Proof. Set . We first prove that S has a unique fixed point and then deduce the corresponding statement for ℵ itself and the convergence of iterates.
- 1.
Uniqueness of a fixed point of S.
Assume
are fixed points of
S, i.e.,
and
. Substituting
in (
4) gives
Because
, we obtain
Since
implies
, we have
, hence,
and
. Thus,
S has at most one fixed point.
- 2.
Existence of a fixed point of S (construction by iteration).
Let
be arbitrary and define the sequence
For convenience define
Apply (
4) with
and
. We obtain
But
and
, so
Rearranging yields
Define
From
, we obtain
, hence,
. Therefore,
and by induction
Now for
,
Since
, the right-hand side tends to 0 as
. Hence,
is Cauchy. By completeness of
, there exists
such that
.
- 3.
The limit u is a fixed point of S.
We show
. Take
and
in (
4):
As
, we have
and
. Also,
since
. Passing to the limit gives
Because
(again coming from
), we deduce
, hence,
; therefore,
. By the uniqueness proved in step 1, this fixed point is unique.
- 4.
From a fixed point of S to a fixed point of ℵ.
We have shown
u satisfies
. Observe that
so
is also a fixed point of
S. Uniqueness of the fixed point of
S implies
. Thus,
u is a fixed point of
ℵ. If
v is any fixed point of
ℵ, then
, so by uniqueness,
. Hence,
ℵ has exactly one fixed point, and it coincides with the unique fixed point of
S.
- 5.
Convergence of the Picard iteration for ℵ.
We now show that for any initial
, the sequence
converges to
u. Fix
. For each
n,
We have already shown that
as
. Since
is continuous along the sequence of iterates (because it is an isometry of the discrete iteration structure; more directly, the sequence
is a subsequence of iterates of
S composed with finitely many applications of
ℵ), it follows that every subsequence
converges to
u. Therefore, all subsequences of the form
converge to the same limit
u, and hence the full sequence
converges to
u. In particular, for the Picard iteration
with arbitrary
, we have
.
This completes the proof. □
Proof. Set . We first show that S has a unique fixed point and then deduce the corresponding statements for ℵ and the convergence of iterates.
- 1.
Uniqueness of a fixed point of .
Assume
are fixed points of
S, i.e.,
and
. Applying (
6) with
and
gives
Because
, we obtain
Since
for all
, the preceding inequality forces
. Hence,
, so
S (and consequently
ℵ) has at most one fixed point.
- 2.
Iterative construction and monotonicity of successive differences.
Fix an arbitrary
and define
For
, set
Apply (
6) with
and
. Note that
for the indices where the points are distinct; if two consecutive points coincide, then the sequence is eventually constant and the claim is trivial. For the nontrivial case, we obtain
But
and
, hence,
Rearranging yields
Define the function
Because
for every
, the denominator
is positive and
for all
. Thus,
Since
, it follows from (2) that
so the sequence
is strictly decreasing (or eventually nonincreasing) and bounded below by 0. Therefore, it has a limit
We show
. Suppose, for contradiction, that
. Because
, the sequence
approaches
L from above and, by monotonicity of
, the one-sided limits
,
,
exist in
. Passing to the limit superior in the inequality
and using that
and
, we obtain
i.e.,
But for every
, we have
; taking the right-hand limit as
(possible because the functions are monotone) preserves the strict inequality, so
Hence, the factor
is strictly positive, which forces
. This contradiction shows
.
- 4.
Cauchyness of the sequence .
Since
and
is nonincreasing, for
, we have
However, one can obtain summability because
with
. Iterating (2) yields
The infinite product
converges to 0 because
whenever
(a standard argument for products of terms in
when the terms do not stay too close to 1); because
and
is a function bounded away from 1 on any interval
for
, there exists
and a constant
such that for all
,
. Consequently, for
, we have
, leading to a geometric estimate
. This directly implies that the tail sum
converges to zero as
.
Consequently, the tail sums tend to 0 as . Concretely, because and is bounded away from 1 on any interval with , one can split the sum into finitely many initial terms (where a geometric estimate holds) and a tail where terms are arbitrarily small. Thus, as . Therefore, is a Cauchy sequence. (An alternative concise route is from (2) we have for some once is small enough; this yields a geometric tail and summability.) By completeness of , there exists such that .
- 5.
The limit u is a fixed point of .
Fix
n and apply (
6) with
and
(when
):
Letting
, we have
,
, and
. Using the one-sided limits of the monotone coefficients at
(they are bounded and take values in
), the right-hand side tends to
. Hence,
Also using (
6) with
and
and taking the limit as
, we obtain
where
(the limit exists due to monotonicity). If
, this implies
. However, since the functions
satisfy
for all
, taking the limit as
yields
. Given that all terms are non-negative,
. The case
would force
. Let us re-examine the contractive inequality for
and
before taking the limit
If
and
, then for large
n, the right-hand side is approximately
, while the left-hand side
tends to
from below due to the strict inequality. This is only possible if
. Therefore, we conclude
in all cases. By the uniqueness in step 1, this fixed point is unique.
- 6.
From fixed point of to fixed point of ℵ and global convergence.
As , we have , so is another fixed point of S. Uniqueness of the fixed point of S implies . Hence, u is a fixed point of ℵ. If v is any fixed point of ℵ, then , so v is a fixed point of S; therefore, . Thus, ℵ has exactly one fixed point. Finally, for an arbitrary starting point , the sequence converges to u by the same argument as above. For each fixed , the subsequence , therefore, also converges to u. Since every term of the full sequence belongs to one of these p subsequences, the full sequence converges to u as well. In particular, the Picard iteration with arbitrary converges to the unique fixed point u.
This completes the proof. □
Proof. Set . We first prove uniqueness of a fixed point of S and then produce a limit for the iterates of S and finally lift the result to ℵ.
Assume
are fixed points of
S. Applying (
7) with
and
gives
Since
u and
v are fixed points of
S, we have
, hence,
which is impossible unless
. Therefore,
. Thus
S (and so
ℵ) admits at most one fixed point.
- 2.
Iteration and monotonicity of successive differences.
Choose an arbitrary
and define the sequence
For
, put
Apply (
7) with
and
(for those
n where
); we obtain
But
and
, while
. Hence,
This strict inequality implies that the maximum on the right-hand side cannot be
; therefore, it must be
, and consequently
(for indices where
; if some equality
occurs, then from that index onward the sequence is constant and the conclusion is immediate). Thus, the sequence
is strictly decreasing while bounded below by 0. Hence, it converges to some limit
We show
. Suppose, for contradiction, that
. Since
, there exists
N, such that for all
, we have
. Fix
and any
. Repeated application of (
7) (applied to appropriate pairs) yields the simple and useful bound
where
is strictly decreasing (as shown in Step 2). (This is proved by induction on
: the case
is trivial; assume true for a given gap and apply (
7) to obtain the step for one larger gap.) In particular, for all
and
, we obtain
Now apply (
7) to the pair
with
:
Define
. From (
8) and the decreasing monotonicity of
, we have
. Now, applying (
7) to the pair
for any
gives
Taking the limit superior as
on both sides yields
But the left-hand side is exactly
. Thus,
.
Since and , both sequences converge to limits and . Taking the limit inferior as of the inequality gives . However, from the definition of , we must also have (consider ). Hence, .
Now, from the strict inequality and the convergence , the only consistent possibility is . For if , then would be bounded below by L and could not satisfy for all sufficiently large n while converging to L.
The strict inequality in (
7) cannot persist in the limit unless the quantity on the right-hand side tends to a strictly larger number than
L. However, by the boundedness above, we see the right-hand side tends to
L. Thus, the only consistent possibility is
. (Intuitively, a strictly decreasing positive sequence cannot be preserved under the strict max-contraction; the successive strict drops force the limit to be 0.) Therefore,
.
- 4.
Cauchyness and existence of a limit for .
From (
8), we have for every
Since
as
, the right-hand side tends to 0 when
uniformly in
. Hence,
is a Cauchy sequence. By completeness of
, there exists
such that
.
- 5.
The limit is a fixed point of .
We show
. Apply (
7) to
and
(for large
n with
):
Passing to the limit
gives (because
and
and
)
The strict inequality becomes a non-strict one in the limit; hence, we deduce
. Therefore,
. By part 1, the fixed point of
S is unique.
- 6.
From fixed point of to fixed point of ℵ and convergence of the Picard iteration.
Since
, we have
so
is another fixed point of
S. By uniqueness of the fixed point of
S, we must have
, i.e.,
u is a fixed point of
ℵ itself. If
v is any fixed point of
ℵ, then
, so by uniqueness,
. Hence,
ℵ has exactly one fixed point. Finally, for an arbitrary starting point
, the sequence
converges to
u by the preceding argument. For each residue class
, the subsequence
, therefore, also converges to
u. Since every term of the full sequence
belongs to one of these
p subsequences, the full sequence
converges to
u. In particular, the Picard iteration
with arbitrary
converges to the unique fixed point
u.
This completes the proof. □