → 0. We know that
ka
n
(f)k
p
≤ k∆
n
fk.
So it suffices to show that
k
∆
n
fk →
0. Since
k
∆
k ≤
1, we know
k
∆
n
fk
is
monotonically decreasing. So it suffices to find a subsequence that tends to 0.
To do so, we simply apply the lemma repeatedly to get k
1
, k
2
, ··· such that
∆
p
k
1
+...+k
n
≤
1
p
n
kfk.
This gives the desired sequence.
Note that
a
n
(f)
p
≤ k∆
n
k ≤ kfk.
So we know
k(a
n
(f))
n
k = max a
n
(f)
p
≤ kfk.
So the map is normdecreasing. Linearity follows from linearity of ∆. To finish,
we have to prove injectivity.
Suppose a
n
(f) = 0 for all n ≥ 0. Then
a
0
(f) = f(0) = 0,
and by induction,we have that
f(n) = ∆
k
f(0) = a
n
(f) = 0.
for all
n ≥
0. So
f
is constantly zero on
Z
≥0
. By continuity, it must be zero
everywhere on Z
p
.
We are almost at Mahler’s theorem. We have found some coefficients already,
and we want to see that it works. We start by proving a small, familiar, lemma.
Lemma. We have
x
n
+
x
n − 1
=
x + 1
n
for all n ∈ Z
≥1
and x ∈ Z
p
.
Proof.
It is well known that this is true when
x ∈ Z
≥n
. Since the expressions
are polynomials in
x
, them agreeing on infinitely many values implies that they
are indeed the same.
Proposition. Let a = (a
n
)
∞
n=0
∈ c
0
. We define f
a
: Z
p
→ Q
p
by
f
a
(x) =
∞
X
n=0
a
n
x
n
.
This defines a normdecreasing linear map
c
0
→ C
(
Z
p
, Q
p
). Moreover
a
n
(
f
a
) =
a
n
for all n ≥ 0.
Proof. Linearity is clear. Normdecreasing follows from
f
a
(x) =
X
a
n
x
n
≤ sup
n
a
n

p
x
n
p
≤ sup
n
a
n

p
= ka
n
k,
where we used the fact that
x
n
∈ Z
p
, hence
x
n
p
≤ 1.
Taking the supremum, we know that
kf
a
k ≤ kak.
For the last statement, for all k ∈ Z
≥0
, we define
a
(k)
= (a
k
, a
k+1
, a
k+1
, ···).
Then we have
∆f
a
(x) = f
a
(x + 1) − f
a
(x)
=
∞
X
n=1
a
n
x + 1
n
−
x
n
=
∞
X
n=1
a
n
x
n − 1
=
∞
X
n=0
a
n+1
x
n
= f
a
(
1)
(x)
Iterating, we have
∆
k
f
a
= f
a
(k)
.
So we have
a
n
(f
a
) = ∆
n
f
a
(0) = f
a
(n)
(0) = a
n
.
Summing up, we now have maps
C(Z
p
, Q
p
) c
0
F
G
with
F (f) = (a
n
(f))
G(a) = f
a
.
We now that
F
is injective and normdecreasing, and
G
is normdecreasing
and
F G
=
id
. It then follows formally that
GF
=
id
and the maps are norm
preserving.
Lemma.
Suppose
V, W
are normed spaces, and
F
:
V → W
,
G
:
W → V
are
maps such that
F
is injective and normdecreasing, and
G
is normdecreasing
and F G = id
W
. Then GF = id
V
and F and G are normpreserving.
Proof. Let v ∈ V . Then
F (v − GF v) = F v − F GF v = (F − F )v = 0.
Since F is injective, we have
v = GF v.
Also, we have
kvk ≥ kF vk ≥ kGF vk = kvk.
So we have equality throughout. Similarly, we have kvk = kGvk.
This finishes the proof Mahler’s theorem, and also finishes this section on
padic analysis.